If nine coins are tossed, what is the probability that the number of heads is even?












17












$begingroup$


So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.



We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$



$n = 9, k = 0$



$$binom{9}{0}bigg(frac{1}{2}bigg)^0bigg(frac{1}{2}bigg)^{9}$$



$n = 9, k = 2$



$$binom{9}{2}bigg(frac{1}{2}bigg)^2bigg(frac{1}{2}bigg)^{7}$$



$n = 9, k = 4$
$$binom{9}{4}bigg(frac{1}{2}bigg)^4bigg(frac{1}{2}bigg)^{5}$$



$n = 9, k = 6$



$$binom{9}{6}bigg(frac{1}{2}bigg)^6bigg(frac{1}{2}bigg)^{3}$$



$n = 9, k = 8$



$$binom{9}{8}bigg(frac{1}{2}bigg)^8bigg(frac{1}{2}bigg)^{1}$$



Add all of these up:



$$=.64$$ so there's a 64% chance of probability?










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$endgroup$








  • 2




    $begingroup$
    Zero is an even number, too.
    $endgroup$
    – saulspatz
    yesterday










  • $begingroup$
    What about $k=0$?
    $endgroup$
    – Dbchatto67
    yesterday






  • 4




    $begingroup$
    Either Heads or Tails but not both must be even, so $.5$
    $endgroup$
    – lulu
    yesterday






  • 3




    $begingroup$
    Your arithmetic is wrong somewhere because it should sum to $0.50$. You can match the chance of $k$ heads with the chance of $9-k$. The powers of $frac 12$ always give $frac 1{2^9}$, so it is probably in your computation of the combination numbers. Your approach is fine, if much harder than the symmetry arguments.
    $endgroup$
    – Ross Millikan
    yesterday








  • 3




    $begingroup$
    Ross Millikan is correct; you need to double-check your arithmetic. The last two terms of each product multiply to 1/512. The first terms of each product are 1, 36, 126, 84 and 9. Those add to 256. So you should be getting 256/512 which is 1/2. Always check your arithmetic.
    $endgroup$
    – Eric Lippert
    yesterday
















17












$begingroup$


So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.



We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$



$n = 9, k = 0$



$$binom{9}{0}bigg(frac{1}{2}bigg)^0bigg(frac{1}{2}bigg)^{9}$$



$n = 9, k = 2$



$$binom{9}{2}bigg(frac{1}{2}bigg)^2bigg(frac{1}{2}bigg)^{7}$$



$n = 9, k = 4$
$$binom{9}{4}bigg(frac{1}{2}bigg)^4bigg(frac{1}{2}bigg)^{5}$$



$n = 9, k = 6$



$$binom{9}{6}bigg(frac{1}{2}bigg)^6bigg(frac{1}{2}bigg)^{3}$$



$n = 9, k = 8$



$$binom{9}{8}bigg(frac{1}{2}bigg)^8bigg(frac{1}{2}bigg)^{1}$$



Add all of these up:



$$=.64$$ so there's a 64% chance of probability?










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Zero is an even number, too.
    $endgroup$
    – saulspatz
    yesterday










  • $begingroup$
    What about $k=0$?
    $endgroup$
    – Dbchatto67
    yesterday






  • 4




    $begingroup$
    Either Heads or Tails but not both must be even, so $.5$
    $endgroup$
    – lulu
    yesterday






  • 3




    $begingroup$
    Your arithmetic is wrong somewhere because it should sum to $0.50$. You can match the chance of $k$ heads with the chance of $9-k$. The powers of $frac 12$ always give $frac 1{2^9}$, so it is probably in your computation of the combination numbers. Your approach is fine, if much harder than the symmetry arguments.
    $endgroup$
    – Ross Millikan
    yesterday








  • 3




    $begingroup$
    Ross Millikan is correct; you need to double-check your arithmetic. The last two terms of each product multiply to 1/512. The first terms of each product are 1, 36, 126, 84 and 9. Those add to 256. So you should be getting 256/512 which is 1/2. Always check your arithmetic.
    $endgroup$
    – Eric Lippert
    yesterday














17












17








17


5



$begingroup$


So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.



We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$



$n = 9, k = 0$



$$binom{9}{0}bigg(frac{1}{2}bigg)^0bigg(frac{1}{2}bigg)^{9}$$



$n = 9, k = 2$



$$binom{9}{2}bigg(frac{1}{2}bigg)^2bigg(frac{1}{2}bigg)^{7}$$



$n = 9, k = 4$
$$binom{9}{4}bigg(frac{1}{2}bigg)^4bigg(frac{1}{2}bigg)^{5}$$



$n = 9, k = 6$



$$binom{9}{6}bigg(frac{1}{2}bigg)^6bigg(frac{1}{2}bigg)^{3}$$



$n = 9, k = 8$



$$binom{9}{8}bigg(frac{1}{2}bigg)^8bigg(frac{1}{2}bigg)^{1}$$



Add all of these up:



$$=.64$$ so there's a 64% chance of probability?










share|cite|improve this question











$endgroup$




So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.



We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$



$n = 9, k = 0$



$$binom{9}{0}bigg(frac{1}{2}bigg)^0bigg(frac{1}{2}bigg)^{9}$$



$n = 9, k = 2$



$$binom{9}{2}bigg(frac{1}{2}bigg)^2bigg(frac{1}{2}bigg)^{7}$$



$n = 9, k = 4$
$$binom{9}{4}bigg(frac{1}{2}bigg)^4bigg(frac{1}{2}bigg)^{5}$$



$n = 9, k = 6$



$$binom{9}{6}bigg(frac{1}{2}bigg)^6bigg(frac{1}{2}bigg)^{3}$$



$n = 9, k = 8$



$$binom{9}{8}bigg(frac{1}{2}bigg)^8bigg(frac{1}{2}bigg)^{1}$$



Add all of these up:



$$=.64$$ so there's a 64% chance of probability?







probability discrete-mathematics






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share|cite|improve this question













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edited yesterday







Stuy

















asked yesterday









StuyStuy

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355211








  • 2




    $begingroup$
    Zero is an even number, too.
    $endgroup$
    – saulspatz
    yesterday










  • $begingroup$
    What about $k=0$?
    $endgroup$
    – Dbchatto67
    yesterday






  • 4




    $begingroup$
    Either Heads or Tails but not both must be even, so $.5$
    $endgroup$
    – lulu
    yesterday






  • 3




    $begingroup$
    Your arithmetic is wrong somewhere because it should sum to $0.50$. You can match the chance of $k$ heads with the chance of $9-k$. The powers of $frac 12$ always give $frac 1{2^9}$, so it is probably in your computation of the combination numbers. Your approach is fine, if much harder than the symmetry arguments.
    $endgroup$
    – Ross Millikan
    yesterday








  • 3




    $begingroup$
    Ross Millikan is correct; you need to double-check your arithmetic. The last two terms of each product multiply to 1/512. The first terms of each product are 1, 36, 126, 84 and 9. Those add to 256. So you should be getting 256/512 which is 1/2. Always check your arithmetic.
    $endgroup$
    – Eric Lippert
    yesterday














  • 2




    $begingroup$
    Zero is an even number, too.
    $endgroup$
    – saulspatz
    yesterday










  • $begingroup$
    What about $k=0$?
    $endgroup$
    – Dbchatto67
    yesterday






  • 4




    $begingroup$
    Either Heads or Tails but not both must be even, so $.5$
    $endgroup$
    – lulu
    yesterday






  • 3




    $begingroup$
    Your arithmetic is wrong somewhere because it should sum to $0.50$. You can match the chance of $k$ heads with the chance of $9-k$. The powers of $frac 12$ always give $frac 1{2^9}$, so it is probably in your computation of the combination numbers. Your approach is fine, if much harder than the symmetry arguments.
    $endgroup$
    – Ross Millikan
    yesterday








  • 3




    $begingroup$
    Ross Millikan is correct; you need to double-check your arithmetic. The last two terms of each product multiply to 1/512. The first terms of each product are 1, 36, 126, 84 and 9. Those add to 256. So you should be getting 256/512 which is 1/2. Always check your arithmetic.
    $endgroup$
    – Eric Lippert
    yesterday








