Infamous Birthday probability

[Jabba]

Indeed, roughly 27% of classes of 31 people will not contain two people sharing a birthday.

[Triskadecamus]

I think the chances are smaller than that.

Tris

[Cabbage]

No, Jabba is correct.

As I mentioned above, to find the probability that no two people share a birthday among n people:

365!

(365-n)! * 365^n

For n=31, this is indeed about 27% (26.95…)

To find the probability that at least two share a birthday, subtract this from one.

[GSV_Consolation_of_Dreams]

This thread in general and Cabbage’sposts in particular have reminded me just how little I know about probability. :o

It’s been a few years since my Chemistry degree but still, I really need to brush up. Anyone in a similar situation might consider taking a gander at this Basic Concepts page.It’s designed for Liberal Arts students so it should be pretty gentle.

[GSV_Consolation_of_Dreams]

Dammit, 10 minutes and I think I may have picked the wrong paper.

Hence, if you are betting $1 that four heads will appear in a four-toss sequence of coins and if your opponent is unwilling to put up $15, you should not undertake the bet (the probabilities are against you). If she is willing to put up more than $15, then you should take the bet (the probabilities are in your favour).

But maybe I’m just nit-picking. The probabilities are still against you, but if you keep it up, your opponents betting should result in you being “ahead” at some point in the future.

/hijack.