Birthdays on Consecutive Days












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You are allowed to select as many random people as you like, and place them in a room. How many do you need for the chances of at least 2 people having a birthday on the same day, or consecutive days, to be greater than 50%? Please explain why this so so.



Note; You may ignore leap years if it interferes with the answer, which I'm pretty sure it would.



CLARIFICATIONS:




  1. the birthday references the day and month. Years do not count.

  2. In this case, random means they have a 1/365 chance of having a birthday on Jan. 1, or any other day.






mathematics probability







share|improve this question











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  • 3




    $begingroup$
    en.wikipedia.org/wiki/Birthday_problem
    $endgroup$
    – Victor Stafusa
    Dec 24 '14 at 14:24






  • 1




    $begingroup$
    @Victor Don't dare ruin my puzzle with the Wikipedia!
    $endgroup$
    – warspyking
    Dec 24 '14 at 14:31










  • $begingroup$
    Besides @Victor this includes consecutive birthdays too.
    $endgroup$
    – warspyking
    Dec 24 '14 at 14:32
















1












$begingroup$


You are allowed to select as many random people as you like, and place them in a room. How many do you need for the chances of at least 2 people having a birthday on the same day, or consecutive days, to be greater than 50%? Please explain why this so so.



Note; You may ignore leap years if it interferes with the answer, which I'm pretty sure it would.



CLARIFICATIONS:




  1. the birthday references the day and month. Years do not count.

  2. In this case, random means they have a 1/365 chance of having a birthday on Jan. 1, or any other day.






mathematics probability







share|improve this question











$endgroup$








  • 3




    $begingroup$
    en.wikipedia.org/wiki/Birthday_problem
    $endgroup$
    – Victor Stafusa
    Dec 24 '14 at 14:24






  • 1




    $begingroup$
    @Victor Don't dare ruin my puzzle with the Wikipedia!
    $endgroup$
    – warspyking
    Dec 24 '14 at 14:31










  • $begingroup$
    Besides @Victor this includes consecutive birthdays too.
    $endgroup$
    – warspyking
    Dec 24 '14 at 14:32














1












1








1





$begingroup$


You are allowed to select as many random people as you like, and place them in a room. How many do you need for the chances of at least 2 people having a birthday on the same day, or consecutive days, to be greater than 50%? Please explain why this so so.



Note; You may ignore leap years if it interferes with the answer, which I'm pretty sure it would.



CLARIFICATIONS:




  1. the birthday references the day and month. Years do not count.

  2. In this case, random means they have a 1/365 chance of having a birthday on Jan. 1, or any other day.






mathematics probability







share|improve this question











$endgroup$




You are allowed to select as many random people as you like, and place them in a room. How many do you need for the chances of at least 2 people having a birthday on the same day, or consecutive days, to be greater than 50%? Please explain why this so so.



Note; You may ignore leap years if it interferes with the answer, which I'm pretty sure it would.



CLARIFICATIONS:




  1. the birthday references the day and month. Years do not count.

  2. In this case, random means they have a 1/365 chance of having a birthday on Jan. 1, or any other day.






mathematics probability




mathematics probability






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 24 '14 at 14:16







warspyking

















asked Dec 24 '14 at 14:08









warspykingwarspyking

7,425854129




7,425854129








  • 3




    $begingroup$
    en.wikipedia.org/wiki/Birthday_problem
    $endgroup$
    – Victor Stafusa
    Dec 24 '14 at 14:24






  • 1




    $begingroup$
    @Victor Don't dare ruin my puzzle with the Wikipedia!
    $endgroup$
    – warspyking
    Dec 24 '14 at 14:31










  • $begingroup$
    Besides @Victor this includes consecutive birthdays too.
    $endgroup$
    – warspyking
    Dec 24 '14 at 14:32














  • 3




    $begingroup$
    en.wikipedia.org/wiki/Birthday_problem
    $endgroup$
    – Victor Stafusa
    Dec 24 '14 at 14:24






  • 1




    $begingroup$
    @Victor Don't dare ruin my puzzle with the Wikipedia!
    $endgroup$
    – warspyking
    Dec 24 '14 at 14:31










  • $begingroup$
    Besides @Victor this includes consecutive birthdays too.
    $endgroup$
    – warspyking
    Dec 24 '14 at 14:32








3




3




$begingroup$
en.wikipedia.org/wiki/Birthday_problem
$endgroup$
– Victor Stafusa
Dec 24 '14 at 14:24




$begingroup$
en.wikipedia.org/wiki/Birthday_problem
$endgroup$
– Victor Stafusa
Dec 24 '14 at 14:24




1




1




$begingroup$
@Victor Don't dare ruin my puzzle with the Wikipedia!
$endgroup$
– warspyking
Dec 24 '14 at 14:31




$begingroup$
@Victor Don't dare ruin my puzzle with the Wikipedia!
$endgroup$
– warspyking
Dec 24 '14 at 14:31












$begingroup$
Besides @Victor this includes consecutive birthdays too.
$endgroup$
– warspyking
Dec 24 '14 at 14:32




$begingroup$
Besides @Victor this includes consecutive birthdays too.
$endgroup$
– warspyking
Dec 24 '14 at 14:32










4 Answers
4






active

oldest

votes


















6












$begingroup$

Let's suppose there are $r$ people in the room. We want to find the probability that at least two have the same birthday or adjacent birthdays. It seems easier to find the complementary probability, i.e. the probability that no two have the same or adjacent birthdays.



