Difficult Coin Problem












4












$begingroup$



Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?



Official answer: 2/3, but I cannot arrive at it.




Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.



When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.



However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.






probability







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$endgroup$












  • $begingroup$
    What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
    $endgroup$
    – Pakk
    7 hours ago










  • $begingroup$
    I have understood the flaw in my method by understanding the correct answer, but thank you anyways
    $endgroup$
    – Victor S.
    6 hours ago
















4












$begingroup$



Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?



Official answer: 2/3, but I cannot arrive at it.




Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.



When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.



However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
    $endgroup$
    – Pakk
    7 hours ago










  • $begingroup$
    I have understood the flaw in my method by understanding the correct answer, but thank you anyways
    $endgroup$
    – Victor S.
    6 hours ago














4












4








4





$begingroup$



Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?



Official answer: 2/3, but I cannot arrive at it.




Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.



When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.



However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.






probability







share|cite|improve this question









$endgroup$





Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?



Official answer: 2/3, but I cannot arrive at it.




Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.



When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.



However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 12 hours ago









Victor S.Victor S.

897




897












  • $begingroup$
    What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
    $endgroup$
    – Pakk
    7 hours ago










  • $begingroup$
    I have understood the flaw in my method by understanding the correct answer, but thank you anyways
    $endgroup$
    – Victor S.
    6 hours ago


















  • $begingroup$
    What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
    $endgroup$
    – Pakk
    7 hours ago










  • $begingroup$
    I have understood the flaw in my method by understanding the correct answer, but thank you anyways
    $endgroup$
    – Victor S.
    6 hours ago
















$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
7 hours ago




$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
7 hours ago












$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
6 hours ago




$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
6 hours ago










5 Answers
5






active

oldest

votes


















10












$begingroup$

So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
$$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



where $p$ is probability that head comes in one toss and $q=1-p$.






share|cite|improve this answer









$endgroup$





















    20












    $begingroup$

    Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$






    share|cite|improve this answer









    $endgroup$





















      6












      $begingroup$

      Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
      $$p = .5+ .25p$$
      which yields $p=2/3$.






      share|cite|improve this answer









      $endgroup$





















        6












        $begingroup$

        Here's another approach.



        Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



        In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



        But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



        This gives A a $frac23$ chance of winning to $frac13$ for B.






        share|cite|improve this answer









        $endgroup$





















          0












          $begingroup$

          P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




          A loses first round, B loses second round, A wins the third round (0.5)^3




          And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




          P(A winning) = 0.5/(1-0.25)







          share|cite|improve this answer








          New contributor




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






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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
            $$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



            where $p$ is probability that head comes in one toss and $q=1-p$.






            share|cite|improve this answer









            $endgroup$


















              10












              $begingroup$

              So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
              $$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



              where $p$ is probability that head comes in one toss and $q=1-p$.






              share|cite|improve this answer









              $endgroup$
















                10












                10








                10





                $begingroup$

                So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
                $$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



                where $p$ is probability that head comes in one toss and $q=1-p$.






                share|cite|improve this answer









                $endgroup$



                So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
                $$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



                where $p$ is probability that head comes in one toss and $q=1-p$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 12 hours ago









                greedoidgreedoid

                41.8k1152105




                41.8k1152105























                    20












                    $begingroup$

                    Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$






                    share|cite|improve this answer









                    $endgroup$


















                      20












                      $begingroup$

                      Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$






                      share|cite|improve this answer









                      $endgroup$
















                        20












                        20








                        20





                        $begingroup$

                        Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$






                        share|cite|improve this answer









                        $endgroup$



                        Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 12 hours ago









                        saulspatzsaulspatz

                        15k21330




                        15k21330























                            6












                            $begingroup$

                            Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
                            $$p = .5+ .25p$$
                            which yields $p=2/3$.






                            share|cite|improve this answer









                            $endgroup$


















                              6












                              $begingroup$

                              Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
                              $$p = .5+ .25p$$
                              which yields $p=2/3$.






                              share|cite|improve this answer









                              $endgroup$
















                                6












                                6








                                6





                                $begingroup$

                                Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
                                $$p = .5+ .25p$$
                                which yields $p=2/3$.






                                share|cite|improve this answer









                                $endgroup$



                                Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
                                $$p = .5+ .25p$$
                                which yields $p=2/3$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 12 hours ago









                                pwerthpwerth

                                3,233417




                                3,233417























                                    6












                                    $begingroup$

                                    Here's another approach.



                                    Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



                                    In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



                                    But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



                                    This gives A a $frac23$ chance of winning to $frac13$ for B.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      6












                                      $begingroup$

                                      Here's another approach.



                                      Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



                                      In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



                                      But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



                                      This gives A a $frac23$ chance of winning to $frac13$ for B.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        6












                                        6








                                        6





                                        $begingroup$

                                        Here's another approach.



                                        Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



                                        In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



                                        But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



                                        This gives A a $frac23$ chance of winning to $frac13$ for B.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Here's another approach.



                                        Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



                                        In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



                                        But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



                                        This gives A a $frac23$ chance of winning to $frac13$ for B.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 11 hours ago









                                        paw88789paw88789

                                        29.1k12349




                                        29.1k12349























                                            0












                                            $begingroup$

                                            P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




                                            A loses first round, B loses second round, A wins the third round (0.5)^3




                                            And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




                                            P(A winning) = 0.5/(1-0.25)







                                            share|cite|improve this answer








                                            New contributor




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






                                            $endgroup$


















                                              0












                                              $begingroup$

                                              P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




                                              A loses first round, B loses second round, A wins the third round (0.5)^3




                                              And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




                                              P(A winning) = 0.5/(1-0.25)







                                              share|cite|improve this answer








                                              New contributor




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






                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




                                                A loses first round, B loses second round, A wins the third round (0.5)^3




                                                And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




                                                P(A winning) = 0.5/(1-0.25)







                                                share|cite|improve this answer








                                                New contributor




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






                                                $endgroup$



                                                P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




                                                A loses first round, B loses second round, A wins the third round (0.5)^3




                                                And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




                                                P(A winning) = 0.5/(1-0.25)








                                                share|cite|improve this answer








                                                New contributor




                                                Shreyas 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






                                                New contributor




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









                                                answered 11 hours ago









                                                Shreyas Shreyas

                                                812




                                                812




                                                New contributor




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





                                                New contributor





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






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






























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                                                    Tabula Rosettana

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                                                    Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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