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Suppose, I am playing a poker match. I have a pair of 5 in hand and see that the there is no 5 in the first 3 drawn cards. What is the probability that there will be at least one 5 in the next two cards.

My solution: {P(5 in the fourth card and a different card in the fifth) = 2/47*45/46}+ {P(different card in fourth and 5 in the fifth)=45/47*2/46}+ {P(5 in both cards)=2/47*1/46}= 0.084

Is it correct? Does it depend on the number of opponents playing?

enamul17
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  • Ask it here https://math.stackexchange.com/ – lurning too koad Feb 16 '20 at 05:40
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    but meanwhile, probably easier is 1-(45/47)(44/46). Take the chances of not getting a 5 either time. – Jeremy Kahan Feb 16 '20 at 05:42
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    Probability is a measure of *information*. It depends only on the number of cards you have seen vs. those you have not. So, assuming you've seen exactly 5 cards, there are 47 unseen. The probability that the next two you see will NOT be 5s is (45/47)(44/46), or about 0.91581864... The probability that at least one of them will be a 5 is therefore 1.0 - that number, or 0.08418..., aka about 8 1/2%, or about 11 to 1. – Lee Daniel Crocker Feb 18 '20 at 17:10

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