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Consider a group of k people. Assume that each person's birthday is drawn uniformly at random from the 365 possibilities. (And ignore leap years.) What is the smallest value of ksuch that the expected number of pairs of distinct people with the same birthday is at least one?

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Total number of people  = k

Total pairs that can be formed: $\frac{k(k-1)}{2}$

Probability that a given pair of people have same birthday = $\frac{1}{365}$

Expected number of pairs having same birthday = $\frac{1}{365}*\frac{k(k-1)}{2}$

According to the question

$\frac{1}{365}*\frac{k(k-1)}{2} \ge 1$

after solving k = 27, 28 (after some approximations).  

Putting k = 27 in  $\frac{1}{365}*\frac{k(k-1)}{2}$ gives a values less than 1 (may be because of approximations or I made some mistake), hence 28 is the answer.

I am sure someone can explain this better than me. Any corrections or modifications are very much welcome.
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