Out of 10 people in a room, probability that exactly 2 of them share the same birthday

Question

I am considering the following problem: Out of 10 people in a room, find the probability that exactly 2 of them share the same birthday. I fully understand the simpler problem, that is, out of 10 people, probability that at least 2 share the same birthday. An easier way to solve this problem than solving for the probability that any 2 or more of the 10 people share a birthday is to solve for the probability that all of the people have unique birthdays (non-shared) and subtracting that from 100%.

The first person has 365 possible days that could be their birthday. Then, the second person has 364 days that could be their birthday without sharing with someone else, because person 1’s birthday is one of those days. This process goes on for all of the 10 people, until the 10th person has 356 possible days that could be their birthday. The easiest way to express the total possible combinations of days in which none of the 10 people share a birthday is $365 \times 364 \times 363 \times … 356$, that is, $\frac{365!}{355!}$.

To solve for the total probability that, out of the original $365^{10}$ days, there are $\frac{365!}{355!}$ where no one shares a birthday, we have:

$\frac{(\frac{365!}{355!})}{365^{10}}$

Then we subtract this from $100%$ to get the complementary probability. But this will give us the probability that at least 2 people share the same birthday. To find the same but for exactly 2 people, how do we approach it? My approach is as follows: First, let's calculate the total number of possible outcomes, which is the total number of ways to assign birthdays to 10 people without restrictions. Since there are 365 days in a non-leap year, each person can have a birthday on any of these days. Therefore, the total number of possible outcomes is $365^{10}$. Then we calculate the number of favorable outcomes, which is the number of ways to choose 2 people out of 10 to share the same birthday, multiplied by the number of ways to choose a specific birthday for them: The number of ways to choose 2 people out of 10 is: ${10 \choose 2}=\frac{10!}{(10−2)!2!}$. For each pair of people, there are 365 possible birthdays they could share.

So, the number of favorable outcomes is $45 \times 365$. Then, we calculate the probability: $\frac{(45\times 365)}{365^{10}}$. Is it correct?

Answer 1

Note that your answer is negligibly small. While “in reality” this should happen quite often.

The number of combinations where exactly two people share a birthday is $$\binom{10}2\cdot 365 \cdot 364 \cdot 363 \cdot 362 \cdot 361 \cdot 360 \cdot 359 \cdot 358 \cdot 357.$$

Indeed, we choose two people who will share their birthday in $\binom{10}2$ ways and then distribute birthdays between $9$ distinguishable entities.

The probability is then $$\frac{\binom{10}2\cdot 365 \cdot 364 \cdot 363 \cdot 362 \cdot 361 \cdot 360 \cdot 359 \cdot 358 \cdot 357}{365^{10}}.$$

This is approximately $0.11$, or $11\%$.