how many flop combinations are there?

[blaze666]

i thought it was 19600, how is that worked out? preflop, you already know 2 cards, so i thought it would have been 50*49*48. why am i wrong?

[Jordan Olsommer]

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i thought it was 19600, how is that worked out? preflop, you already know 2 cards, so i thought it would have been 50*49*48. why am i wrong?

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Because your calculation takes into account the order in which the cards come - for example, in your calculation, a flop of A/images/graemlins/club.gifK/images/graemlins/club.gifQ/images/graemlins/club.gif is counted as being different than a flop of Q/images/graemlins/club.gifK/images/graemlins/club.gifA/images/graemlins/club.gif. In other words, 50*49*48 gives you the number of possible flop permutations, not combinations, which is what you want.

So the correct answer (assuming you know 2 cards) is 50 choose 3 or 19,600.,

[BruceZ]

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i thought it was 19600, how is that worked out? preflop, you already know 2 cards, so i thought it would have been 50*49*48. why am i wrong?

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50*49*48/6 = 19600. You have to divide by 6 or else you will count each combination 6 times for 6 different ways each combination can be ordered.

[blaze666]

so, am i right in thinking that there are 663 possible starting hands, because 52*51/4=663?

and while on the subject, am i right in thinking that if i hold 2 random cards, the chances of pairing one of them on the flop, and nothing else, is 2(3/50+3/49+3/48)= 0.36744, or 36.744%?

[Jordan Olsommer]

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so, am i right in thinking that there are 663 possible starting hands, because 52*51/4=663?

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Right thinking, wrong denominator - there are only two combinations per holdem hand - you get card A first, then card B, or you get card B first, then card A. so 52*51/2 = 1326, which is equal to 52 choose 2.

[BruceZ]

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so, am i right in thinking that there are 663 possible starting hands, because 52*51/4=663?

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52*51/2 = 1326.


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and while on the subject, am i right in thinking that if i hold 2 random cards, the chances of pairing one of them on the flop, and nothing else, is 2(3/50+3/49+3/48)= 0.36744, or 36.744%?

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No. You can't add these 3 probabilities because this over counts the times you flop trips or quads, and you can't multiply by 2 since this over counts the times you pair both cards. If you hold a non-pair, the number of flops that pair exactly 1 card, with no trips or quads, is 6*44*43/2, so the probability is 6*44*43/2/19600 = 29.0%. This includes times that the board pairs. If you want to exclude the board pairing, that would be 6*44*40/2/19600 = 26.9%. The probability that you pair at least one card, including possible 2-pair, trips, or quads, is 1 minus the probability of no pairs, or 1 - 44*43*42/6/19600 or 1 - 44*43*42/(50*49*48) = 32.4%.

[AaronBrown]

BruceZ and Jason Olsommer gave you excellent answers. I would add that this stuff can seem very confusing. There are a lot of ways of calculating the same thing, and it sometimes seems that whatever you do is wrong. In fact, the things you did were almost right, and because you're thinking about them, you're way ahead of people who have memorized one right way to do each calculation.

When enumerating flops, you could use the total number of permutations, 50*49*48. Dividing by 6 is convenient because the order doesn't matter, but as long as you're consistent you can do it either way. When you want to figure the probability that a suited hand will produce a flush on the flop, you figure there are 11 unknown cards that match the suit so it's 11*10*9. That makes the answer 11*10*9/(50*49*48). People usually divide the numerator and denominator by 6, but you get the same answer either way.

For pairing, 2*(3/50 + 3/50 + 3/50) = 18/50 is the expected number of cards in the flop that pair with something in your hand. You could get zero (no pair), one (one pair), two (two pair or three of a kind) or three (full house or four of a kind). Expected values add, probabilities don't.

The denominator stays at 50. When you deal the second card, it's true that the first card has already been dealt, but we're not using that information, so it's just the same as a card that has been dealt to an opponent. It's been dealt, but since we don't know what it is, it doesn't matter. For the second card you could instead do:

(47/50)*(3/49) + (3/50)*(2/49)

That is, 47 times out of 50 there are still three pairing cards available out of 49, and 3 times out of 50 there are 2 available out of 49. In this case we do use the information about the first card, so the denominator goes down to 49. But this is equal to 3/50.

The 18/50 (36%) breaks down into 5,676/19,600 (28.96%) of getting one match, 660/19,600 (3.37%) of getting two matches and 20/19,600 (0.10%) of getting three. 5,676 + 2*660 + 3*20 = 7,056. 7,056/19,600 = 18/50. But the chance of getting exactly one pair is 5,676/19,600 and the chance of getting at least one of your two cards paired is 5,676 + 660 + 20 = 6,356/19,600 = 32.43%.