the birthday paradox for $p=0.5$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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What is the minimum number of people such that the probability for some of them to have a common birthday is $frac12$?




So looking at the problem as "balls and bins". We have $n = 365$ bins and $m$ balls (people).



There's a probability of $1/n$ for two balls to be on the same bin.



Hence, the expected number of couples to be on the same bin is $fracm(m-1)2n$



How can I leverage this information to conclude that the desired $m$ is $23$?




probability birthday




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  • More than two people can have a birthday on the same day.
    – Avnish Kabaj
    Aug 20 at 8:49






  • 1




    Joriki has given a nice approximation, now you can look for the smallest value of $m$ such that $dfrac365times364timescdots(365-m+1)365^m<dfrac12$. Find why.
    – Jean-Claude Arbaut
    Aug 20 at 9:51











  • @Jean-ClaudeArbaut, thanks!
    – Elimination
    Aug 20 at 10:26














up vote
1
down vote

favorite













What is the minimum number of people such that the probability for some of them to have a common birthday is $frac12$?




So looking at the problem as "balls and bins". We have $n = 365$ bins and $m$ balls (people).



There's a probability of $1/n$ for two balls to be on the same bin.



Hence, the expected number of couples to be on the same bin is $fracm(m-1)2n$



How can I leverage this information to conclude that the desired $m$ is $23$?




probability birthday




share|cite|improve this question




















  • More than two people can have a birthday on the same day.
    – Avnish Kabaj
    Aug 20 at 8:49






  • 1




    Joriki has given a nice approximation, now you can look for the smallest value of $m$ such that $dfrac365times364timescdots(365-m+1)365^m<dfrac12$. Find why.
    – Jean-Claude Arbaut
    Aug 20 at 9:51











  • @Jean-ClaudeArbaut, thanks!
    – Elimination
    Aug 20 at 10:26












up vote
1
down vote

favorite









up vote
1
down vote

favorite












What is the minimum number of people such that the probability for some of them to have a common birthday is $frac12$?




So looking at the problem as "balls and bins". We have $n = 365$ bins and $m$ balls (people).



There's a probability of $1/n$ for two balls to be on the same bin.



Hence, the expected number of couples to be on the same bin is $fracm(m-1)2n$



How can I leverage this information to conclude that the desired $m$ is $23$?




probability birthday




share|cite|improve this question













What is the minimum number of people such that the probability for some of them to have a common birthday is $frac12$?




So looking at the problem as "balls and bins". We have $n = 365$ bins and $m$ balls (people).



There's a probability of $1/n$ for two balls to be on the same bin.



Hence, the expected number of couples to be on the same bin is $fracm(m-1)2n$



How can I leverage this information to conclude that the desired $m$ is $23$?





probability birthday





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 20 at 8:34









Elimination

1,389822




1,389822











  • More than two people can have a birthday on the same day.
    – Avnish Kabaj
    Aug 20 at 8:49






  • 1




    Joriki has given a nice approximation, now you can look for the smallest value of $m$ such that $dfrac365times364timescdots(365-m+1)365^m<dfrac12$. Find why.
    – Jean-Claude Arbaut
    Aug 20 at 9:51











  • @Jean-ClaudeArbaut, thanks!
    – Elimination
    Aug 20 at 10:26
















  • More than two people can have a birthday on the same day.
    – Avnish Kabaj
    Aug 20 at 8:49






  • 1




    Joriki has given a nice approximation, now you can look for the smallest value of $m$ such that $dfrac365times364timescdots(365-m+1)365^m<dfrac12$. Find why.
    – Jean-Claude Arbaut
    Aug 20 at 9:51











  • @Jean-ClaudeArbaut, thanks!
    – Elimination
    Aug 20 at 10:26















More than two people can have a birthday on the same day.
– Avnish Kabaj
Aug 20 at 8:49




More than two people can have a birthday on the same day.
– Avnish Kabaj
Aug 20 at 8:49