There are $365^r$ possible arrangements of birthdays for the $r$ people, all of which we assume are equally likely. The number of arrangements in which the birthdays are distinct and non-adjacent can be obtained from a well-known combinatorial formula, as follows. There are $binom{365-r+1}{r}$ ways to pick $r$ integers from 1 to 365 with no two adjacent, where $binom{n}{m}$ is a binomial coefficient: $binom{n}{m} = frac{n!}{m! (n-m)!}$. But we need to multiply by $r!$, because we are considering the people to be distinguishable, so the order of the $r$ integers is significant. Hence the probability that no two people have the same or adjacent birthdays is
$$f(r) = frac{binom{365-r+1}{r} r!}{365^r}$$



The probability that at least two people do have the same or adjacent birthdays is $1-f(r)$. If we do the computations, we find $1-f(13) = 0.4822$ and $1-f(14) = 0.5368$, so the answer to your question is that you need 14 people.



I should add that we're not considering 31 December and 1 January to be adjacent. If we did, it would change the probabilities a little bit, but I don't think it would change the final answer of 14 people.






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Nice answer. In case you're interested, if we count December 31 and January 1 as adjacent, the number of ways to pick $r$ days with no two adjacent is ${365-r+1choose r}-{365-r-1choose r-2}$. I checked, and you are correct that this doesn't change the answer of 14.
    $endgroup$
    – Julian Rosen
    Dec 24 '14 at 17:43



















4












$begingroup$

This is more complicated than simply the birthday problem times three, because there's the potential for overlap. For instance, if person A is born on Jan 5, and person B is born on Jan 7, then person C only has a 5/365 chance to meet the criteria.



Rather than do the math, I wrote a script to solve it, ignoring leap years (although it turns out that the margin between N=16 and N=17 is large enough that I expect the same answer holds if leap years are accounted for).



With 16 people, you will satisfy the criteria 48.3% of the time, and with 17 people, you satisfy them 52.6% of the time, so the answer is 17. I put the script up at http://pastebin.com/YjNzECrP if anyone would like to check my work.



EDIT: Thanks to Lopsy for pointing out a mistake in the script, fixing that causes it to produce 14 people as the answer.






share|improve this answer











$endgroup$













  • $begingroup$
    thank you for right approach :) (im too lazy to calculate that) It seems OP does not know the answer, so he accepted other one.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:25










  • $begingroup$
    Your script has a typo. b.count([i]) > 1 should be b.count(i) > 1. As a result, it only counts consecutive birthdays, not equal birthdays. The correct answer is 14, as was proven by awkward.
    $endgroup$
    – Lopsy
    Dec 24 '14 at 17:47












  • $begingroup$
    I knew it was 14...
    $endgroup$
    – warspyking
    Dec 25 '14 at 1:40



















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Like the birthday problem, but with triple the likelihood, I guess...



So, person 1 would have a birthday, doesn't matter when, won't be before or after anyone else.



Person 2 could have their birthday on any of the 365 days (lets ignore leap years, since the question does), chance of the 2nd person being within a day of the first person is 3/365.



If they don't (362/365) then the next person has a 6/365 chance to be within a day of the first two, and so on.



So we have 1 * 362/365 * 359/365 * ... First time this drops below 50% is 365/365 * 362/365 * 359/365 * 356/365 * 353/365 * 350/365 * 347/365 * 344/365 * 341/365 * 338/365 * 335/365 * 332/365 * 329/365 * 326/365 = 0.4596676, so 14 people.



Edit: Typed too fast and fat fingers took over :)






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    actually with your thinking/calculation it should be 14 people with prob is equal to 0.4596676222355141. You jumped from 365 to 352. However still this isnt the right answer for question.
    $endgroup$
    – shyos
    Dec 24 '14 at 14:48












  • $begingroup$
    D'oh, thanks for spotting that - updated now.
    $endgroup$
    – CaffeinatedDave
    Dec 24 '14 at 14:52










  • $begingroup$
    this is not the right answer. why accepted?
    $endgroup$
    – shyos
    Dec 24 '14 at 15:21












  • $begingroup$
    @warspyking Then your source is wrong, this answer does not include overlapping days.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:28












  • $begingroup$
    @warspyking I can't reach his script from work, but his approach is right, if code is also right then it is the right answer.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:32



















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The solution:




If we consider Jan 1 and Dec 31 not adjacent is given above. In the comment of the accepted answer, it is indicated that if we consider Jan 1 and Dec 31 adjacent, the number of ways to arrange the birthdays is $binom{365-r+1}{r}-binom{365-r-1}{r-2}$. I'm having some trouble understanding this.




Could anyone points out where I'm doing wrong?



I think this is equivalent to ask:




How many ways can we arrange 365-r $0$s and r $1$s such that the ones are not adjacent. First, for the case treating Jan 1 and Dec 31 non adjacent, I paired the $r$ ones in the following way:
$$10,10,...,10,1$$
Therefore, there are $365-2r+1$ 0s left. Then the number of ways satisfying the non-adjacency condition is $binom{365-2r+1+r}{r}=binom{365-r+1}{r}$, which agrees with the accepted answer.




