# Probability at least one person receives exactly two aces out of five cards

Three players are each dealt, in a random manner, five cards from a deck containing 52 cards. Four of the 52 cards are aces. What is the probability that at least one person receives exactly two aces in their five cards?

Let $A_i$ represent the player $i$ with two aces where $i = 1,2,3$. The probability a player receives two aces is the following. $$P(A_i) = \frac{{4 \choose 2}{48 \choose 3}}{{52 \choose 5}} \approx .0399$$

Then the probability at least one person receives exactly two aces is the following. $$3 \cdot .0399 - 3 \cdot .0399^2 \approx .1149$$

Is this correct?

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What is your reasoning? And what about the probability that three people receive two aces? Is that $0.0399^3$? –  Hurkyl Dec 24 '12 at 8:21

You have used Inclusion/Exclusion. That is a good idea, but the exclusion part needs to be recomputed. Let us, for example, calculate the probability that A and B each get exactly $2$ Aces. This is the probability A gets $2$ Aces, times the probability B gets $2$ Aces given that A got $2$. So B is getting her $5$ cards from a deck that has $2$ Aces and $45$ non-Aces. The probability she gets $2$ Aces given that A got $2$ is $\binom{45}{3}/\binom{47}{5}$. Quite small.
The order of dealing is irrelevant. I calculated $\Pr(A\cap B)$ by conditioning on A getting $2$ Aces. There would be other ways of calculating, like finding the probability that between them they have $4$ Aces, and then given that, the probability these Aces are split $2$-$2$. The point is that when you squared, you were implicitly assuming that the events A has $2$ Aces and B has $2$ Aces are independent. That would be true only if they were getting their cards from different decks. –  André Nicolas Dec 24 '12 at 8:42