2




2




$begingroup$
Zero is an even number, too.
$endgroup$
– saulspatz
yesterday




$begingroup$
Zero is an even number, too.
$endgroup$
– saulspatz
yesterday












$begingroup$
What about $k=0$?
$endgroup$
– Dbchatto67
yesterday




$begingroup$
What about $k=0$?
$endgroup$
– Dbchatto67
yesterday




4




4




$begingroup$
Either Heads or Tails but not both must be even, so $.5$
$endgroup$
– lulu
yesterday




$begingroup$
Either Heads or Tails but not both must be even, so $.5$
$endgroup$
– lulu
yesterday




3




3




$begingroup$
Your arithmetic is wrong somewhere because it should sum to $0.50$. You can match the chance of $k$ heads with the chance of $9-k$. The powers of $frac 12$ always give $frac 1{2^9}$, so it is probably in your computation of the combination numbers. Your approach is fine, if much harder than the symmetry arguments.
$endgroup$
– Ross Millikan
yesterday






$begingroup$
Your arithmetic is wrong somewhere because it should sum to $0.50$. You can match the chance of $k$ heads with the chance of $9-k$. The powers of $frac 12$ always give $frac 1{2^9}$, so it is probably in your computation of the combination numbers. Your approach is fine, if much harder than the symmetry arguments.
$endgroup$
– Ross Millikan
yesterday






3




3




$begingroup$
Ross Millikan is correct; you need to double-check your arithmetic. The last two terms of each product multiply to 1/512. The first terms of each product are 1, 36, 126, 84 and 9. Those add to 256. So you should be getting 256/512 which is 1/2. Always check your arithmetic.
$endgroup$
– Eric Lippert
yesterday




$begingroup$
Ross Millikan is correct; you need to double-check your arithmetic. The last two terms of each product multiply to 1/512. The first terms of each product are 1, 36, 126, 84 and 9. Those add to 256. So you should be getting 256/512 which is 1/2. Always check your arithmetic.
$endgroup$
– Eric Lippert
yesterday










9 Answers
9






active

oldest

votes


















58












$begingroup$

The probability is $frac{1}{2}$ because the last flip determines it.






share|cite|improve this answer









$endgroup$









  • 25




    $begingroup$
    Though this answer is correct for this specific question, it is interesting to consider why it works. Consider: an unfair coin comes up heads 75% of the time, tails 25% of the time. We flip it nine times. Again, the last flip determines whether the total is odd or even. In this case, is it even 75% of the time, because the last flip is heads 75% of the time, or is it even 50% of the time, as it was before? Or is it even some other percentage of the time? Before you do the math, what's your gut feeling? And can you explain why your argument is only valid for a fair coin?
    $endgroup$
    – Eric Lippert
    22 hours ago










  • $begingroup$
    Interesting question Eric. Formally I would use recursion, let us define $O_n$ as the probability that we get an odd number of heads when flipping $n$ coins, and let $p$ be the probability the coin comes up as heads. Then the previous reasoning turns into the recurrence relation $O_n=(1-p)O_{n-1} + p(1-O_{n-1})=(1-2p)O_{n-1}+p$.
    $endgroup$
    – Jorge Fernández Hidalgo
    21 hours ago








  • 14




    $begingroup$
    That's a good analysis. My reason for bringing it up is: plainly the OP is relatively new to probability, so I think it is important to call out the assumptions when giving "shortcut" reasoning like this answer. The underlying assumption is that with eight coins, an even number of heads or tails is also 1/2. I think it is worthwhile calling that out. I wouldn't want the OP to deduce that for more general problems, they only need to consider the probability of the final flip.
    $endgroup$
    – Eric Lippert
    21 hours ago






  • 2




    $begingroup$
    Yes you are right, I was a bit careless when I wrote it, thank you for pointing out that detail!
    $endgroup$
    – Jorge Fernández Hidalgo
    21 hours ago








  • 3




    $begingroup$
    I disagree that the underlying assumption is "with eight coins, an even number of heads occurs with probability $1/2$" -- that assumption is not necessary for the answer at all. If you flip 8 coins of varying biases, but the 9th and final one is a fair coin, then this answer is still correct.
    $endgroup$
    – Mees de Vries
    9 hours ago



















31












$begingroup$

If there are an even number of heads then there must be an odd number of tails. But heads and tails are symmetrical, so the probability must be $1/2$.






share|cite|improve this answer









$endgroup$





















    15












    $begingroup$

    Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is
    $$binom{9}{0}+binom{9}{2}+binom{9}{4}+binom{9}{6}+binom{9}{8}=1+36+126+84+9=256$$
    The number of all possible outcomes is $512$ thus the probability to get even number of heads is $0.5$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Comparing the number of favorable outcomes with all possible outcomes in this way only works for a fair (50% heads/50% tails) coin.
      $endgroup$
      – Richard Ward
      4 hours ago



















    14












    $begingroup$

    All the possible sequences of tosses (which are all equally likely) may be divided into the two categories "even number of heads" and "odd number of heads". You can then pair up each sequence in one category with the "flipped" version (flip every coin in the sequence around) in the other category. This shows that there are equally many sequences in each of the two categories. So the probability of landing in a specific one of them must be $frac12$.






    share|cite|improve this answer









    $endgroup$





















      8












      $begingroup$

      There are two cases here:




      • There's an even number of heads: 0, 2, 4, 6 or 8 heads

      • There's an odd number of heads: 1, 3, 5, 7 or 9 heads


      But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:




      • There's an even number of tails: 0, 2, 4, 6 or 8 tails


      Since heads and tails are equally probable, we can, by symmetry, see that these two cases have the same probability. Therefore each must have probability $1/2$.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Ethan Bolker already gave this explanation 2 hours ago.
        $endgroup$
        – Paul Sinclair
        yesterday






      • 3




        $begingroup$
        @PaulSinclair Yes, he did. But I found this explanation clearer than Ethan's, which skips a number of steps.
        $endgroup$
        – CJ Dennis
        20 hours ago










      • $begingroup$
        @PaulSinclair are they required to have read Ethan's answer first before going here?
        $endgroup$
        – The Great Duck
        18 hours ago










      • $begingroup$
        I didn't understand Ethan's answer and I understood this one perfectly.
        $endgroup$
        – Todd Wilcox
        6 hours ago



















      7












      $begingroup$

      The easiest way to see this : Consider the number of heads we have in the first $8$ coins.




      • If this number is even, we need a tail, we have probability $frac{1}{2}$

      • If this number is odd, we need a head, we have probability $frac{1}{2}$


      Hence no matter what the $8$ coins delivered, we have probability $frac{1}{2}$ , which is the answer.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        This is , by the way, true for EVERY number of coins (even for one coin).
        $endgroup$
        – Peter
        yesterday






      • 6




        $begingroup$
        Pedantic point: If zero coins are tossed, the probability of an even number of heads is 100%.
        $endgroup$
        – Brian
        yesterday






      • 3




        $begingroup$
        This answer presupposes that having an even number of heads and an odd number of heads in eight flips is equally likely, but we haven't proved that lemma.
        $endgroup$
        – Eric Lippert
        22 hours ago






      • 1




        $begingroup$
        @EricLippert: No it doesn't. If the probability in both cases is the same, then the probability overall is the same...
        $endgroup$
        – user21820
        14 hours ago






      • 2




        $begingroup$
        @EricLippert No. If the probability of an even number of heads in 8 tosses is p (and the probability of an odd number of heads is 1-p), without assuming the value of p, the probability of an even number in all 9 tosses is p·(1/2) + (1-p)(1/2), which turns out to be exactly 1/2.
        $endgroup$
        – dgstranz
        12 hours ago



















      4












      $begingroup$

      There's a way to do it with barely any maths:



      It's clear that if there's an odd number of heads, there's an even number of tails and vice versa, so P(even number of heads) + P(even number of tails) = 1.



      Formally rename "heads" to "tails". The problem remains unchanged.



      So P(even number of heads) = P(even number of tails) = 1/2.






      share|cite|improve this answer










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      Remellion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        3












        $begingroup$

        The probability generating function of a Binomiall random variable $Xsim text{Bin}(n, 1/2)$ with probability of success $1/2$ is given by
        $$
        g_{X}(t)=Et^X=sum_{k=0}^nP(X=k)t^k=sum_{k=0}^nbinom{n}{k}frac{t^k}{2^n}=frac{1}{2^n}(1+t)^n
        $$

        In particular the probability that $X$ is even is given by
        $$
        sum_{0leq kleq n, k,{text{even}}}P(X=k)=frac{g(1)+g(-1)}{2}=frac{1+0}{2}=frac{1}{2}.
        $$






        share|cite|improve this answer









        $endgroup$





















          2












          $begingroup$

          If $h$ denotes getting a heads and $p$ denotes getting tails, let's write $frac{1}{2}h+frac{1}{2}p$ for the notion of a fair coin: half the time it turns up heads, and half the time tails.



