Question I play poker regularly. In Hold'em there is what is called the 'flop', three community cards which come out together after the initial betting round. I like side bets, so my question is, what is the probability that any of three cards will come out on this three card flop. For example, what are the chances that a King, a Ten, or a Four will come out on the next flop. Assumption, 52 card deck, four of every denomination, Ace through King, four suits. Initial thought by most would be 12 out of 52, which is wrong, second thought is 12/52 + 12/51 + 12/50, but I am not certain that is correct either. Please advise.
Answer rick -
i understand your question to mean: what is the chance that at least one of the three
cards {K,10,4} will appear in the flop.
if the flop were dealt first, before any other cards, the answer would be
1 - (52-12)/52 x (51-12)/51 x (50-12)/50 = 1 - 40/52 x 39/51 x 38/50.
here, the product being subtracted from 1 is the chance of dealing 3 consecutive flop cards, none of which are {K,10,4}. for the first flop card, there are 52 cards in the deck, of which 52-12 = 40 cards are not {K,10,4}. so the chance of not getting any of {K,10,4} for the first flop card is 40/52.
for for the second flop card, 51 cards remain in the deck, of which 51-12=39 are not {K,10,4}, so the chance that the second flop card will also not be any of {K,10,4} is 39/51.
then the chance that neither of the first two flop cards will be {K,10,4} is the product
40/52 x 39/51.
you continue the product for as many terms as there are flop cards - in this case, to the
third term as above.
as the product is the chance of not getting any of {K,10,4} in the flop, subtracting from 1 gives the chance that at least one of {K,10,4} will be in the flop.
you can refine this answer a bit:
suppose the flop is dealt after each player receives 5 cards, say, and you have one each of {K,10,4} in your hand and two other cards that are not {K,10,4} - (perhaps a pair of sixes). then there are 52 - 5 = 47 cards unaccounted for (by you), of which 12 - 3 = 9 are {K,10,4} and 47 - 9 (or 40 - 2) = 38 are other than {K,10,4}. so the 3-term product above is replaced by 38/47 x 37/46 x 36/45 and then subtracted from 1.
you can see that the actual answer depends on how many cards you can see before the flop is dealt.
