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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!

Henry
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Sal.Cognato
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  • 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. – lulu Nov 03 '20 at 09:20
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    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. – Shubham Johri Nov 03 '20 at 09:26

1 Answers1

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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 $57.1$

Henry
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  • Henry: For the 1st case (probability someone shares his birthday with no other people): Isn't it zero? – Sal.Cognato Nov 03 '20 at 10:15
  • @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 $\left(\frac{364}{365}\right)^{449} \approx 0.29176$ – Henry Nov 03 '20 at 10:25
  • Thank you! So the expected number of people who share their birthday with 1 more, is $\frac {1}{1-0.29176}$ = 1.41? – Sal.Cognato Nov 03 '20 at 11:02
  • 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? – Tom Galle Nov 03 '20 at 11:17
  • @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 – Henry Nov 03 '20 at 11:23
  • How about the second case "at least two people"? Is it $\left(\frac{363}{365}\right)^{448}$ ? – Sal.Cognato Nov 03 '20 at 12:28
  • @TomGalle: No. Just guessing a pattern will not work. Following the hints in order would make this easier – Henry Nov 03 '20 at 12:35
  • I don't understand why you people are so sparing in giving a complete solution to a problem that is not abc! If someone (like me!) is interested in reading the solution, he has to go through a hundred of comments!! – Juan Manuel Prada Nov 03 '20 at 13:06
  • @JuanManuelPrada Because sometimes we would prefer people to learn rather than doing their homework for them. This kind of question has been asked before, for example at https://math.stackexchange.com/questions/35791/birthday-problem-expected-number-of-collisions – Henry Nov 03 '20 at 13:10
  • @Henry, agree but this doesn't seem to be homework (and I've seen this many other times). If you read the OP profile, it says "Energy commodities trader. Like to (try to) solve math problems, puzzles & riddles in my free time" :)) – Juan Manuel Prada Nov 03 '20 at 13:16