          If we expand out the following product while keeping track of multiplication order,
          $$left(frac{1}{2}h+frac{1}{2}pright)left(frac{1}{2}h+frac{1}{2}pright)=frac{1}{4}hh+frac{1}{4}hp+frac{1}{4}ph+frac{1}{4}pp,$$
          we see that the sequences $hh$, $hp$, $ph$, and $pp$ are equally likely. Forgetting about multiplication order, which we want to do because we only care about how many times heads or tails showed up, corresponds to just treating this like a polynomial:
          $$=frac{1}{4}h^2+frac{1}{2}hp+frac{1}{4}p^2.$$
          We can keep multiplying copies of a fair coin together to find the probability that a certain number of heads or tails happened, where the coefficient in front of $h^kp^ell$ is the probability of $k$ heads and $ell$ tails.



          Nine coins is the expansion
          $$left(frac{1}{2}h+frac{1}{2}pright)^9=sum_{k=0}^9binom{9}{k}left(frac{1}{2}hright)^kleft(frac{1}{2}pright)^{9-k}=sum_{k=0}^9frac{1}{2^9}binom{9}{k}h^kp^{9-k}.$$
          So far, all this has done is explain why you were adding up $2^{-9}binom{9}{k}$ for $k=0,2,4,dots,8$. Here, now, is a nice trick. If we formally set $h=1$ and $p=1$, then we get
          $$1=sum_{k=0}^9frac{1}{2^9}binom{9}{k},$$
          and if we formally set $h=-1$ and $p=1$, then we get
          $$0=sum_{k=0}^9frac{1}{2^9}binom{9}{k}(-1)^k.$$
          The average of these two equations is
          $$frac{1}{2}=sum_{k=0,ktext{ even}}^9frac{1}{2^9}binom{9}{k},$$
          since $frac{1}{2}(1+(-1)^k)$ is $1$ or $0$ depending on whether $k$ is even or odd. Thus the probability of an even number of heads is $frac{1}{2}$.



          Notice that this did not use the fact nine coins were tossed at any point! (Other than the fact that at least one coin was tossed. In the case of tossing no coins, an even number of heads happens with probability $1$. What part of my argument goes wrong for the case of zero coins?)






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            9 Answers
            9






            active

            oldest

            votes








            9 Answers
            9






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            58












            $begingroup$

            The probability is $frac{1}{2}$ because the last flip determines it.






            share|cite|improve this answer









            $endgroup$









            • 25




              $begingroup$
              Though this answer is correct for this specific question, it is interesting to consider why it works. Consider: an unfair coin comes up heads 75% of the time, tails 25% of the time. We flip it nine times. Again, the last flip determines whether the total is odd or even. In this case, is it even 75% of the time, because the last flip is heads 75% of the time, or is it even 50% of the time, as it was before? Or is it even some other percentage of the time? Before you do the math, what's your gut feeling? And can you explain why your argument is only valid for a fair coin?
              $endgroup$
              – Eric Lippert
              22 hours ago










            • $begingroup$
              Interesting question Eric. Formally I would use recursion, let us define $O_n$ as the probability that we get an odd number of heads when flipping $n$ coins, and let $p$ be the probability the coin comes up as heads. Then the previous reasoning turns into the recurrence relation $O_n=(1-p)O_{n-1} + p(1-O_{n-1})=(1-2p)O_{n-1}+p$.
              $endgroup$
              – Jorge Fernández Hidalgo
              21 hours ago








            • 14




              $begingroup$
              That's a good analysis. My reason for bringing it up is: plainly the OP is relatively new to probability, so I think it is important to call out the assumptions when giving "shortcut" reasoning like this answer. The underlying assumption is that with eight coins, an even number of heads or tails is also 1/2. I think it is worthwhile calling that out. I wouldn't want the OP to deduce that for more general problems, they only need to consider the probability of the final flip.
              $endgroup$
              – Eric Lippert
              21 hours ago






            • 2




              $begingroup$
              Yes you are right, I was a bit careless when I wrote it, thank you for pointing out that detail!
              $endgroup$
              – Jorge Fernández Hidalgo
              21 hours ago








            • 3




              $begingroup$
              I disagree that the underlying assumption is "with eight coins, an even number of heads occurs with probability $1/2$" -- that assumption is not necessary for the answer at all. If you flip 8 coins of varying biases, but the 9th and final one is a fair coin, then this answer is still correct.
              $endgroup$
              – Mees de Vries
              9 hours ago
















            58












            $begingroup$

            The probability is $frac{1}{2}$ because the last flip determines it.






            share|cite|improve this answer









            $endgroup$









            • 25




              $begingroup$
              Though this answer is correct for this specific question, it is interesting to consider why it works. Consider: an unfair coin comes up heads 75% of the time, tails 25% of the time. We flip it nine times. Again, the last flip determines whether the total is odd or even. In this case, is it even 75% of the time, because the last flip is heads 75% of the time, or is it even 50% of the time, as it was before? Or is it even some other percentage of the time? Before you do the math, what's your gut feeling? And can you explain why your argument is only valid for a fair coin?
              $endgroup$
              – Eric Lippert
              22 hours ago










            • $begingroup$
              Interesting question Eric. Formally I would use recursion, let us define $O_n$ as the probability that we get an odd number of heads when flipping $n$ coins, and let $p$ be the probability the coin comes up as heads. Then the previous reasoning turns into the recurrence relation $O_n=(1-p)O_{n-1} + p(1-O_{n-1})=(1-2p)O_{n-1}+p$.
              $endgroup$
              – Jorge Fernández Hidalgo
              21 hours ago








            • 14




              $begingroup$
              That's a good analysis. My reason for bringing it up is: plainly the OP is relatively new to probability, so I think it is important to call out the assumptions when giving "shortcut" reasoning like this answer. The underlying assumption is that with eight coins, an even number of heads or tails is also 1/2. I think it is worthwhile calling that out. I wouldn't want the OP to deduce that for more general problems, they only need to consider the probability of the final flip.
              $endgroup$
              – Eric Lippert
              21 hours ago






            • 2




              $begingroup$
              Yes you are right, I was a bit careless when I wrote it, thank you for pointing out that detail!
              $endgroup$
              – Jorge Fernández Hidalgo
              21 hours ago








            • 3




              $begingroup$
              I disagree that the underlying assumption is "with eight coins, an even number of heads occurs with probability $1/2$" -- that assumption is not necessary for the answer at all. If you flip 8 coins of varying biases, but the 9th and final one is a fair coin, then this answer is still correct.
              $endgroup$
              – Mees de Vries
              9 hours ago














            58












            58








            58





            $begingroup$

            The probability is $frac{1}{2}$ because the last flip determines it.






            share|cite|improve this answer









            $endgroup$



            The probability is $frac{1}{2}$ because the last flip determines it.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            Jorge Fernández HidalgoJorge Fernández Hidalgo

            76k1193195




            76k1193195








            • 25




              $begingroup$
              Though this answer is correct for this specific question, it is interesting to consider why it works. Consider: an unfair coin comes up heads 75% of the time, tails 25% of the time. We flip it nine times. Again, the last flip determines whether the total is odd or even. In this case, is it even 75% of the time, because the last flip is heads 75% of the time, or is it even 50% of the time, as it was before? Or is it even some other percentage of the time? Before you do the math, what's your gut feeling? And can you explain why your argument is only valid for a fair coin?
              $endgroup$
              – Eric Lippert
              22 hours ago










            • $begingroup$
              Interesting question Eric. Formally I would use recursion, let us define $O_n$ as the probability that we get an odd number of heads when flipping $n$ coins, and let $p$ be the probability the coin comes up as heads. Then the previous reasoning turns into the recurrence relation $O_n=(1-p)O_{n-1} + p(1-O_{n-1})=(1-2p)O_{n-1}+p$.
              $endgroup$
              – Jorge Fernández Hidalgo
              21 hours ago








