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Let $X$ be the number of distinct birthdays in a group of $110$ people (i.e., the number of days in a year such that at least one person in the group has that birthday). Under the usual assumptions (no Feb 29, all the other 365 days of the year are equally likely, and the day when one person is born is independent of the days when the other people are born).

Find the mean and variance of $X$.

I know how to get the mean but I'm having problems getting the variance. I totally understand the book solution (see below).

My question: What's wrong with my approach below to get $Var(X)$?

Unlike the book, to find $Var(X)$ I want to use this formula,

$$Var(X) = 2E\left[{X \choose 2} \right] + E[X] - (E[X])^2$$

$$E\left[{X \choose 2} \right] = E \left[ Y_1 + ... + Y_{{365 \choose 2}} \right]$$

where $Y_i = 1$ if the $i$th pair of days both have at least one birthday. So,

$$= {365 \choose 2} E[Y_1] = {365 \choose 2} P(Y_1 = 1) = {365 \choose 2} (1- P(Y_1 = 0)) $$

The $P(Y_1 = 0)$ is the probability that either the first day doesn't have a birthday or the second day doesn't have a birthday, or both days don't have birthdays. Therefore using inclusion exclusion,

$$P(Y_1 = 0) = 2\left( \frac{364}{365} \right)^{110} - \left( \frac{363}{365} \right)^{110} \approx .93259$$

Therefore $P(Y_1 = 1) \approx .067407$ and $E\left[{X \choose 2} \right] = 365(364)(.067407)/2 \approx 4477$

$$Var(X) = 2(4477) + 95 - 95*95 \approx 25 $$

but the correct answer is about $10$.

Again I do understand the book solution, but can't figure out what I'm doing wrong. Thank you for your help and patience.


Book solution

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Rounding at the very end is causing this issue. While $$2\times4477+95-95^2 = 24$$ you would do better with $$2\times 4477.872+95.083-95.083^2\approx 10.05$$ and it is the rounding of $95.083^2$ which is having the substantial effect

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  • $\begingroup$ What the hell... you're right the $.083$ has a very dramatic effect - or dramatic compared to my incorrect expectations. Sheesh... well this is a good mistake for me to learn from. Sorry to put everyone through that. I really thank you for your help Henry I kept thinking over and over about this problem. $\endgroup$ – HJ_beginner Jan 27 '19 at 19:36

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