# What's the expected number of days of the year on which at least 2 people in a group of 10 have the same birthday?

This question assumes that a year has 365 days, and each day of the year is equally likely to be a birthday for someone.

At first I thought I simply had to select a day and then 2 individuals ($\frac{\binom{10}{2}}{365}$, but then this can't make sense because I am seeking how many days, not people.

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You have 365 slots and will throw balls randomly and want to know how many slots will be occupied by two balls. – xavierm02 Oct 3 '12 at 21:04
+1. This is a nice take on the old "birthday paradox", as it's sometimes called. – Rick Decker Oct 4 '12 at 0:30

The probability a given day has no birthdays is $\left(1-\frac{1}{365}\right)^{10}$ so the expected number of days with no birthday is $365$ times this, about $355.1224$.
The probability a given day has exactly one birthday is $10 \times \frac{1}{365} \times \left(1-\frac{1}{365}\right)^9$ so the expected number of days with exactly one birthday is $365$ times this, about $9.7561$.
Subtract these two figures from $365$ and you get a result of about $0.1215$ for the expected number of days with two or more birthdays.
Ah, thanks. Each day independently has a $(\frac{364}{365})^{10}$ probability of having no birthdays... and similar for exactly 1 birthday. Thanks! That makes sense. – David Faux Oct 3 '12 at 23:10