Flop odds

[IIAce]

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No, if you want 3 different suits on the flop it would have to be 3*2*11/19600 = 0.34%.

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Ok I see why you have 2 instead of 3 but I don't get the 11 part. Let's say you have 67 of clubs, the flop could contain clubs, diamonds, or hearts. 3 ways to pick the first card (assume it happens to be a diamond), 2 ways to pick the second card (assume it happens to be a heart), 22 ways to pick the last card (all other cards that are not 6, 7, diamonds, or hearts). Right?

[BruceZ]

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No, if you want 3 different suits on the flop it would have to be 3*2*11/19600 = 0.34%.

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Ok I see why you have 2 instead of 3 but I don't get the 11 part. Let's say you have 67 of clubs, the flop could contain clubs, diamonds, or hearts. 3 ways to pick the first card (assume it happens to be a diamond), 2 ways to pick the second card (assume it happens to be a heart), 22 ways to pick the last card (all other cards that are not 6, 7, diamonds, or hearts). Right?

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I thought you wanted 3 different suits on the flop that were different from the suit in your hand. Then if you have 67c in your hand, the flop would be spades, hearts, and diamonds. 3 choices for the six (say spades), 2 choices for the 7 (say hearts), and 11 choices for the last suit (diamonds) since it can't be 6 or 7.

If you just want a rainbow flop which can match your hole cards, then the last card could be 11 diamonds or 11 clubs, and the probability would be 3*2*22/19600 = 0.67%.

[MarkD]

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Note that here we need the extra term. We cannot just add 1 to the 44 in the first term to get 2*C(3,2)*45 as this would have the effect of counting the quads 3 times. This problem comes about when the cards we are adding are the same rank as some cards already on the board. If you can just understand that, you'll be light years ahead of most people.


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You lost me here. Sorry, but I'm not sure what you are talking about.

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Right, and we could just as well have added 3 to the first term to get 47 instead of 44. There are many cases where you would need the separate term to avoid over counting, so I think it is best not to get in the habit of combining terms like this until you are sure you understand when it is appropriate.


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Ok, I think I might understand a bit now. I really do prefer to break out the individual terms. I like doing that in algebra as well - I always solved problems in a lot of steps (but very fast).

[BruceZ]

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Note that here we need the extra term. We cannot just add 1 to the 44 in the first term to get 2*C(3,2)*45 as this would have the effect of counting the quads 3 times. This problem comes about when the cards we are adding are the same rank as some cards already on the board. If you can just understand that, you'll be light years ahead of most people.


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You lost me here. Sorry, but I'm not sure what you are talking about.

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I'm just saying that when we counted the number of flops that made trips, we got 2*C(3,2)*44. Now suppose we want to add the flops that make quads. Many would be tempted to simply add the 1 extra card that makes quads on to the 44, to get 2*C(3,2)*45. That's a big no-no. Look how many flops we added, 2*C(3,2)*1 = 6. How many should we have added? Only 2. We are actually counting the quad flops 3 times! This will happen whenever we have a situation like this where the board is paired, and instead of handling this as a separate case, we try to handle the paired rank like it is any other rank. In this example it is ridiculously easy to see what the problem is, but even advanced people make exactly this kind of error all the time in cases where it is not so obvious, and sometimes they can't see it even when it is pointed out to them. A popular book on hold'em odds by Mike Petriv if full of these types of errors.