If two cards are dealt to each of the 6 players from a 52 deck of cards what is the probability that there is at least one ace dealt?
I've tried to solve this using the following reasoning:
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In total there are $ {52 \choose 12 } \cdot {12 \choose 6}$ ways to deal 12 cards to the 6 players.
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Count the number of ways to deal with at least one ace.
To count 2 notice that there is in general ${4 \choose N} \cdot 48 \cdot 47 \cdots (48 – (12 – N)) $ ways to choose $N$ aces in the $12$ dealt cards. For each of these combinations there are also ${12 \choose 6}$ ways to distribute two cards to each player. Thus in general the probability would be, following the above reasoning:
$$ P(\text{at least one A}) = \frac{{4 \choose N} \cdot 48 \cdot 47 \cdots (48 – (12 – N))}{ {52 \choose 12 } \cdot {12 \choose 6}^2 }$$
However, this does not give a sensible answer.
Can anyone see where the flaw in the above reasoning is?
Edit:
Thanks to those who have answered below – I can see that taking the inverse of the question by asking how to deal no aces is much easier. However, can anyone show me how to do it by count the positive cases as I have tried above?
Best Answer
The probability of at least one ace is 1 minus the probability of no ace: $$1-\frac{\binom{4}{0}\binom{48}{12}}{\binom{52}{12}}=\frac{2759}{4165},$$ so it is about twice as likely to have at least one ace than no ace.
If you prefer to count positive cases: $$\sum_{k=1}^4 \frac{\binom{4}{k}\binom{48}{12-k}}{\binom{52}{12}}=\frac{2759}{4165}$$