To consider the case treating Jan 1 and Dec 31 adjacent. I do the following:




$$10,10,...,10$$
and there are then $365-2r$ 0s left. Similar argument gives the number of ways being $binom{365-r}{r}$.




We can also think about the second case by subtracting the number of arrangements satisfying condition in the first case and Jan 1 and Dec 31 are occupied by 1.
I do the following:




$$1,01,01,...,01,0,1$$
There are $r-2$ 01 pairs and one 0 in the middle, a total of $r-1$ spots. There are still $365-2r+1$ 0s out there. Therefore, the number the ways fixing a one on Jan 1 and another one on Dec 31 while all ones in between satisfying the non-adjacency condition is $binom{365-2r+1+r-1}{r-1}=binom{365-r}{r-1}$. Therefore the final result is $binom{365-r+1}{r}-binom{365-r}{r-1}=binom{365-r}{r}$, which agrees with my previous argument.




Could anyone points out where is wrong?






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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Let's suppose there are $r$ people in the room. We want to find the probability that at least two have the same birthday or adjacent birthdays. It seems easier to find the complementary probability, i.e. the probability that no two have the same or adjacent birthdays.



There are $365^r$ possible arrangements of birthdays for the $r$ people, all of which we assume are equally likely. The number of arrangements in which the birthdays are distinct and non-adjacent can be obtained from a well-known combinatorial formula, as follows. There are $binom{365-r+1}{r}$ ways to pick $r$ integers from 1 to 365 with no two adjacent, where $binom{n}{m}$ is a binomial coefficient: $binom{n}{m} = frac{n!}{m! (n-m)!}$. But we need to multiply by $r!$, because we are considering the people to be distinguishable, so the order of the $r$ integers is significant. Hence the probability that no two people have the same or adjacent birthdays is
$$f(r) = frac{binom{365-r+1}{r} r!}{365^r}$$



The probability that at least two people do have the same or adjacent birthdays is $1-f(r)$. If we do the computations, we find $1-f(13) = 0.4822$ and $1-f(14) = 0.5368$, so the answer to your question is that you need 14 people.



I should add that we're not considering 31 December and 1 January to be adjacent. If we did, it would change the probabilities a little bit, but I don't think it would change the final answer of 14 people.






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Nice answer. In case you're interested, if we count December 31 and January 1 as adjacent, the number of ways to pick $r$ days with no two adjacent is ${365-r+1choose r}-{365-r-1choose r-2}$. I checked, and you are correct that this doesn't change the answer of 14.
    $endgroup$
    – Julian Rosen
    Dec 24 '14 at 17:43
















6












$begingroup$

Let's suppose there are $r$ people in the room. We want to find the probability that at least two have the same birthday or adjacent birthdays. It seems easier to find the complementary probability, i.e. the probability that no two have the same or adjacent birthdays.



There are $365^r$ possible arrangements of birthdays for the $r$ people, all of which we assume are equally likely. The number of arrangements in which the birthdays are distinct and non-adjacent can be obtained from a well-known combinatorial formula, as follows. There are $binom{365-r+1}{r}$ ways to pick $r$ integers from 1 to 365 with no two adjacent, where $binom{n}{m}$ is a binomial coefficient: $binom{n}{m} = frac{n!}{m! (n-m)!}$. But we need to multiply by $r!$, because we are considering the people to be distinguishable, so the order of the $r$ integers is significant. Hence the probability that no two people have the same or adjacent birthdays is
$$f(r) = frac{binom{365-r+1}{r} r!}{365^r}$$



The probability that at least two people do have the same or adjacent birthdays is $1-f(r)$. If we do the computations, we find $1-f(13) = 0.4822$ and $1-f(14) = 0.5368$, so the answer to your question is that you need 14 people.



I should add that we're not considering 31 December and 1 January to be adjacent. If we did, it would change the probabilities a little bit, but I don't think it would change the final answer of 14 people.






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Nice answer. In case you're interested, if we count December 31 and January 1 as adjacent, the number of ways to pick $r$ days with no two adjacent is ${365-r+1choose r}-{365-r-1choose r-2}$. I checked, and you are correct that this doesn't change the answer of 14.
    $endgroup$
    – Julian Rosen
    Dec 24 '14 at 17:43














6












6








6





$begingroup$

Let's suppose there are $r$ people in the room. We want to find the probability that at least two have the same birthday or adjacent birthdays. It seems easier to find the complementary probability, i.e. the probability that no two have the same or adjacent birthdays.



There are $365^r$ possible arrangements of birthdays for the $r$ people, all of which we assume are equally likely. The number of arrangements in which the birthdays are distinct and non-adjacent can be obtained from a well-known combinatorial formula, as follows. There are $binom{365-r+1}{r}$ ways to pick $r$ integers from 1 to 365 with no two adjacent, where $binom{n}{m}$ is a binomial coefficient: $binom{n}{m} = frac{n!}{m! (n-m)!}$. But we need to multiply by $r!$, because we are considering the people to be distinguishable, so the order of the $r$ integers is significant. Hence the probability that no two people have the same or adjacent birthdays is
$$f(r) = frac{binom{365-r+1}{r} r!}{365^r}$$



The probability that at least two people do have the same or adjacent birthdays is $1-f(r)$. If we do the computations, we find $1-f(13) = 0.4822$ and $1-f(14) = 0.5368$, so the answer to your question is that you need 14 people.



