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I know this has been posted several times and I have gone through most of the relevant posts. Here is one which I am having a difficult time to solve:

There are 450 people in a room; (1) how many of them are expected to have the same birthday with some other person in the room, (2) with at least 2 other people in the room and (3) with at least 3.

(1) is easy - by the pigeonhole principle, 450-365 (or 366) = 85 people are expected to have the same birthday.

How do we do (2) and (3)?

I am thinking that in 85 people we have $\frac {85*84} {2} = 3570$ possible pairs so the probability for a 3rd person to share one of their birthdays is $1-\frac {364}{365}^{85}$. And then how do we find the expected number of people for each case?

Any help is greatly appreciated! Thank you!

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  • $\begingroup$ I don't understand your use of the pigeonhole principle. Use Linearity of expectation instead, with an indicator variable for each person in the room. $\endgroup$
    – lulu
    Commented Nov 3, 2020 at 9:20
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    $\begingroup$ From the pigeonhole principle you get that at-least $86$ people share their birthdays (not necessarily on the same day). You could also have $100$ people sharing a birthday with some probability. You need to find the expected number of people who share a birthday, not the least. $\endgroup$
    – Shubham Johri
    Commented Nov 3, 2020 at 9:26

1 Answer 1

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Hints:

  • Picking a particular individual, what is the probability that person shares their birthday:

    • with no other people
    • with exactly one person
    • with exactly two people?

  • So for that individual, what is the probability that person shares their birthday:

    • with at least one person
    • with at least two people
    • with at least three people?

  • And using the linearity of expectation, what is the expected number of people that share their birthday

    • with at least one person (much more than or $85$ or $86$)
    • with at least two people
    • with at least three people?

Following the hints:

  • Picking a particular individual, the probability that person shares their birthday:

    • with no other people is $\frac{364^{449}}{365^{449}}$
    • with exactly one person is ${449 \choose 1}\frac{364^{448}}{365^{449}}$
    • with exactly two people is ${449 \choose 2}\frac{364^{447}}{365^{449}}$

  • So for that individual, the probability that person shares their birthday:

    • with at least one person is $1-\frac{364^{449}}{365^{449}}$
    • with at least two people is $1-\frac{364^{449}}{365^{449}}-{449 \choose 1}\frac{364^{448}}{365^{449}}$
    • with at least three people is $1-\frac{364^{449}}{365^{449}}-{449 \choose 1}\frac{364^{448}}{365^{449}} - {449 \choose 2}\frac{364^{447}}{365^{449}}$

  • And using the linearity of expectation, the expected number of people that share their birthday

    • with at least one person is $450\left(1-\frac{364^{449}}{365^{449}} \right)$
    • with at least two people is $450\left(1-\frac{364^{449}}{365^{449}}-{449 \choose 1}\frac{364^{448}}{365^{449}} \right)$
    • with at least three people is $450\left(1-\frac{364^{449}}{365^{449}}-{449 \choose 1}\frac{364^{448}}{365^{449}} - {449 \choose 2}\frac{364^{447}}{365^{449}} \right)$

and these values are about $318.7$ (much more than or $85$ or $86$) and $156.8$ and $57.1$

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  • $\begingroup$ Henry: For the 1st case (probability someone shares his birthday with no other people): Isn't it zero? $\endgroup$
    – Sal.Cognato
    Commented Nov 3, 2020 at 10:15
  • $\begingroup$ @Sal.Cognato It is much more than zero: the first three sub-bullets come from a binomial distribution with parameters $450-1=449$ and $\frac{1}{365}$, so the probability a particular person shares their birthday with no other people is $\left(\frac{364}{365}\right)^{449} \approx 0.29176$ $\endgroup$
    – Henry
    Commented Nov 3, 2020 at 10:25
  • $\begingroup$ Thank you! So the expected number of people who share their birthday with 1 more, is $\frac {1}{1-0.29176}$ = 1.41? $\endgroup$
    – Sal.Cognato
    Commented Nov 3, 2020 at 11:02
  • $\begingroup$ Strange! I would also expect that in any group of people >365, the probability is 100% but I think the tricky point is that this must apply for any person, right? $\endgroup$
    – Tom Galle
    Commented Nov 3, 2020 at 11:17
  • $\begingroup$ @TomGalle No - the probability a individual shares their birthday with at least one other person is $1-0.29176$ and the expected number of people who share their with at least one other person is $450$ times this $\endgroup$
    – Henry
    Commented Nov 3, 2020 at 11:23

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