three birthdays and a numeral

[This article was first published on R – Xi’an’s Og, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

Want to share your content on R-bloggers? click here if you have a blog, or here if you don’t.

The riddle of the week on The Riddler was to find the size n of an audience for at least a 50% chance of observing at least one triplet of people sharing a birthday, as is the case in the present U.S. Senate. The question is much harder to solve than for a pair of people but the formula exists!, as detailed on this blog entry, this X validated entry, or my friend Anirban Das Gupta’s review of birthday problems. If W is the number of triplets among n people,

\mathbb P(W =0) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{365! n!}{i! (n-2i)! (365-n+i)! 2^i 365^n}

\mathbb P(W =0) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{365! n!}{i! (n-2i)! (365-n+i)! 2^i 365^n}

which returns n=88 as the smallest population size for which P(W=0|n=88)=0.4889349, while P(W=0|n=87)=0.5005451. A simulation based on 10⁶ draws confirms this boundary value, P(W=0|n=88)≈0.4890849 and P(W=0|n=87)≈0.5006471.

If you got this far, why not subscribe for updates from the site? Choose your flavor: e-mail, twitter, RSS, or facebook


Leave a Comment