            • 14




              $begingroup$
              That's a good analysis. My reason for bringing it up is: plainly the OP is relatively new to probability, so I think it is important to call out the assumptions when giving "shortcut" reasoning like this answer. The underlying assumption is that with eight coins, an even number of heads or tails is also 1/2. I think it is worthwhile calling that out. I wouldn't want the OP to deduce that for more general problems, they only need to consider the probability of the final flip.
              $endgroup$
              – Eric Lippert
              21 hours ago






            • 2




              $begingroup$
              Yes you are right, I was a bit careless when I wrote it, thank you for pointing out that detail!
              $endgroup$
              – Jorge Fernández Hidalgo
              21 hours ago








            • 3




              $begingroup$
              I disagree that the underlying assumption is "with eight coins, an even number of heads occurs with probability $1/2$" -- that assumption is not necessary for the answer at all. If you flip 8 coins of varying biases, but the 9th and final one is a fair coin, then this answer is still correct.
              $endgroup$
              – Mees de Vries
              9 hours ago














            • 25




              $begingroup$
              Though this answer is correct for this specific question, it is interesting to consider why it works. Consider: an unfair coin comes up heads 75% of the time, tails 25% of the time. We flip it nine times. Again, the last flip determines whether the total is odd or even. In this case, is it even 75% of the time, because the last flip is heads 75% of the time, or is it even 50% of the time, as it was before? Or is it even some other percentage of the time? Before you do the math, what's your gut feeling? And can you explain why your argument is only valid for a fair coin?
              $endgroup$
              – Eric Lippert
              22 hours ago










            • $begingroup$
              Interesting question Eric. Formally I would use recursion, let us define $O_n$ as the probability that we get an odd number of heads when flipping $n$ coins, and let $p$ be the probability the coin comes up as heads. Then the previous reasoning turns into the recurrence relation $O_n=(1-p)O_{n-1} + p(1-O_{n-1})=(1-2p)O_{n-1}+p$.
              $endgroup$
              – Jorge Fernández Hidalgo
              21 hours ago








            • 14




              $begingroup$
              That's a good analysis. My reason for bringing it up is: plainly the OP is relatively new to probability, so I think it is important to call out the assumptions when giving "shortcut" reasoning like this answer. The underlying assumption is that with eight coins, an even number of heads or tails is also 1/2. I think it is worthwhile calling that out. I wouldn't want the OP to deduce that for more general problems, they only need to consider the probability of the final flip.
              $endgroup$
              – Eric Lippert
              21 hours ago






            • 2




              $begingroup$
              Yes you are right, I was a bit careless when I wrote it, thank you for pointing out that detail!
              $endgroup$
              – Jorge Fernández Hidalgo
              21 hours ago








            • 3




              $begingroup$
              I disagree that the underlying assumption is "with eight coins, an even number of heads occurs with probability $1/2$" -- that assumption is not necessary for the answer at all. If you flip 8 coins of varying biases, but the 9th and final one is a fair coin, then this answer is still correct.
              $endgroup$
              – Mees de Vries
              9 hours ago








            25




            25




            $begingroup$
            Though this answer is correct for this specific question, it is interesting to consider why it works. Consider: an unfair coin comes up heads 75% of the time, tails 25% of the time. We flip it nine times. Again, the last flip determines whether the total is odd or even. In this case, is it even 75% of the time, because the last flip is heads 75% of the time, or is it even 50% of the time, as it was before? Or is it even some other percentage of the time? Before you do the math, what's your gut feeling? And can you explain why your argument is only valid for a fair coin?
            $endgroup$
            – Eric Lippert
            22 hours ago




            $begingroup$
            Though this answer is correct for this specific question, it is interesting to consider why it works. Consider: an unfair coin comes up heads 75% of the time, tails 25% of the time. We flip it nine times. Again, the last flip determines whether the total is odd or even. In this case, is it even 75% of the time, because the last flip is heads 75% of the time, or is it even 50% of the time, as it was before? Or is it even some other percentage of the time? Before you do the math, what's your gut feeling? And can you explain why your argument is only valid for a fair coin?
            $endgroup$
            – Eric Lippert
            22 hours ago












            $begingroup$
            Interesting question Eric. Formally I would use recursion, let us define $O_n$ as the probability that we get an odd number of heads when flipping $n$ coins, and let $p$ be the probability the coin comes up as heads. Then the previous reasoning turns into the recurrence relation $O_n=(1-p)O_{n-1} + p(1-O_{n-1})=(1-2p)O_{n-1}+p$.
            $endgroup$
            – Jorge Fernández Hidalgo
            21 hours ago






            $begingroup$
            Interesting question Eric. Formally I would use recursion, let us define $O_n$ as the probability that we get an odd number of heads when flipping $n$ coins, and let $p$ be the probability the coin comes up as heads. Then the previous reasoning turns into the recurrence relation $O_n=(1-p)O_{n-1} + p(1-O_{n-1})=(1-2p)O_{n-1}+p$.
            $endgroup$
            – Jorge Fernández Hidalgo
            21 hours ago






            14




            14




            $begingroup$
            That's a good analysis. My reason for bringing it up is: plainly the OP is relatively new to probability, so I think it is important to call out the assumptions when giving "shortcut" reasoning like this answer. The underlying assumption is that with eight coins, an even number of heads or tails is also 1/2. I think it is worthwhile calling that out. I wouldn't want the OP to deduce that for more general problems, they only need to consider the probability of the final flip.
            $endgroup$
            – Eric Lippert
            21 hours ago




            $begingroup$
            That's a good analysis. My reason for bringing it up is: plainly the OP is relatively new to probability, so I think it is important to call out the assumptions when giving "shortcut" reasoning like this answer. The underlying assumption is that with eight coins, an even number of heads or tails is also 1/2. I think it is worthwhile calling that out. I wouldn't want the OP to deduce that for more general problems, they only need to consider the probability of the final flip.
            $endgroup$
            – Eric Lippert
            21 hours ago




            2




            2




            $begingroup$
            Yes you are right, I was a bit careless when I wrote it, thank you for pointing out that detail!
            $endgroup$
            – Jorge Fernández Hidalgo
            21 hours ago






            $begingroup$
            Yes you are right, I was a bit careless when I wrote it, thank you for pointing out that detail!
            $endgroup$
            – Jorge Fernández Hidalgo
            21 hours ago






            3




            3




            $begingroup$
            I disagree that the underlying assumption is "with eight coins, an even number of heads occurs with probability $1/2$" -- that assumption is not necessary for the answer at all. If you flip 8 coins of varying biases, but the 9th and final one is a fair coin, then this answer is still correct.
            $endgroup$
            – Mees de Vries
            9 hours ago




            $begingroup$
            I disagree that the underlying assumption is "with eight coins, an even number of heads occurs with probability $1/2$" -- that assumption is not necessary for the answer at all. If you flip 8 coins of varying biases, but the 9th and final one is a fair coin, then this answer is still correct.
            $endgroup$
            – Mees de Vries
            9 hours ago











            31












            $begingroup$

            If there are an even number of heads then there must be an odd number of tails. But heads and tails are symmetrical, so the probability must be $1/2$.






            share|cite|improve this answer









            $endgroup$


















              31












              $begingroup$

              If there are an even number of heads then there must be an odd number of tails. But heads and tails are symmetrical, so the probability must be $1/2$.






              share|cite|improve this answer









              $endgroup$
















                31












                31








                31





                $begingroup$

                If there are an even number of heads then there must be an odd number of tails. But heads and tails are symmetrical, so the probability must be $1/2$.






                share|cite|improve this answer









                $endgroup$



                If there are an even number of heads then there must be an odd number of tails. But heads and tails are symmetrical, so the probability must be $1/2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                Ethan BolkerEthan Bolker

                44.3k553118




                44.3k553118























                    15












                    $begingroup$

                    Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is
                    $$binom{9}{0}+binom{9}{2}+binom{9}{4}+binom{9}{6}+binom{9}{8}=1+36+126+84+9=256$$
                    The number of all possible outcomes is $512$ thus the probability to get even number of heads is $0.5$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Comparing the number of favorable outcomes with all possible outcomes in this way only works for a fair (50% heads/50% tails) coin.
                      $endgroup$
                      – Richard Ward
                      4 hours ago
















