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Problem: A deck of 52 cards is dealt to 4 players, each receiving 13 cards. What is the probability each player gets an ace.

I understand the combinatoric approach, but am unsure why the following approach using the inclusion-exclusion principle gets 0.09375 instead of 0.10550.

Let $1, 2, 3, 4$ be random variables representing the event that player $i$ gets an ace. Thus we are finding $P(1 \cap 2 \cap 3 \cap 4) = 1 - P(1^c \cup 2^c \cup 3^c \cup 4^c)$. We can solve for this by using inclusion exclusion principle.

$1 - P(1^c \cup 2^c \cup 3^c \cup 4^c) = 1 - (\binom{4}{1} (\frac{3}{4})^4 - \binom{4}{2} (\frac{2}{4})^4 + \binom{4}{3} (\frac{1}{4})^4) = 0.09375$

Each of the $\cap$ probabilities in the inclusion-exclusion expansion is from the fact that all $4$ aces must go to the other players. For example, $P(2^c \cap 4^c) = \frac{2}{4}^4$ since each ace must go to players $1$ and $2$, which has $\frac{2}{4}$ probability.

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    $\begingroup$ I haven't checked all the math but one mistake I see is that with inclusion-exclusion the sign of the terms alternate between + and - but you have consecutive terms that are negative. $\endgroup$
    – Michael R. Chernick
    Commented Aug 28, 2019 at 16:32
  • $\begingroup$ Good observation @Michael. However, the result of 0.09375 is obtained only by using the signs correctly, so we must presume there are just some typos in the post. Rob: it's unclear what your random variables really are. Please note that the "combinatorial approach" yields $4! 13^4 / (52\times 51\times 50\times 49)$ as an answer, yet there's no way prime factors like $7$ or $17$ can appear in the denominator of your PIE calculation. This shows that you aren't accounting correctly for the actual counts in the problems. $\endgroup$
    – whuber ♦
    Commented Aug 28, 2019 at 20:11

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To clear up your notation, define $A_k$ for $k=1,2,3,4$ to be the event that player $k$ receives at least one ace. (Let's not use integers to represent events; and events are not random variables.) Then the desired probability is $$P(A_1\cap A_2\cap A_3\cap A_4) = 1 - P(A_1^c\cup A_2^c \cup A_3 ^c\cup A_4 ^c).\tag1 $$ You are correct in applying inclusion-exclusion and symmetry to simplify (1) to: $$ 1 - \left [{4\choose 1}P(A_1 ^c) -{4\choose 2}P(A_1^c A_2^c) + {4\choose 3}P(A_1 ^c A_2 ^c A_3^c)\right];\tag2 $$ notice that we've omitted the final probability $P(A_1 ^cA_2 ^cA_3 ^c A_4^c)$, since the event that nobody gets any aces has zero probability.

So far so good. Your error occurs in your calculation of the probabilities in (2). To calculate the probability that player 1 gets none of the aces, imagine dealing a pile of 13 cards to player 1. There are $52 \choose 13$ ways to do this. If we require all 13 cards to be non-aces, there are $48\choose 13$ ways to do this, so $P(A_1^c)$ is ${48\choose 13}/{52 \choose 13}$. Similarly if we require all aces to not go to players 1 and 2, then they are getting all their 26 cards from a subset of 48; this has probability ${48\choose 26}/{52\choose 26}$. In the same manner $P(A_1 ^cA_2 ^c A_3^c)={48\choose 39}/{52\choose 39}$. Plugging all this into (2) gives the answer $0.1054982$.

Your approach gives the wrong answer because it's assuming that aces are independently assigned; i.e., that if player 1 already has an ace, then the probability is still $1/4$ that she receives another ace. This is incorrect because the cards are not being dealt with replacement. (Imagine repeating the experiment with a deck of 8 cards, four of which are aces. Your approach wouldn't account for the fact that no player can receive more than two aces.)

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