I should add that we're not considering 31 December and 1 January to be adjacent. If we did, it would change the probabilities a little bit, but I don't think it would change the final answer of 14 people.






share|improve this answer











$endgroup$



Let's suppose there are $r$ people in the room. We want to find the probability that at least two have the same birthday or adjacent birthdays. It seems easier to find the complementary probability, i.e. the probability that no two have the same or adjacent birthdays.



There are $365^r$ possible arrangements of birthdays for the $r$ people, all of which we assume are equally likely. The number of arrangements in which the birthdays are distinct and non-adjacent can be obtained from a well-known combinatorial formula, as follows. There are $binom{365-r+1}{r}$ ways to pick $r$ integers from 1 to 365 with no two adjacent, where $binom{n}{m}$ is a binomial coefficient: $binom{n}{m} = frac{n!}{m! (n-m)!}$. But we need to multiply by $r!$, because we are considering the people to be distinguishable, so the order of the $r$ integers is significant. Hence the probability that no two people have the same or adjacent birthdays is
$$f(r) = frac{binom{365-r+1}{r} r!}{365^r}$$



The probability that at least two people do have the same or adjacent birthdays is $1-f(r)$. If we do the computations, we find $1-f(13) = 0.4822$ and $1-f(14) = 0.5368$, so the answer to your question is that you need 14 people.



I should add that we're not considering 31 December and 1 January to be adjacent. If we did, it would change the probabilities a little bit, but I don't think it would change the final answer of 14 people.







share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 24 '14 at 21:16

























answered Dec 24 '14 at 17:21









awkwardawkward

762




762








  • 1




    $begingroup$
    Nice answer. In case you're interested, if we count December 31 and January 1 as adjacent, the number of ways to pick $r$ days with no two adjacent is ${365-r+1choose r}-{365-r-1choose r-2}$. I checked, and you are correct that this doesn't change the answer of 14.
    $endgroup$
    – Julian Rosen
    Dec 24 '14 at 17:43














  • 1




    $begingroup$
    Nice answer. In case you're interested, if we count December 31 and January 1 as adjacent, the number of ways to pick $r$ days with no two adjacent is ${365-r+1choose r}-{365-r-1choose r-2}$. I checked, and you are correct that this doesn't change the answer of 14.
    $endgroup$
    – Julian Rosen
    Dec 24 '14 at 17:43








1




1




$begingroup$
Nice answer. In case you're interested, if we count December 31 and January 1 as adjacent, the number of ways to pick $r$ days with no two adjacent is ${365-r+1choose r}-{365-r-1choose r-2}$. I checked, and you are correct that this doesn't change the answer of 14.
$endgroup$
– Julian Rosen
Dec 24 '14 at 17:43




$begingroup$
Nice answer. In case you're interested, if we count December 31 and January 1 as adjacent, the number of ways to pick $r$ days with no two adjacent is ${365-r+1choose r}-{365-r-1choose r-2}$. I checked, and you are correct that this doesn't change the answer of 14.
$endgroup$
– Julian Rosen
Dec 24 '14 at 17:43











4












$begingroup$

This is more complicated than simply the birthday problem times three, because there's the potential for overlap. For instance, if person A is born on Jan 5, and person B is born on Jan 7, then person C only has a 5/365 chance to meet the criteria.



Rather than do the math, I wrote a script to solve it, ignoring leap years (although it turns out that the margin between N=16 and N=17 is large enough that I expect the same answer holds if leap years are accounted for).



With 16 people, you will satisfy the criteria 48.3% of the time, and with 17 people, you satisfy them 52.6% of the time, so the answer is 17. I put the script up at http://pastebin.com/YjNzECrP if anyone would like to check my work.



EDIT: Thanks to Lopsy for pointing out a mistake in the script, fixing that causes it to produce 14 people as the answer.






share|improve this answer











$endgroup$













  • $begingroup$
    thank you for right approach :) (im too lazy to calculate that) It seems OP does not know the answer, so he accepted other one.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:25










  • $begingroup$
    Your script has a typo. b.count([i]) > 1 should be b.count(i) > 1. As a result, it only counts consecutive birthdays, not equal birthdays. The correct answer is 14, as was proven by awkward.
    $endgroup$
    – Lopsy
    Dec 24 '14 at 17:47












  • $begingroup$
    I knew it was 14...
    $endgroup$
    – warspyking
    Dec 25 '14 at 1:40
















4












$begingroup$

This is more complicated than simply the birthday problem times three, because there's the potential for overlap. For instance, if person A is born on Jan 5, and person B is born on Jan 7, then person C only has a 5/365 chance to meet the criteria.



Rather than do the math, I wrote a script to solve it, ignoring leap years (although it turns out that the margin between N=16 and N=17 is large enough that I expect the same answer holds if leap years are accounted for).



With 16 people, you will satisfy the criteria 48.3% of the time, and with 17 people, you satisfy them 52.6% of the time, so the answer is 17. I put the script up at http://pastebin.com/YjNzECrP if anyone would like to check my work.



