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Kevin
The following table shows the probability of 0 to 4 aces in a totally random hand.
Ace Probabilities - Random Hand
Aces Formula Combinations Probability
0 Combin(48,5) 1712304 0.658842
1 combin(4,1)*combin(48,4) 778320 0.299474
2 combin(4,2)*combin(48,3) 103776 0.03993
3 combin(4,3)*combin(48,2) 4512 0.001736
4 combin(4,4)*combin(48,1) 48 0.000018
Total 2598960 1
Take the sum for 1 to 4 aces we see the probability of at least one ace is 0.341158. The probability of two or more aces is 0.041684.
The probability of there being at least one more ace given there is at least one can be restated per Bayes' theorem as probability(two more aces given at least one ace) = probability (two or more aces)/probability(at least one ace) = 0.041684/ 0.341158 = 0.122185.
For those rusty on Bayes' Theorem is states the probability of A given B equals the probability of A and B divided by the probability of B, or P(A given B) = P(A and B)/P(B).
The next table shows the combinations and probability for each number of other aces given that the ace of spades was removed from the deck.
Ace Probabilities - Ace Removed Hand
Aces Formula Combinations Probability
0 combin(3,0)*combin(48,4) 194580 0.778631
1 combin(3,1)*combin(48,3) 51888 0.207635
2 combin(3,2)*combin(48,2) 3384 0.013541
3 combin(3,3)*combin(48,1) 48 0.000192
Total 249900 1
This shows the probability of at least one more ace is 0.221369.
However suppose you want to solve it the same way as hand A, using Bayes' Theorem. The probability of at least one additional ace, given the hand contains the ace of spades, could be rewritten as probability(at least two aces given ace of spades is in hand). According to Bayes' theorem this equals P(hand contains ace of spades and at least one more ace) / P(hand contains ace of spades). We can break up the numerator as P(2 aces including ace of spades) + P(3 aces including ace of spades) + P(4 aces). Using the first table this equals 0.039930*(2/4) + 0.001736*(3/4) + 0.000018 = 0.021285. The probability of the ace of spades is 5/52 = 0.096154. So the probability of at least two aces given the aces of spades is 0.021285/0.096154 = 0.221369.
So the probability of two or more aces given at least one ace is 12.22% and given the ace of spades is 22.14%.
Let's look at another simpler situation. Suppose woman A says "I have two kids at least one is a boy." Woman B says "I have two kids and one is named John. We can assume nobody named John is a girl and no woman gives the same name to more than one kid. Using conditional probability the probability of both kids being a boy of woman A is P(both boys)/P(at least one boy) = P(both boys)/(1-P(both girls)) = (1/4)/(1-(1/4)) = (1/4)/(3/4) = 1/3. However the probability woman B's other kid is a boy, or both kids are boys, is ½, because saying one kid is named John tells us nothing about the other kid. Likewise a test for any ace will probably turn up the only ace, while a test for the ace of spades does not check for the other three suits, leaving them more likely to be aces. |
