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birthday paradox, one by one

An easy riddle, riding the birthday paradox. Namely how many people on average have to sequentially enter a room for a double birthday to occur?

The answer is straightforward,

\sum_{n=2}^{366}n\prod_{m=1}^{n-2}\left(1-\frac{m}{365}\right)\frac{n-1}{365}

as only the last person in can share a birthday with those already in. This sum equals 24.62. Since the “magic number” is 23, this may sound wrong but the median attached to the above distribution is truly 23. My R code is

q=rep(1,a<-366)/365
for(n in 3:a)q[n]=q[n-1]*(1+1/(n-2)-(n-1)/365)
sum(q[-1]*(2:a))

This entry was posted on November 1, 2022 at 12:22 am and is filed under Books, Kids, pictures, Statistics with tags , , , , , . You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “birthday paradox, one by one”

  1. Robin Ryder Says:
    November 1, 2022 at 11:52 am

    R has the neat `pbirthday` function, which does exactly what you think it does. Unfortunately it isn’t vectorized. You can get the same answer with
    n = 1:365
    sum(n * (1-sapply(n-1, pbirthday)) * (n-1)/365)

    Reply

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