EDIT: Thanks to Lopsy for pointing out a mistake in the script, fixing that causes it to produce 14 people as the answer.






share|improve this answer











$endgroup$













  • $begingroup$
    thank you for right approach :) (im too lazy to calculate that) It seems OP does not know the answer, so he accepted other one.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:25










  • $begingroup$
    Your script has a typo. b.count([i]) > 1 should be b.count(i) > 1. As a result, it only counts consecutive birthdays, not equal birthdays. The correct answer is 14, as was proven by awkward.
    $endgroup$
    – Lopsy
    Dec 24 '14 at 17:47












  • $begingroup$
    I knew it was 14...
    $endgroup$
    – warspyking
    Dec 25 '14 at 1:40














4












4








4





$begingroup$

This is more complicated than simply the birthday problem times three, because there's the potential for overlap. For instance, if person A is born on Jan 5, and person B is born on Jan 7, then person C only has a 5/365 chance to meet the criteria.



Rather than do the math, I wrote a script to solve it, ignoring leap years (although it turns out that the margin between N=16 and N=17 is large enough that I expect the same answer holds if leap years are accounted for).



With 16 people, you will satisfy the criteria 48.3% of the time, and with 17 people, you satisfy them 52.6% of the time, so the answer is 17. I put the script up at http://pastebin.com/YjNzECrP if anyone would like to check my work.



EDIT: Thanks to Lopsy for pointing out a mistake in the script, fixing that causes it to produce 14 people as the answer.






share|improve this answer











$endgroup$



This is more complicated than simply the birthday problem times three, because there's the potential for overlap. For instance, if person A is born on Jan 5, and person B is born on Jan 7, then person C only has a 5/365 chance to meet the criteria.



Rather than do the math, I wrote a script to solve it, ignoring leap years (although it turns out that the margin between N=16 and N=17 is large enough that I expect the same answer holds if leap years are accounted for).



With 16 people, you will satisfy the criteria 48.3% of the time, and with 17 people, you satisfy them 52.6% of the time, so the answer is 17. I put the script up at http://pastebin.com/YjNzECrP if anyone would like to check my work.



EDIT: Thanks to Lopsy for pointing out a mistake in the script, fixing that causes it to produce 14 people as the answer.







share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 24 '14 at 18:05









Community♦

1




1










answered Dec 24 '14 at 15:22









Ninety-ThreeNinety-Three

1,454712




1,454712












  • $begingroup$
    thank you for right approach :) (im too lazy to calculate that) It seems OP does not know the answer, so he accepted other one.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:25










  • $begingroup$
    Your script has a typo. b.count([i]) > 1 should be b.count(i) > 1. As a result, it only counts consecutive birthdays, not equal birthdays. The correct answer is 14, as was proven by awkward.
    $endgroup$
    – Lopsy
    Dec 24 '14 at 17:47












  • $begingroup$
    I knew it was 14...
    $endgroup$
    – warspyking
    Dec 25 '14 at 1:40


















  • $begingroup$
    thank you for right approach :) (im too lazy to calculate that) It seems OP does not know the answer, so he accepted other one.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:25










  • $begingroup$
    Your script has a typo. b.count([i]) > 1 should be b.count(i) > 1. As a result, it only counts consecutive birthdays, not equal birthdays. The correct answer is 14, as was proven by awkward.
    $endgroup$
    – Lopsy
    Dec 24 '14 at 17:47












  • $begingroup$
    I knew it was 14...
    $endgroup$
    – warspyking
    Dec 25 '14 at 1:40
















$begingroup$
thank you for right approach :) (im too lazy to calculate that) It seems OP does not know the answer, so he accepted other one.
$endgroup$
– shyos
Dec 24 '14 at 15:25




$begingroup$
thank you for right approach :) (im too lazy to calculate that) It seems OP does not know the answer, so he accepted other one.
$endgroup$
– shyos
Dec 24 '14 at 15:25












$begingroup$
Your script has a typo. b.count([i]) > 1 should be b.count(i) > 1. As a result, it only counts consecutive birthdays, not equal birthdays. The correct answer is 14, as was proven by awkward.
$endgroup$
– Lopsy
Dec 24 '14 at 17:47






$begingroup$
Your script has a typo. b.count([i]) > 1 should be b.count(i) > 1. As a result, it only counts consecutive birthdays, not equal birthdays. The correct answer is 14, as was proven by awkward.
$endgroup$
– Lopsy
Dec 24 '14 at 17:47














$begingroup$
I knew it was 14...
$endgroup$
– warspyking
Dec 25 '14 at 1:40




$begingroup$
I knew it was 14...
$endgroup$
– warspyking
Dec 25 '14 at 1:40











2












$begingroup$

Like the birthday problem, but with triple the likelihood, I guess...



So, person 1 would have a birthday, doesn't matter when, won't be before or after anyone else.



Person 2 could have their birthday on any of the 365 days (lets ignore leap years, since the question does), chance of the 2nd person being within a day of the first person is 3/365.



If they don't (362/365) then the next person has a 6/365 chance to be within a day of the first two, and so on.



So we have 1 * 362/365 * 359/365 * ... First time this drops below 50% is 365/365 * 362/365 * 359/365 * 356/365 * 353/365 * 350/365 * 347/365 * 344/365 * 341/365 * 338/365 * 335/365 * 332/365 * 329/365 * 326/365 = 0.4596676, so 14 people.



