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A dealer is using a standard deck of 52 cards. One extra ace of spades is put into the deck. So now he got 53 cards in the deck with two ace of spades in total.

The dealer deals 4 hands with 5 cards each. What is the probabilty that one 5-card hand contains the two ace of spades?

I got some work done on this already. The probabilty of two ace of spades in the first hand must be C(2,2)*C(52,3)/C(53,5). C(n,k) is binomial numbers.

How do i expand this to four hands?

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There are $\binom{51}{18}$ sets of $20$ cards that include both of the aces of spades, so the probability that both aces of spades are in play is $$\frac{\binom{51}{18}}{\binom{53}{20}}=\left(\frac{51!}{18!33!}\right)\left(\frac{20!33!}{53!}\right)=\frac{20\cdot19}{53\cdot52}=\frac{95}{689}\;.$$

Pretend that one of the aces is marked. The marked ace can be in any of the four hands. The other ace is equally likely to be any of the other $19$ cards that are in play, so the probability that the other ace is in the same hand is $\frac4{19}$. Thus, the probability that both aces are in play and in the same hand is $$\frac{95}{689}\cdot\frac4{19}=\frac{20}{689}\approx0.02903\;.$$

Added: You can also get the $\frac4{19}$ by counting the deals that put both aces of spades into the same hand, but I don’t recommend doing it that way. If the two aces are in the first hand, there are $\binom{18}3$ ways to pick the rest of that hand and $\binom{15}5\binom{10}5\binom55$ ways to deal out the other three hands, so for a given set of $20$ cards that includes both aces of spades there are $\binom{18}3\binom{15}5\binom{10}5\binom55$ ways to deal the cards so that both aces are in the first hand. There are just as many ways with the aces in the second hand, the third, and the fourth, so the total number of deals of those $20$ cards that give both aces to the same person is $4\binom{18}3\binom{15}5\binom{10}5\binom55$. There are altogether $\binom{20}5\binom{15}5\binom{10}5\binom55$ ways to deal the $20$ cards into four hands of $5$ cards each, so the probability that both aces of spades end up in the same hand is

$$\left(\frac{4\cdot18\cdot17\cdot16\cdot15!}{3!\,5!\,5!\,5!}\right)\left(\frac{5!\,5!\,5!\,5!}{20!}\right)=\frac{4\cdot5!}{3!\cdot20\cdot19}=\frac4{19}\;.$$

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Thanks Brian! I was trying something along the things you showed in the added note. But got lost somewhere on the way. Thanks again. –  Roger T Jun 11 '12 at 12:22

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