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#1
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[ QUOTE ]
This should be easy. First, how many flops contain quads? Well, two of the cards are already determined. That only leaves the final card, which can be any of the remaining 48 cards in the deck. So 48/(50 choose 3) is the chance of quads on the flop. On the turn we can make our quads 48*47 possible ways, since we have 2 free cards which may vary. So 48*47/(50 choose 4) is the chance of quads by the turn. River is similar. gm [/ QUOTE ] i think it should be (48 choose 2)/(50 choose 4) for the turn, which is half of your number. |
#2
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[ QUOTE ]
i think it should be (48 choose 2)/(50 choose 4) for the turn, which is half of your number. [/ QUOTE ] Yep. You're right. gm |
#3
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Hmmm...I'm always trying to make these more difficult than they are, I guess. The flop was easy. And the turn turned out to be OK too.
I can see now, based on your explanation, that the river is just C(48,3)/C(50,5)...CORRECT? But when I was thinking about it (quads by the river) prior to seeing your explanation, I was saying to myself: I could get one of my paired cards on the flop, miss on the turn, and get it on the river. Or I could get none of my paired cards on the flop, but get one on the turn and the river, etc. Then I started the "Are they mutually exclusive" stuff, and then I got twisted up in my shorts...This probably just comes with practice, but if you have any "mental process" that you can come up with, it would be appreciated. I realize that's a pretty nebulous request, but I guess it can't hurt to ask. I suppose that, after seeing your formulas, what I should have said to myself (about the river) was: a. How many 5 card combinations are there in the unseen 50 cards? b. Two of the 5 cards must be my pocket denomination. So that leaves 48 cards to be chosen 3 ways. |
#4
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Dave,
Yes, your river answer is correct. As far as mental process, I think you're getting it. There's no magic formula except for seeing lots of problems solved -- after that, you'll start to see recurring patterns. With poker calculations, I would say there are easily less than 10 "types" of problems, all requiring only basic combinatorics. Just keep messing around with problems, and they'll get easier and easier. I can already tell you know alot more than you did after your first post. gm |
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