Edit: Typed too fast and fat fingers took over :)






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    actually with your thinking/calculation it should be 14 people with prob is equal to 0.4596676222355141. You jumped from 365 to 352. However still this isnt the right answer for question.
    $endgroup$
    – shyos
    Dec 24 '14 at 14:48












  • $begingroup$
    D'oh, thanks for spotting that - updated now.
    $endgroup$
    – CaffeinatedDave
    Dec 24 '14 at 14:52










  • $begingroup$
    this is not the right answer. why accepted?
    $endgroup$
    – shyos
    Dec 24 '14 at 15:21












  • $begingroup$
    @warspyking Then your source is wrong, this answer does not include overlapping days.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:28












  • $begingroup$
    @warspyking I can't reach his script from work, but his approach is right, if code is also right then it is the right answer.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:32
















2












$begingroup$

Like the birthday problem, but with triple the likelihood, I guess...



So, person 1 would have a birthday, doesn't matter when, won't be before or after anyone else.



Person 2 could have their birthday on any of the 365 days (lets ignore leap years, since the question does), chance of the 2nd person being within a day of the first person is 3/365.



If they don't (362/365) then the next person has a 6/365 chance to be within a day of the first two, and so on.



So we have 1 * 362/365 * 359/365 * ... First time this drops below 50% is 365/365 * 362/365 * 359/365 * 356/365 * 353/365 * 350/365 * 347/365 * 344/365 * 341/365 * 338/365 * 335/365 * 332/365 * 329/365 * 326/365 = 0.4596676, so 14 people.



Edit: Typed too fast and fat fingers took over :)






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    actually with your thinking/calculation it should be 14 people with prob is equal to 0.4596676222355141. You jumped from 365 to 352. However still this isnt the right answer for question.
    $endgroup$
    – shyos
    Dec 24 '14 at 14:48












  • $begingroup$
    D'oh, thanks for spotting that - updated now.
    $endgroup$
    – CaffeinatedDave
    Dec 24 '14 at 14:52










  • $begingroup$
    this is not the right answer. why accepted?
    $endgroup$
    – shyos
    Dec 24 '14 at 15:21












  • $begingroup$
    @warspyking Then your source is wrong, this answer does not include overlapping days.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:28












  • $begingroup$
    @warspyking I can't reach his script from work, but his approach is right, if code is also right then it is the right answer.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:32














2












2








2





$begingroup$

Like the birthday problem, but with triple the likelihood, I guess...



So, person 1 would have a birthday, doesn't matter when, won't be before or after anyone else.



Person 2 could have their birthday on any of the 365 days (lets ignore leap years, since the question does), chance of the 2nd person being within a day of the first person is 3/365.



If they don't (362/365) then the next person has a 6/365 chance to be within a day of the first two, and so on.



So we have 1 * 362/365 * 359/365 * ... First time this drops below 50% is 365/365 * 362/365 * 359/365 * 356/365 * 353/365 * 350/365 * 347/365 * 344/365 * 341/365 * 338/365 * 335/365 * 332/365 * 329/365 * 326/365 = 0.4596676, so 14 people.



Edit: Typed too fast and fat fingers took over :)






share|improve this answer











$endgroup$



Like the birthday problem, but with triple the likelihood, I guess...



So, person 1 would have a birthday, doesn't matter when, won't be before or after anyone else.



Person 2 could have their birthday on any of the 365 days (lets ignore leap years, since the question does), chance of the 2nd person being within a day of the first person is 3/365.



If they don't (362/365) then the next person has a 6/365 chance to be within a day of the first two, and so on.



So we have 1 * 362/365 * 359/365 * ... First time this drops below 50% is 365/365 * 362/365 * 359/365 * 356/365 * 353/365 * 350/365 * 347/365 * 344/365 * 341/365 * 338/365 * 335/365 * 332/365 * 329/365 * 326/365 = 0.4596676, so 14 people.



Edit: Typed too fast and fat fingers took over :)







share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 24 '14 at 14:54









Victor Stafusa

6,39512249




6,39512249










answered Dec 24 '14 at 14:37









CaffeinatedDaveCaffeinatedDave

692




692








  • 1




    $begingroup$
    actually with your thinking/calculation it should be 14 people with prob is equal to 0.4596676222355141. You jumped from 365 to 352. However still this isnt the right answer for question.
    $endgroup$
    – shyos
    Dec 24 '14 at 14:48












  • $begingroup$
    D'oh, thanks for spotting that - updated now.
    $endgroup$
    – CaffeinatedDave
    Dec 24 '14 at 14:52










  • $begingroup$
    this is not the right answer. why accepted?
    $endgroup$
    – shyos
    Dec 24 '14 at 15:21












  • $begingroup$
    @warspyking Then your source is wrong, this answer does not include overlapping days.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:28












  • $begingroup$
    @warspyking I can't reach his script from work, but his approach is right, if code is also right then it is the right answer.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:32














  • 1




    $begingroup$
    actually with your thinking/calculation it should be 14 people with prob is equal to 0.4596676222355141. You jumped from 365 to 352. However still this isnt the right answer for question.
    $endgroup$
    – shyos
    Dec 24 '14 at 14:48












  • $begingroup$
    D'oh, thanks for spotting that - updated now.
    $endgroup$
    – CaffeinatedDave
    Dec 24 '14 at 14:52










  • $begingroup$
    this is not the right answer. why accepted?
    $endgroup$
    – shyos
    Dec 24 '14 at 15:21












