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?
Best Answer
Your approach and result are correct, and the answer indeed doesn’t depend on the number of opponents playing. It would have been considerably more efficient, though, to calculate the probability of the complementary event that there is no $5$ in the next two cards to obtain
$$ 1-\frac{45\cdot44}{47\cdot46}=\frac{91}{1081}\approx0.084\;. $$