suppose you and I each have a box of 600 marbles. In my box, 4 of the marbles are black, while 3 of your marbles are black. we each draw 300 marbles with replacement from our own boxes.Approximately, what's the chance that you and I draw the same number of black marbles?
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A Poisson approximation will need this: http://wolframalpha.com/input/?i=sum+x%5Ek%2F%28k%21%5E2%29+from+0+to+infinity – J.G. Oct 04 '18 at 22:41
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You have two binomial distributions $B(n, p_1)$ and $B(n, p_2)$. Approximate them by Poisson distributions with $\lambda_i = n p_i$. Then $$\operatorname P(\tilde X_1 = \tilde X_2) = \sum_{k = 0}^\infty \operatorname P(\tilde X_1 = \tilde X_2 = k) = e^{-\lambda_1 - \lambda_2} \sum_{k = 0}^\infty \frac {(\lambda_1 \lambda_2)^k} {(k!)^2}.$$ Take the first few terms to get a good approximation for $\operatorname P(X_1 = X_2)$.
Maxim
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