  • $begingroup$
    @warspyking Then your source is wrong, this answer does not include overlapping days.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:28












  • $begingroup$
    @warspyking I can't reach his script from work, but his approach is right, if code is also right then it is the right answer.
    $endgroup$
    – shyos
    Dec 24 '14 at 15:32








1




1




$begingroup$
actually with your thinking/calculation it should be 14 people with prob is equal to 0.4596676222355141. You jumped from 365 to 352. However still this isnt the right answer for question.
$endgroup$
– shyos
Dec 24 '14 at 14:48






$begingroup$
actually with your thinking/calculation it should be 14 people with prob is equal to 0.4596676222355141. You jumped from 365 to 352. However still this isnt the right answer for question.
$endgroup$
– shyos
Dec 24 '14 at 14:48














$begingroup$
D'oh, thanks for spotting that - updated now.
$endgroup$
– CaffeinatedDave
Dec 24 '14 at 14:52




$begingroup$
D'oh, thanks for spotting that - updated now.
$endgroup$
– CaffeinatedDave
Dec 24 '14 at 14:52












$begingroup$
this is not the right answer. why accepted?
$endgroup$
– shyos
Dec 24 '14 at 15:21






$begingroup$
this is not the right answer. why accepted?
$endgroup$
– shyos
Dec 24 '14 at 15:21














$begingroup$
@warspyking Then your source is wrong, this answer does not include overlapping days.
$endgroup$
– shyos
Dec 24 '14 at 15:28






$begingroup$
@warspyking Then your source is wrong, this answer does not include overlapping days.
$endgroup$
– shyos
Dec 24 '14 at 15:28














$begingroup$
@warspyking I can't reach his script from work, but his approach is right, if code is also right then it is the right answer.
$endgroup$
– shyos
Dec 24 '14 at 15:32




$begingroup$
@warspyking I can't reach his script from work, but his approach is right, if code is also right then it is the right answer.
$endgroup$
– shyos
Dec 24 '14 at 15:32











2












$begingroup$

The solution:




If we consider Jan 1 and Dec 31 not adjacent is given above. In the comment of the accepted answer, it is indicated that if we consider Jan 1 and Dec 31 adjacent, the number of ways to arrange the birthdays is $binom{365-r+1}{r}-binom{365-r-1}{r-2}$. I'm having some trouble understanding this.




Could anyone points out where I'm doing wrong?



I think this is equivalent to ask:




How many ways can we arrange 365-r $0$s and r $1$s such that the ones are not adjacent. First, for the case treating Jan 1 and Dec 31 non adjacent, I paired the $r$ ones in the following way:
$$10,10,...,10,1$$
Therefore, there are $365-2r+1$ 0s left. Then the number of ways satisfying the non-adjacency condition is $binom{365-2r+1+r}{r}=binom{365-r+1}{r}$, which agrees with the accepted answer.




To consider the case treating Jan 1 and Dec 31 adjacent. I do the following:




$$10,10,...,10$$
and there are then $365-2r$ 0s left. Similar argument gives the number of ways being $binom{365-r}{r}$.




We can also think about the second case by subtracting the number of arrangements satisfying condition in the first case and Jan 1 and Dec 31 are occupied by 1.
I do the following:




$$1,01,01,...,01,0,1$$
There are $r-2$ 01 pairs and one 0 in the middle, a total of $r-1$ spots. There are still $365-2r+1$ 0s out there. Therefore, the number the ways fixing a one on Jan 1 and another one on Dec 31 while all ones in between satisfying the non-adjacency condition is $binom{365-2r+1+r-1}{r-1}=binom{365-r}{r-1}$. Therefore the final result is $binom{365-r+1}{r}-binom{365-r}{r-1}=binom{365-r}{r}$, which agrees with my previous argument.




Could anyone points out where is wrong?






share|improve this answer










New contributor




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






$endgroup$













  • $begingroup$
    Hi, welcome to puzzling SE! You can add spoiler tags by adding >! in front of your text!
    $endgroup$
    – North
    55 mins ago
















2












$begingroup$

The solution:




If we consider Jan 1 and Dec 31 not adjacent is given above. In the comment of the accepted answer, it is indicated that if we consider Jan 1 and Dec 31 adjacent, the number of ways to arrange the birthdays is $binom{365-r+1}{r}-binom{365-r-1}{r-2}$. I'm having some trouble understanding this.




Could anyone points out where I'm doing wrong?



I think this is equivalent to ask:




How many ways can we arrange 365-r $0$s and r $1$s such that the ones are not adjacent. First, for the case treating Jan 1 and Dec 31 non adjacent, I paired the $r$ ones in the following way:
$$10,10,...,10,1$$
Therefore, there are $365-2r+1$ 0s left. Then the number of ways satisfying the non-adjacency condition is $binom{365-2r+1+r}{r}=binom{365-r+1}{r}$, which agrees with the accepted answer.




To consider the case treating Jan 1 and Dec 31 adjacent. I do the following:




$$10,10,...,10$$
and there are then $365-2r$ 0s left. Similar argument gives the number of ways being $binom{365-r}{r}$.




