Can You Plan The Party?

Riddler Classic

Today I happen to be celebrating the birthday of a family member, which got me wondering about how likely it is for two people in a room to have the same birthday.

Suppose people walk into a room, one at a time. Their birthdays happen to be randomly distributed throughout the 365 days of the year (and no one was born on a leap day). The moment two people in the room have the same birthday, no more people enter the room and everyone inside celebrates by eating cake, regardless of whether that common birthday happens to be today.

On average, what is the expected number of people in the room when they eat cake?

Extra credit: Suppose everyone eats cake the moment three people in the room have the same birthday. On average, what is this expected number of people?

Solution

Assume the next person that comes in shares a birthday with someone in the room.

• Let $s$ be the number of of people in the room. $1 \le s \le 365$.
• There are ${365 \choose s}$ ways to choose $s$ days out of the year.
• There are $s!$ orders for the $s$ people to arrive in the room.
• There are $365^{s}$ total ordered birthdays among $s$ people.
• The probability that the next person arriving forms a pair is $\dfrac{s}{365}$. The number of people eating cake is $s+1$.

This can be represented as:
$$EV = \sum \limits_{s = 1}^{365} \dfrac{{365 \choose s} \cdot s!}{365^{s}} \cdot \dfrac{s}{365} \cdot (s+1)$$

$$= \sum \limits_{s = 1}^{365} \dfrac{365!\cdot s \cdot(s+1)}{365^{s+1}\cdot(365-s)!}$$

Answer

I ran a loop and arrived at:

$$EV \approx 24.617$$

Extra Credit

Assume the next person that comes in shares a birthday with two others in the room.

• Let $p$ be the number of pairs of people in the room. $1 \le p \le 365$.
• Let $s$ be the number of of people in the room not sharing a birthday. $0 \le s \le 365-p$.
• There are ${365 \choose p+s}$ ways to choose $p+s$ days out of the year.
• There are ${p+s \choose s}$ ways to set $p$ pairs and $s$ singles among the $p+s$ days.
• There are $\dfrac{(2p+s)!}{2!^p \cdot 1!^s}$ orders for the $2p+s$ people to arrive in the room.
• There are $365^{2p+s}$ total ordered birthdays among $2p+s$ people.
• The probability that the next person arriving forms a triple is $\dfrac{p}{365}$. The number of people eating cake is $2p+s+1$.

This can be represented as:
$$EV = \sum \limits_{p = 1}^{365} \sum \limits_{s = 0}^{365-p} \dfrac{{365 \choose p+s} \cdot {p+s \choose p} \cdot \dfrac{(2p+s)!}{2!^p \cdot 1!^s}}{365^{2p+s}} \cdot \dfrac{p}{365} \cdot (2p+s+1)$$

$$= \sum \limits_{p = 1}^{365} \sum \limits_{s = 0}^{365-p} \dfrac{\dfrac{365!}{(p+s)!\cdot(365-p-s)!} \cdot \dfrac{(p+s)!}{p!\cdot s!} \cdot \dfrac{(2p+s)!}{2^p} \cdot p \cdot (2p+s+1)}{365^{2p+s}\cdot 365}$$

$$= \sum \limits_{p = 1}^{365} \sum \limits_{s = 0}^{365-p} \dfrac{365!\cdot(2p+s)!\cdot(2p+s+1)}{365^{2p+s+1}\cdot(365-p-s)!\cdot(p-1)!\cdot s!\cdot 2^p }$$
I ran a double loop and arrived at:

$$EV \approx 88.738$$

Rohan Lewis

2022.10.16

Code can be found here.