Doubts about the Birthday Problem

Question

I was thinking about the birthday problem, in a slight different version. In the birthday problem, no date is chosen in advance. The variant I'm thinking of is then "same birthday as you", as in the wikipedia page

https://en.wikipedia.org/wiki/Birthday_problem#Same_birthday_as_you

I still have to fully understand the general problem, but let's focus on this variant: can you pleas explain my (naive) reasoning is wrong?

This: if I am in a room and another person comes in, then that person will have a probabilty of $1/365$ (assuming non leap year) of being born my same day. If $23$ people come in, then the possibilities will be $23/365$, right?

In the general problem, the person $P$ born in the day $D$ will always have the probabilities I stated above, I mean: the person will always be born in a day which will have $D/365$ possibilities of finding another person with the same date...

Thank you and sorry! I just want to get this.

Answer 1

Yes this is a different problem. Even then, the probability you surmise is only an upper bound and wrong as soon as $D$ is large enough.

If $D$ people arrive, and $D$ is small, the probability of one of them having the same birthday as you is not exactly $D/365$ since their birthdays may collide even though this is relatively unlikely if $D$ is much smaller than $23.$

Recognising this may help you understand the general principle. The idea is that given $D$ people, there are $\binom{D}{2}=D(D-1)/2$ pairs of people thus $D(D-1)/2$ possibilities of a collision while sampling a set of size $365.$ It is the fact that $D(D-1)/2\approx D^2/2$ is roughly the same as $365$ which makes the probability of collision be significant, around $1/2.$

Answer 2

Your issue is you're assuming all $23$ people have different birthdays. We cannot assume this. If we do, then after $365$ people would would have probability $1$, but that isn't true.

Instead, we do what wikipedia says, and find that for $n$ people the probability is: $$1 - \left( \frac{364}{365} \right)^n$$ Since each person has a probability of $\frac{364}{365}$ of not having your birthday, and we repeat that event $n$ times.

Answer 3

Alternative Explanation:

This: if I am in a room and another person comes in, then that person will have a probabilty of 1/365 (assuming non leap year) of being born my same day. If 23 people come in, then the possibilities will be 23/365, right?

Your question is actually focusing on the distinction between probability and expected value. It is somewhat subtle in the specific scenario of your posted question, because it would be unusual for more than one person (out of 23 people) to have the same birthday as you.

Perhaps an easier way to illustrate the distinction between probability and expected value is as follows:

You are being offered an unusual chance to try to win money, playing a game. A bin has 10 pieces of paper in it, and each piece of paper has a distinct number written on it, from the set $~\{1,2,3,\cdots,10\}.~$

Rules:

• If you draw the number $~1,~$ you win $~5\$$.
• If you draw the number $~2,~$ you win $~3\$$.
• If you draw the number $~3,~$ you win $~1\$$.
• If you draw any other number, you win nothing.

The probability of your winning a prize is $~\dfrac{3}{10}.$

The expected value of what you will win, in dollars, is computed as

$$\left[ ~\frac{1}{10} ~\times 5 ~\right] + \left[ ~\frac{1}{10} ~\times 3 ~\right] + \left[ ~\frac{1}{10} ~\times 1 ~\right] = \frac{9}{10}.$$

So, here the expected value of what you will win, $~\dfrac{9}{10},~$ is greater than the probability of your winning a prize, $~\dfrac{3}{10},~$ specifically because some of the prizes are larger than $~1\$$.

In the birthday scenario of $~23~$ people entering the room, you have a somewhat similar analysis. If 1 person comes in to the room, the expected value of this person having the same birthday as you is 1/365. If instead, 23 people come in to the room, the expected value of the number of people that will share a birthday with you is 23/365, just as your intuition suggested.

As other responses have indicated, the probability that none of the people share a birthday with you is $~\displaystyle ~\left( ~\frac{364}{365} ~\right)^{23}.~$ Then, the probability that at least one person shares a birthday with you is $~\displaystyle 1 - \left[ ~\left( ~\frac{364}{365} ~\right)^{23} ~\right].$

Similar to the game described at the start of this answer, the probability of a successful event (i.e. at least one person matching your birthday) is lower than the expected value of the number of matching birthdays.

Just as with the game, the reason is that sometimes, the successful event will result in more than one person matching your birthday.