                    15












                    $begingroup$

                    Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is
                    $$binom{9}{0}+binom{9}{2}+binom{9}{4}+binom{9}{6}+binom{9}{8}=1+36+126+84+9=256$$
                    The number of all possible outcomes is $512$ thus the probability to get even number of heads is $0.5$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Comparing the number of favorable outcomes with all possible outcomes in this way only works for a fair (50% heads/50% tails) coin.
                      $endgroup$
                      – Richard Ward
                      4 hours ago














                    15












                    15








                    15





                    $begingroup$

                    Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is
                    $$binom{9}{0}+binom{9}{2}+binom{9}{4}+binom{9}{6}+binom{9}{8}=1+36+126+84+9=256$$
                    The number of all possible outcomes is $512$ thus the probability to get even number of heads is $0.5$.






                    share|cite|improve this answer









                    $endgroup$



                    Your approach is good also, you probably made a mistake in calculations. The number of favorable outcomes is
                    $$binom{9}{0}+binom{9}{2}+binom{9}{4}+binom{9}{6}+binom{9}{8}=1+36+126+84+9=256$$
                    The number of all possible outcomes is $512$ thus the probability to get even number of heads is $0.5$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    VasyaVasya

                    3,5671517




                    3,5671517












                    • $begingroup$
                      Comparing the number of favorable outcomes with all possible outcomes in this way only works for a fair (50% heads/50% tails) coin.
                      $endgroup$
                      – Richard Ward
                      4 hours ago


















                    • $begingroup$
                      Comparing the number of favorable outcomes with all possible outcomes in this way only works for a fair (50% heads/50% tails) coin.
                      $endgroup$
                      – Richard Ward
                      4 hours ago
















                    $begingroup$
                    Comparing the number of favorable outcomes with all possible outcomes in this way only works for a fair (50% heads/50% tails) coin.
                    $endgroup$
                    – Richard Ward
                    4 hours ago




                    $begingroup$
                    Comparing the number of favorable outcomes with all possible outcomes in this way only works for a fair (50% heads/50% tails) coin.
                    $endgroup$
                    – Richard Ward
                    4 hours ago











                    14












                    $begingroup$

                    All the possible sequences of tosses (which are all equally likely) may be divided into the two categories "even number of heads" and "odd number of heads". You can then pair up each sequence in one category with the "flipped" version (flip every coin in the sequence around) in the other category. This shows that there are equally many sequences in each of the two categories. So the probability of landing in a specific one of them must be $frac12$.






                    share|cite|improve this answer









                    $endgroup$


















                      14












                      $begingroup$

                      All the possible sequences of tosses (which are all equally likely) may be divided into the two categories "even number of heads" and "odd number of heads". You can then pair up each sequence in one category with the "flipped" version (flip every coin in the sequence around) in the other category. This shows that there are equally many sequences in each of the two categories. So the probability of landing in a specific one of them must be $frac12$.






                      share|cite|improve this answer









                      $endgroup$
















                        14












                        14








                        14





                        $begingroup$

                        All the possible sequences of tosses (which are all equally likely) may be divided into the two categories "even number of heads" and "odd number of heads". You can then pair up each sequence in one category with the "flipped" version (flip every coin in the sequence around) in the other category. This shows that there are equally many sequences in each of the two categories. So the probability of landing in a specific one of them must be $frac12$.






                        share|cite|improve this answer









                        $endgroup$



                        All the possible sequences of tosses (which are all equally likely) may be divided into the two categories "even number of heads" and "odd number of heads". You can then pair up each sequence in one category with the "flipped" version (flip every coin in the sequence around) in the other category. This shows that there are equally many sequences in each of the two categories. So the probability of landing in a specific one of them must be $frac12$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered yesterday









                        ArthurArthur

                        117k7116200




                        117k7116200























                            8












                            $begingroup$

                            There are two cases here:




                            • There's an even number of heads: 0, 2, 4, 6 or 8 heads

                            • There's an odd number of heads: 1, 3, 5, 7 or 9 heads


                            But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:




                            • There's an even number of tails: 0, 2, 4, 6 or 8 tails


                            Since heads and tails are equally probable, we can, by symmetry, see that these two cases have the same probability. Therefore each must have probability $1/2$.






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              Ethan Bolker already gave this explanation 2 hours ago.
                              $endgroup$
                              – Paul Sinclair
                              yesterday






                            • 3




                              $begingroup$
                              @PaulSinclair Yes, he did. But I found this explanation clearer than Ethan's, which skips a number of steps.
                              $endgroup$
                              – CJ Dennis
                              20 hours ago










                            • $begingroup$
                              @PaulSinclair are they required to have read Ethan's answer first before going here?
                              $endgroup$
                              – The Great Duck
                              18 hours ago










                            • $begingroup$
                              I didn't understand Ethan's answer and I understood this one perfectly.
                              $endgroup$
                              – Todd Wilcox
                              6 hours ago
















                            8












                            $begingroup$

                            There are two cases here:




                            • There's an even number of heads: 0, 2, 4, 6 or 8 heads

                            • There's an odd number of heads: 1, 3, 5, 7 or 9 heads


                            But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:




                            • There's an even number of tails: 0, 2, 4, 6 or 8 tails


                            Since heads and tails are equally probable, we can, by symmetry, see that these two cases have the same probability. Therefore each must have probability $1/2$.






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              Ethan Bolker already gave this explanation 2 hours ago.
                              $endgroup$
                              – Paul Sinclair
                              yesterday






                            • 3




                              $begingroup$
                              @PaulSinclair Yes, he did. But I found this explanation clearer than Ethan's, which skips a number of steps.
                              $endgroup$
                              – CJ Dennis
                              20 hours ago










                            • $begingroup$
                              @PaulSinclair are they required to have read Ethan's answer first before going here?
                              $endgroup$
                              – The Great Duck
                              18 hours ago










                            • $begingroup$
                              I didn't understand Ethan's answer and I understood this one perfectly.
                              $endgroup$
                              – Todd Wilcox
                              6 hours ago














                            8












                            8








                            8





                            $begingroup$

                            There are two cases here:




                            • There's an even number of heads: 0, 2, 4, 6 or 8 heads

                            • There's an odd number of heads: 1, 3, 5, 7 or 9 heads


                            But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:




                            • There's an even number of tails: 0, 2, 4, 6 or 8 tails


                            Since heads and tails are equally probable, we can, by symmetry, see that these two cases have the same probability. Therefore each must have probability $1/2$.






                            share|cite|improve this answer









                            $endgroup$



                            There are two cases here:




                            • There's an even number of heads: 0, 2, 4, 6 or 8 heads

                            • There's an odd number of heads: 1, 3, 5, 7 or 9 heads


                            But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:




                            • There's an even number of tails: 0, 2, 4, 6 or 8 tails


                            Since heads and tails are equally probable, we can, by symmetry, see that these two cases have the same probability. Therefore each must have probability $1/2$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            FrxstremFrxstrem

                            4631813




                            4631813








                            • 1




                              $begingroup$
                              Ethan Bolker already gave this explanation 2 hours ago.
                              $endgroup$
                              – Paul Sinclair
                              yesterday






                            • 3




                              $begingroup$
                              @PaulSinclair Yes, he did. But I found this explanation clearer than Ethan's, which skips a number of steps.
                              $endgroup$
                              – CJ Dennis
                              20 hours ago










                            • $begingroup$
                              @PaulSinclair are they required to have read Ethan's answer first before going here?
                              $endgroup$
                              – The Great Duck
                              18 hours ago










                            • $begingroup$
                              I didn't understand Ethan's answer and I understood this one perfectly.
                              $endgroup$
                              – Todd Wilcox
                              6 hours ago














                            • 1




                              $begingroup$
                              Ethan Bolker already gave this explanation 2 hours ago.
                              $endgroup$
                              – Paul Sinclair
                              yesterday






                            • 3




                              $begingroup$
                              @PaulSinclair Yes, he did. But I found this explanation clearer than Ethan's, which skips a number of steps.
                              $endgroup$
                              – CJ Dennis
                              20 hours ago










                            • $begingroup$
                              @PaulSinclair are they required to have read Ethan's answer first before going here?
                              $endgroup$
                              – The Great Duck
                              18 hours ago