1




1




Joriki has given a nice approximation, now you can look for the smallest value of $m$ such that $dfrac365times364timescdots(365-m+1)365^m<dfrac12$. Find why.
– Jean-Claude Arbaut
Aug 20 at 9:51





Joriki has given a nice approximation, now you can look for the smallest value of $m$ such that $dfrac365times364timescdots(365-m+1)365^m<dfrac12$. Find why.
– Jean-Claude Arbaut
Aug 20 at 9:51













@Jean-ClaudeArbaut, thanks!
– Elimination
Aug 20 at 10:26




@Jean-ClaudeArbaut, thanks!
– Elimination
Aug 20 at 10:26










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










You can't conclude any exact result from this, since the expected number includes contributions from cases with more than one collision. However, when the probability of at least one collision is $frac12$, the probability of more than one collision is still rather low, so you can get an estimate by taking the expected number as an approximation of the probability of at least one collision. Then



$$
fracm(m-1)2ngefrac12
$$



yields



$$
m^2-m-nge0
$$



and thus



$$
mgefrac12+sqrtn+frac14;,
$$



which for $n=365$ yields $mge20$, not a bad estimate.






share|cite|improve this answer




















  • Thank you joriki!
    – Elimination
    Aug 20 at 10:27










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










You can't conclude any exact result from this, since the expected number includes contributions from cases with more than one collision. However, when the probability of at least one collision is $frac12$, the probability of more than one collision is still rather low, so you can get an estimate by taking the expected number as an approximation of the probability of at least one collision. Then



$$
fracm(m-1)2ngefrac12
$$



yields



$$
m^2-m-nge0
$$



and thus



$$
mgefrac12+sqrtn+frac14;,
$$



which for $n=365$ yields $mge20$, not a bad estimate.






share|cite|improve this answer




















  • Thank you joriki!
    – Elimination
    Aug 20 at 10:27














up vote
2
down vote



accepted










You can't conclude any exact result from this, since the expected number includes contributions from cases with more than one collision. However, when the probability of at least one collision is $frac12$, the probability of more than one collision is still rather low, so you can get an estimate by taking the expected number as an approximation of the probability of at least one collision. Then



$$
fracm(m-1)2ngefrac12
$$



yields



$$
m^2-m-nge0
$$



and thus



$$
mgefrac12+sqrtn+frac14;,
$$



which for $n=365$ yields $mge20$, not a bad estimate.






share|cite|improve this answer




















  • Thank you joriki!
    – Elimination
    Aug 20 at 10:27












up vote
2
down vote



accepted







up vote
2
down vote



accepted






You can't conclude any exact result from this, since the expected number includes contributions from cases with more than one collision. However, when the probability of at least one collision is $frac12$, the probability of more than one collision is still rather low, so you can get an estimate by taking the expected number as an approximation of the probability of at least one collision. Then



$$
fracm(m-1)2ngefrac12
$$



yields



$$
m^2-m-nge0
$$



and thus



$$
mgefrac12+sqrtn+frac14;,
$$



which for $n=365$ yields $mge20$, not a bad estimate.






share|cite|improve this answer












You can't conclude any exact result from this, since the expected number includes contributions from cases with more than one collision. However, when the probability of at least one collision is $frac12$, the probability of more than one collision is still rather low, so you can get an estimate by taking the expected number as an approximation of the probability of at least one collision. Then



$$
fracm(m-1)2ngefrac12
$$



yields



$$
m^2-m-nge0
$$



and thus



$$
mgefrac12+sqrtn+frac14;,
$$



which for $n=365$ yields $mge20$, not a bad estimate.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 20 at 9:43









joriki

166k10180331




166k10180331











  • Thank you joriki!
    – Elimination
    Aug 20 at 10:27
















  • Thank you joriki!
    – Elimination
    Aug 20 at 10:27















Thank you joriki!
– Elimination
Aug 20 at 10:27




Thank you joriki!
– Elimination
Aug 20 at 10:27












 

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