We can also think about the second case by subtracting the number of arrangements satisfying condition in the first case and Jan 1 and Dec 31 are occupied by 1.
I do the following:




$$1,01,01,...,01,0,1$$
There are $r-2$ 01 pairs and one 0 in the middle, a total of $r-1$ spots. There are still $365-2r+1$ 0s out there. Therefore, the number the ways fixing a one on Jan 1 and another one on Dec 31 while all ones in between satisfying the non-adjacency condition is $binom{365-2r+1+r-1}{r-1}=binom{365-r}{r-1}$. Therefore the final result is $binom{365-r+1}{r}-binom{365-r}{r-1}=binom{365-r}{r}$, which agrees with my previous argument.




Could anyone points out where is wrong?






share|improve this answer










New contributor




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






$endgroup$













  • $begingroup$
    Hi, welcome to puzzling SE! You can add spoiler tags by adding >! in front of your text!
    $endgroup$
    – North
    55 mins ago














2












2








2





$begingroup$

The solution:




If we consider Jan 1 and Dec 31 not adjacent is given above. In the comment of the accepted answer, it is indicated that if we consider Jan 1 and Dec 31 adjacent, the number of ways to arrange the birthdays is $binom{365-r+1}{r}-binom{365-r-1}{r-2}$. I'm having some trouble understanding this.




Could anyone points out where I'm doing wrong?



I think this is equivalent to ask:




How many ways can we arrange 365-r $0$s and r $1$s such that the ones are not adjacent. First, for the case treating Jan 1 and Dec 31 non adjacent, I paired the $r$ ones in the following way:
$$10,10,...,10,1$$
Therefore, there are $365-2r+1$ 0s left. Then the number of ways satisfying the non-adjacency condition is $binom{365-2r+1+r}{r}=binom{365-r+1}{r}$, which agrees with the accepted answer.




To consider the case treating Jan 1 and Dec 31 adjacent. I do the following:




$$10,10,...,10$$
and there are then $365-2r$ 0s left. Similar argument gives the number of ways being $binom{365-r}{r}$.




We can also think about the second case by subtracting the number of arrangements satisfying condition in the first case and Jan 1 and Dec 31 are occupied by 1.
I do the following:




$$1,01,01,...,01,0,1$$
There are $r-2$ 01 pairs and one 0 in the middle, a total of $r-1$ spots. There are still $365-2r+1$ 0s out there. Therefore, the number the ways fixing a one on Jan 1 and another one on Dec 31 while all ones in between satisfying the non-adjacency condition is $binom{365-2r+1+r-1}{r-1}=binom{365-r}{r-1}$. Therefore the final result is $binom{365-r+1}{r}-binom{365-r}{r-1}=binom{365-r}{r}$, which agrees with my previous argument.




Could anyone points out where is wrong?






share|improve this answer










New contributor




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






$endgroup$



The solution:




If we consider Jan 1 and Dec 31 not adjacent is given above. In the comment of the accepted answer, it is indicated that if we consider Jan 1 and Dec 31 adjacent, the number of ways to arrange the birthdays is $binom{365-r+1}{r}-binom{365-r-1}{r-2}$. I'm having some trouble understanding this.




Could anyone points out where I'm doing wrong?



I think this is equivalent to ask:




How many ways can we arrange 365-r $0$s and r $1$s such that the ones are not adjacent. First, for the case treating Jan 1 and Dec 31 non adjacent, I paired the $r$ ones in the following way:
$$10,10,...,10,1$$
Therefore, there are $365-2r+1$ 0s left. Then the number of ways satisfying the non-adjacency condition is $binom{365-2r+1+r}{r}=binom{365-r+1}{r}$, which agrees with the accepted answer.




To consider the case treating Jan 1 and Dec 31 adjacent. I do the following:




$$10,10,...,10$$
and there are then $365-2r$ 0s left. Similar argument gives the number of ways being $binom{365-r}{r}$.




We can also think about the second case by subtracting the number of arrangements satisfying condition in the first case and Jan 1 and Dec 31 are occupied by 1.
I do the following:




$$1,01,01,...,01,0,1$$
There are $r-2$ 01 pairs and one 0 in the middle, a total of $r-1$ spots. There are still $365-2r+1$ 0s out there. Therefore, the number the ways fixing a one on Jan 1 and another one on Dec 31 while all ones in between satisfying the non-adjacency condition is $binom{365-2r+1+r-1}{r-1}=binom{365-r}{r-1}$. Therefore the final result is $binom{365-r+1}{r}-binom{365-r}{r-1}=binom{365-r}{r}$, which agrees with my previous argument.




Could anyone points out where is wrong?







share|improve this answer










New contributor




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









share|improve this answer



share|improve this answer








edited 56 mins ago









North

2,0031734




2,0031734






New contributor




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









answered 1 hour ago









nevereverneverneverevernever

211




211




New contributor




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





New contributor





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






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












  • $begingroup$
    Hi, welcome to puzzling SE! You can add spoiler tags by adding >! in front of your text!
    $endgroup$
    – North
    55 mins ago


















  • $begingroup$
    Hi, welcome to puzzling SE! You can add spoiler tags by adding >! in front of your text!
    $endgroup$
    – North
    55 mins ago
















$begingroup$
Hi, welcome to puzzling SE! You can add spoiler tags by adding >! in front of your text!
$endgroup$
– North
55 mins ago




$begingroup$
Hi, welcome to puzzling SE! You can add spoiler tags by adding >! in front of your text!
$endgroup$
– North
55 mins ago


















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