                            • $begingroup$
                              I didn't understand Ethan's answer and I understood this one perfectly.
                              $endgroup$
                              – Todd Wilcox
                              6 hours ago








                            1




                            1




                            $begingroup$
                            Ethan Bolker already gave this explanation 2 hours ago.
                            $endgroup$
                            – Paul Sinclair
                            yesterday




                            $begingroup$
                            Ethan Bolker already gave this explanation 2 hours ago.
                            $endgroup$
                            – Paul Sinclair
                            yesterday




                            3




                            3




                            $begingroup$
                            @PaulSinclair Yes, he did. But I found this explanation clearer than Ethan's, which skips a number of steps.
                            $endgroup$
                            – CJ Dennis
                            20 hours ago




                            $begingroup$
                            @PaulSinclair Yes, he did. But I found this explanation clearer than Ethan's, which skips a number of steps.
                            $endgroup$
                            – CJ Dennis
                            20 hours ago












                            $begingroup$
                            @PaulSinclair are they required to have read Ethan's answer first before going here?
                            $endgroup$
                            – The Great Duck
                            18 hours ago




                            $begingroup$
                            @PaulSinclair are they required to have read Ethan's answer first before going here?
                            $endgroup$
                            – The Great Duck
                            18 hours ago












                            $begingroup$
                            I didn't understand Ethan's answer and I understood this one perfectly.
                            $endgroup$
                            – Todd Wilcox
                            6 hours ago




                            $begingroup$
                            I didn't understand Ethan's answer and I understood this one perfectly.
                            $endgroup$
                            – Todd Wilcox
                            6 hours ago











                            7












                            $begingroup$

                            The easiest way to see this : Consider the number of heads we have in the first $8$ coins.




                            • If this number is even, we need a tail, we have probability $frac{1}{2}$

                            • If this number is odd, we need a head, we have probability $frac{1}{2}$


                            Hence no matter what the $8$ coins delivered, we have probability $frac{1}{2}$ , which is the answer.






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              This is , by the way, true for EVERY number of coins (even for one coin).
                              $endgroup$
                              – Peter
                              yesterday






                            • 6




                              $begingroup$
                              Pedantic point: If zero coins are tossed, the probability of an even number of heads is 100%.
                              $endgroup$
                              – Brian
                              yesterday






                            • 3




                              $begingroup$
                              This answer presupposes that having an even number of heads and an odd number of heads in eight flips is equally likely, but we haven't proved that lemma.
                              $endgroup$
                              – Eric Lippert
                              22 hours ago






                            • 1




                              $begingroup$
                              @EricLippert: No it doesn't. If the probability in both cases is the same, then the probability overall is the same...
                              $endgroup$
                              – user21820
                              14 hours ago






                            • 2




                              $begingroup$
                              @EricLippert No. If the probability of an even number of heads in 8 tosses is p (and the probability of an odd number of heads is 1-p), without assuming the value of p, the probability of an even number in all 9 tosses is p·(1/2) + (1-p)(1/2), which turns out to be exactly 1/2.
                              $endgroup$
                              – dgstranz
                              12 hours ago
















                            7












                            $begingroup$

                            The easiest way to see this : Consider the number of heads we have in the first $8$ coins.




                            • If this number is even, we need a tail, we have probability $frac{1}{2}$

                            • If this number is odd, we need a head, we have probability $frac{1}{2}$


                            Hence no matter what the $8$ coins delivered, we have probability $frac{1}{2}$ , which is the answer.






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              This is , by the way, true for EVERY number of coins (even for one coin).
                              $endgroup$
                              – Peter
                              yesterday






                            • 6




                              $begingroup$
                              Pedantic point: If zero coins are tossed, the probability of an even number of heads is 100%.
                              $endgroup$
                              – Brian
                              yesterday






                            • 3




                              $begingroup$
                              This answer presupposes that having an even number of heads and an odd number of heads in eight flips is equally likely, but we haven't proved that lemma.
                              $endgroup$
                              – Eric Lippert
                              22 hours ago






                            • 1




                              $begingroup$
                              @EricLippert: No it doesn't. If the probability in both cases is the same, then the probability overall is the same...
                              $endgroup$
                              – user21820
                              14 hours ago






                            • 2




                              $begingroup$
                              @EricLippert No. If the probability of an even number of heads in 8 tosses is p (and the probability of an odd number of heads is 1-p), without assuming the value of p, the probability of an even number in all 9 tosses is p·(1/2) + (1-p)(1/2), which turns out to be exactly 1/2.
                              $endgroup$
                              – dgstranz
                              12 hours ago














                            7












                            7








                            7





                            $begingroup$

                            The easiest way to see this : Consider the number of heads we have in the first $8$ coins.




                            • If this number is even, we need a tail, we have probability $frac{1}{2}$

                            • If this number is odd, we need a head, we have probability $frac{1}{2}$


                            Hence no matter what the $8$ coins delivered, we have probability $frac{1}{2}$ , which is the answer.






                            share|cite|improve this answer









                            $endgroup$



                            The easiest way to see this : Consider the number of heads we have in the first $8$ coins.




                            • If this number is even, we need a tail, we have probability $frac{1}{2}$

                            • If this number is odd, we need a head, we have probability $frac{1}{2}$


                            Hence no matter what the $8$ coins delivered, we have probability $frac{1}{2}$ , which is the answer.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            PeterPeter

                            48.2k1139133




                            48.2k1139133








                            • 1




                              $begingroup$
                              This is , by the way, true for EVERY number of coins (even for one coin).
                              $endgroup$
                              – Peter
                              yesterday






                            • 6




                              $begingroup$
                              Pedantic point: If zero coins are tossed, the probability of an even number of heads is 100%.
                              $endgroup$
                              – Brian
                              yesterday






                            • 3




                              $begingroup$
                              This answer presupposes that having an even number of heads and an odd number of heads in eight flips is equally likely, but we haven't proved that lemma.
                              $endgroup$
                              – Eric Lippert
                              22 hours ago






                            • 1




                              $begingroup$
                              @EricLippert: No it doesn't. If the probability in both cases is the same, then the probability overall is the same...
                              $endgroup$
                              – user21820
                              14 hours ago






                            • 2




                              $begingroup$
                              @EricLippert No. If the probability of an even number of heads in 8 tosses is p (and the probability of an odd number of heads is 1-p), without assuming the value of p, the probability of an even number in all 9 tosses is p·(1/2) + (1-p)(1/2), which turns out to be exactly 1/2.
                              $endgroup$
                              – dgstranz
                              12 hours ago














                            • 1




                              $begingroup$
                              This is , by the way, true for EVERY number of coins (even for one coin).
                              $endgroup$
                              – Peter
                              yesterday






                            • 6




                              $begingroup$
                              Pedantic point: If zero coins are tossed, the probability of an even number of heads is 100%.
                              $endgroup$
                              – Brian
                              yesterday






                            • 3




                              $begingroup$
                              This answer presupposes that having an even number of heads and an odd number of heads in eight flips is equally likely, but we haven't proved that lemma.
                              $endgroup$
                              – Eric Lippert
                              22 hours ago






                            • 1




                              $begingroup$
                              @EricLippert: No it doesn't. If the probability in both cases is the same, then the probability overall is the same...
                              $endgroup$
                              – user21820
                              14 hours ago






                            • 2




                              $begingroup$
                              @EricLippert No. If the probability of an even number of heads in 8 tosses is p (and the probability of an odd number of heads is 1-p), without assuming the value of p, the probability of an even number in all 9 tosses is p·(1/2) + (1-p)(1/2), which turns out to be exactly 1/2.
                              $endgroup$
                              – dgstranz
                              12 hours ago








                            1




                            1




                            $begingroup$
                            This is , by the way, true for EVERY number of coins (even for one coin).
                            $endgroup$
                            – Peter
                            yesterday




                            $begingroup$
                            This is , by the way, true for EVERY number of coins (even for one coin).
                            $endgroup$
                            – Peter
                            yesterday




                            6




                            6




                            $begingroup$
                            Pedantic point: If zero coins are tossed, the probability of an even number of heads is 100%.
                            $endgroup$
                            – Brian
                            yesterday




                            $begingroup$
                            Pedantic point: If zero coins are tossed, the probability of an even number of heads is 100%.
                            $endgroup$
                            – Brian
                            yesterday




