Variation of Birthday problem - Group of n people

Question

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!

Answer 1

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?

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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 $57.1$