                            3




                            3




                            $begingroup$
                            This answer presupposes that having an even number of heads and an odd number of heads in eight flips is equally likely, but we haven't proved that lemma.
                            $endgroup$
                            – Eric Lippert
                            22 hours ago




                            $begingroup$
                            This answer presupposes that having an even number of heads and an odd number of heads in eight flips is equally likely, but we haven't proved that lemma.
                            $endgroup$
                            – Eric Lippert
                            22 hours ago




                            1




                            1




                            $begingroup$
                            @EricLippert: No it doesn't. If the probability in both cases is the same, then the probability overall is the same...
                            $endgroup$
                            – user21820
                            14 hours ago




                            $begingroup$
                            @EricLippert: No it doesn't. If the probability in both cases is the same, then the probability overall is the same...
                            $endgroup$
                            – user21820
                            14 hours ago




                            2




                            2




                            $begingroup$
                            @EricLippert No. If the probability of an even number of heads in 8 tosses is p (and the probability of an odd number of heads is 1-p), without assuming the value of p, the probability of an even number in all 9 tosses is p·(1/2) + (1-p)(1/2), which turns out to be exactly 1/2.
                            $endgroup$
                            – dgstranz
                            12 hours ago




                            $begingroup$
                            @EricLippert No. If the probability of an even number of heads in 8 tosses is p (and the probability of an odd number of heads is 1-p), without assuming the value of p, the probability of an even number in all 9 tosses is p·(1/2) + (1-p)(1/2), which turns out to be exactly 1/2.
                            $endgroup$
                            – dgstranz
                            12 hours ago











                            4












                            $begingroup$

                            There's a way to do it with barely any maths:



                            It's clear that if there's an odd number of heads, there's an even number of tails and vice versa, so P(even number of heads) + P(even number of tails) = 1.



                            Formally rename "heads" to "tails". The problem remains unchanged.



                            So P(even number of heads) = P(even number of tails) = 1/2.






                            share|cite|improve this answer










                            New contributor




                            Remellion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            $endgroup$


















                              4












                              $begingroup$

                              There's a way to do it with barely any maths:



                              It's clear that if there's an odd number of heads, there's an even number of tails and vice versa, so P(even number of heads) + P(even number of tails) = 1.



                              Formally rename "heads" to "tails". The problem remains unchanged.



                              So P(even number of heads) = P(even number of tails) = 1/2.






                              share|cite|improve this answer










                              New contributor




                              Remellion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.






                              $endgroup$
















                                4












                                4








                                4





                                $begingroup$

                                There's a way to do it with barely any maths:



                                It's clear that if there's an odd number of heads, there's an even number of tails and vice versa, so P(even number of heads) + P(even number of tails) = 1.



                                Formally rename "heads" to "tails". The problem remains unchanged.



                                So P(even number of heads) = P(even number of tails) = 1/2.






                                share|cite|improve this answer










                                New contributor




                                Remellion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                $endgroup$



                                There's a way to do it with barely any maths:



                                It's clear that if there's an odd number of heads, there's an even number of tails and vice versa, so P(even number of heads) + P(even number of tails) = 1.



                                Formally rename "heads" to "tails". The problem remains unchanged.



                                So P(even number of heads) = P(even number of tails) = 1/2.







                                share|cite|improve this answer










                                New contributor




                                Remellion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.









                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited 14 hours ago





















                                New contributor




                                Remellion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.









                                answered 14 hours ago









                                RemellionRemellion

                                1434




                                1434




                                New contributor




                                Remellion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.





                                New contributor





                                Remellion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                Remellion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.























                                    3












                                    $begingroup$

                                    The probability generating function of a Binomiall random variable $Xsim text{Bin}(n, 1/2)$ with probability of success $1/2$ is given by
                                    $$
                                    g_{X}(t)=Et^X=sum_{k=0}^nP(X=k)t^k=sum_{k=0}^nbinom{n}{k}frac{t^k}{2^n}=frac{1}{2^n}(1+t)^n
                                    $$

                                    In particular the probability that $X$ is even is given by
                                    $$
                                    sum_{0leq kleq n, k,{text{even}}}P(X=k)=frac{g(1)+g(-1)}{2}=frac{1+0}{2}=frac{1}{2}.
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$


















                                      3












                                      $begingroup$

                                      The probability generating function of a Binomiall random variable $Xsim text{Bin}(n, 1/2)$ with probability of success $1/2$ is given by
                                      $$
                                      g_{X}(t)=Et^X=sum_{k=0}^nP(X=k)t^k=sum_{k=0}^nbinom{n}{k}frac{t^k}{2^n}=frac{1}{2^n}(1+t)^n
                                      $$

                                      In particular the probability that $X$ is even is given by
                                      $$
                                      sum_{0leq kleq n, k,{text{even}}}P(X=k)=frac{g(1)+g(-1)}{2}=frac{1+0}{2}=frac{1}{2}.
                                      $$






                                      share|cite|improve this answer









                                      $endgroup$
















                                        3












                                        3








                                        3





                                        $begingroup$

                                        The probability generating function of a Binomiall random variable $Xsim text{Bin}(n, 1/2)$ with probability of success $1/2$ is given by
                                        $$
                                        g_{X}(t)=Et^X=sum_{k=0}^nP(X=k)t^k=sum_{k=0}^nbinom{n}{k}frac{t^k}{2^n}=frac{1}{2^n}(1+t)^n
                                        $$

                                        In particular the probability that $X$ is even is given by
                                        $$
                                        sum_{0leq kleq n, k,{text{even}}}P(X=k)=frac{g(1)+g(-1)}{2}=frac{1+0}{2}=frac{1}{2}.
                                        $$






                                        share|cite|improve this answer









                                        $endgroup$



                                        The probability generating function of a Binomiall random variable $Xsim text{Bin}(n, 1/2)$ with probability of success $1/2$ is given by
                                        $$
                                        g_{X}(t)=Et^X=sum_{k=0}^nP(X=k)t^k=sum_{k=0}^nbinom{n}{k}frac{t^k}{2^n}=frac{1}{2^n}(1+t)^n
                                        $$

                                        In particular the probability that $X$ is even is given by
                                        $$
                                        sum_{0leq kleq n, k,{text{even}}}P(X=k)=frac{g(1)+g(-1)}{2}=frac{1+0}{2}=frac{1}{2}.
                                        $$







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 22 hours ago









                                        Foobaz JohnFoobaz John

                                        22.4k41452




                                        22.4k41452























                                            2












                                            $begingroup$

                                            If $h$ denotes getting a heads and $p$ denotes getting tails, let's write $frac{1}{2}h+frac{1}{2}p$ for the notion of a fair coin: half the time it turns up heads, and half the time tails.



                                            If we expand out the following product while keeping track of multiplication order,
                                            $$left(frac{1}{2}h+frac{1}{2}pright)left(frac{1}{2}h+frac{1}{2}pright)=frac{1}{4}hh+frac{1}{4}hp+frac{1}{4}ph+frac{1}{4}pp,$$
                                            we see that the sequences $hh$, $hp$, $ph$, and $pp$ are equally likely. Forgetting about multiplication order, which we want to do because we only care about how many times heads or tails showed up, corresponds to just treating this like a polynomial:
                                            $$=frac{1}{4}h^2+frac{1}{2}hp+frac{1}{4}p^2.$$
                                            We can keep multiplying copies of a fair coin together to find the probability that a certain number of heads or tails happened, where the coefficient in front of $h^kp^ell$ is the probability of $k$ heads and $ell$ tails.



                                            Nine coins is the expansion
                                            $$left(frac{1}{2}h+frac{1}{2}pright)^9=sum_{k=0}^9binom{9}{k}left(frac{1}{2}hright)^kleft(frac{1}{2}pright)^{9-k}=sum_{k=0}^9frac{1}{2^9}binom{9}{k}h^kp^{9-k}.$$
                                            So far, all this has done is explain why you were adding up $2^{-9}binom{9}{k}$ for $k=0,2,4,dots,8$. Here, now, is a nice trick. If we formally set $h=1$ and $p=1$, then we get
                                            $$1=sum_{k=0}^9frac{1}{2^9}binom{9}{k},$$
                                            and if we formally set $h=-1$ and $p=1$, then we get
                                            $$0=sum_{k=0}^9frac{1}{2^9}binom{9}{k}(-1)^k.$$
                                            The average of these two equations is
                                            $$frac{1}{2}=sum_{k=0,ktext{ even}}^9frac{1}{2^9}binom{9}{k},$$
                                            since $frac{1}{2}(1+(-1)^k)$ is $1$ or $0$ depending on whether $k$ is even or odd. Thus the probability of an even number of heads is $frac{1}{2}$.



                                            Notice that this did not use the fact nine coins were tossed at any point! (Other than the fact that at least one coin was tossed. In the case of tossing no coins, an even number of heads happens with probability $1$. What part of my argument goes wrong for the case of zero coins?)






                                            share|cite|improve this answer









                                            $endgroup$


















                                              2












                                              $begingroup$

                                              If $h$ denotes getting a heads and $p$ denotes getting tails, let's write $frac{1}{2}h+frac{1}{2}p$ for the notion of a fair coin: half the time it turns up heads, and half the time tails.



                                              If we expand out the following product while keeping track of multiplication order,
                                              $$left(frac{1}{2}h+frac{1}{2}pright)left(frac{1}{2}h+frac{1}{2}pright)=frac{1}{4}hh+frac{1}{4}hp+frac{1}{4}ph+frac{1}{4}pp,$$
                                              we see that the sequences $hh$, $hp$, $ph$, and $pp$ are equally likely. Forgetting about multiplication order, which we want to do because we only care about how many times heads or tails showed up, corresponds to just treating this like a polynomial:
                                              $$=frac{1}{4}h^2+frac{1}{2}hp+frac{1}{4}p^2.$$
                                              We can keep multiplying copies of a fair coin together to find the probability that a certain number of heads or tails happened, where the coefficient in front of $h^kp^ell$ is the probability of $k$ heads and $ell$ tails.



                                              Nine coins is the expansion
                                              $$left(frac{1}{2}h+frac{1}{2}pright)^9=sum_{k=0}^9binom{9}{k}left(frac{1}{2}hright)^kleft(frac{1}{2}pright)^{9-k}=sum_{k=0}^9frac{1}{2^9}binom{9}{k}h^kp^{9-k}.$$
                                              So far, all this has done is explain why you were adding up $2^{-9}binom{9}{k}$ for $k=0,2,4,dots,8$. Here, now, is a nice trick. If we formally set $h=1$ and $p=1$, then we get
                                              $$1=sum_{k=0}^9frac{1}{2^9}binom{9}{k},$$
                                              and if we formally set $h=-1$ and $p=1$, then we get
                                              $$0=sum_{k=0}^9frac{1}{2^9}binom{9}{k}(-1)^k.$$
                                              The average of these two equations is
                                              $$frac{1}{2}=sum_{k=0,ktext{ even}}^9frac{1}{2^9}binom{9}{k},$$
                                              since $frac{1}{2}(1+(-1)^k)$ is $1$ or $0$ depending on whether $k$ is even or odd. Thus the probability of an even number of heads is $frac{1}{2}$.



                                              Notice that this did not use the fact nine coins were tossed at any point! (Other than the fact that at least one coin was tossed. In the case of tossing no coins, an even number of heads happens with probability $1$. What part of my argument goes wrong for the case of zero coins?)






                                              share|cite|improve this answer









                                              $endgroup$
















                                                2












                                                2








                                                2





                                                $begingroup$

                                                If $h$ denotes getting a heads and $p$ denotes getting tails, let's write $frac{1}{2}h+frac{1}{2}p$ for the notion of a fair coin: half the time it turns up heads, and half the time tails.



                                                If we expand out the following product while keeping track of multiplication order,
                                                $$left(frac{1}{2}h+frac{1}{2}pright)left(frac{1}{2}h+frac{1}{2}pright)=frac{1}{4}hh+frac{1}{4}hp+frac{1}{4}ph+frac{1}{4}pp,$$
                                                we see that the sequences $hh$, $hp$, $ph$, and $pp$ are equally likely. Forgetting about multiplication order, which we want to do because we only care about how many times heads or tails showed up, corresponds to just treating this like a polynomial:
                                                $$=frac{1}{4}h^2+frac{1}{2}hp+frac{1}{4}p^2.$$
                                                We can keep multiplying copies of a fair coin together to find the probability that a certain number of heads or tails happened, where the coefficient in front of $h^kp^ell$ is the probability of $k$ heads and $ell$ tails.



                                                Nine coins is the expansion
                                                $$left(frac{1}{2}h+frac{1}{2}pright)^9=sum_{k=0}^9binom{9}{k}left(frac{1}{2}hright)^kleft(frac{1}{2}pright)^{9-k}=sum_{k=0}^9frac{1}{2^9}binom{9}{k}h^kp^{9-k}.$$
                                                So far, all this has done is explain why you were adding up $2^{-9}binom{9}{k}$ for $k=0,2,4,dots,8$. Here, now, is a nice trick. If we formally set $h=1$ and $p=1$, then we get
                                                $$1=sum_{k=0}^9frac{1}{2^9}binom{9}{k},$$
                                                and if we formally set $h=-1$ and $p=1$, then we get
                                                $$0=sum_{k=0}^9frac{1}{2^9}binom{9}{k}(-1)^k.$$
                                                The average of these two equations is
                                                $$frac{1}{2}=sum_{k=0,ktext{ even}}^9frac{1}{2^9}binom{9}{k},$$
                                                since $frac{1}{2}(1+(-1)^k)$ is $1$ or $0$ depending on whether $k$ is even or odd. Thus the probability of an even number of heads is $frac{1}{2}$.



                                                Notice that this did not use the fact nine coins were tossed at any point! (Other than the fact that at least one coin was tossed. In the case of tossing no coins, an even number of heads happens with probability $1$. What part of my argument goes wrong for the case of zero coins?)






                                                share|cite|improve this answer









                                                $endgroup$



                                                If $h$ denotes getting a heads and $p$ denotes getting tails, let's write $frac{1}{2}h+frac{1}{2}p$ for the notion of a fair coin: half the time it turns up heads, and half the time tails.



                                                If we expand out the following product while keeping track of multiplication order,
                                                $$left(frac{1}{2}h+frac{1}{2}pright)left(frac{1}{2}h+frac{1}{2}pright)=frac{1}{4}hh+frac{1}{4}hp+frac{1}{4}ph+frac{1}{4}pp,$$
                                                we see that the sequences $hh$, $hp$, $ph$, and $pp$ are equally likely. Forgetting about multiplication order, which we want to do because we only care about how many times heads or tails showed up, corresponds to just treating this like a polynomial:
                                                $$=frac{1}{4}h^2+frac{1}{2}hp+frac{1}{4}p^2.$$
                                                We can keep multiplying copies of a fair coin together to find the probability that a certain number of heads or tails happened, where the coefficient in front of $h^kp^ell$ is the probability of $k$ heads and $ell$ tails.



                                                Nine coins is the expansion
                                                $$left(frac{1}{2}h+frac{1}{2}pright)^9=sum_{k=0}^9binom{9}{k}left(frac{1}{2}hright)^kleft(frac{1}{2}pright)^{9-k}=sum_{k=0}^9frac{1}{2^9}binom{9}{k}h^kp^{9-k}.$$
                                                So far, all this has done is explain why you were adding up $2^{-9}binom{9}{k}$ for $k=0,2,4,dots,8$. Here, now, is a nice trick. If we formally set $h=1$ and $p=1$, then we get
                                                $$1=sum_{k=0}^9frac{1}{2^9}binom{9}{k},$$
                                                and if we formally set $h=-1$ and $p=1$, then we get
                                                $$0=sum_{k=0}^9frac{1}{2^9}binom{9}{k}(-1)^k.$$
                                                The average of these two equations is
                                                $$frac{1}{2}=sum_{k=0,ktext{ even}}^9frac{1}{2^9}binom{9}{k},$$
                                                since $frac{1}{2}(1+(-1)^k)$ is $1$ or $0$ depending on whether $k$ is even or odd. Thus the probability of an even number of heads is $frac{1}{2}$.



                                                Notice that this did not use the fact nine coins were tossed at any point! (Other than the fact that at least one coin was tossed. In the case of tossing no coins, an even number of heads happens with probability $1$. What part of my argument goes wrong for the case of zero coins?)







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered 19 hours ago









                                                Kyle MillerKyle Miller

                                                9,485930




                                                9,485930






























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