Poker Question - What are the odds?
- Thread starter hapaboy
- Start date
i wouldn't worry about the occasional improbable hand, who cares? it's the likely others that will grind you up.
Same goes trading
Same goes trading
Quote from hapaboy:
So last night I'm playing Texas No Limit Hold 'Em (cash game, not tournament).
I have pocket Kings. Flop comes K - 8 - 4.
There are three of us in the hand.
Well, guess what, one of the other guys had pocket 8's, and the other guy had pocket 4's.
In other words, all three of us flopped three of a kind!!
What are the odds of this happening? First time I've ever seen it happen.
And yes, I won a monster pot.
Sorry about the week long lapse here. A friend died last week and now its back to the grinding stone. I looked at what each of us has done and see where we were using different logic. You were pretty much using the logic of one of x qty possible remaining cards out of y qty cards remaining in the deck. Whereas I was using x qty cards to hit the previously dealt cards out of y qty remaining cards. I like your logic better. After all, if there are 2 two's left out of 50 cards, it's 1:25 that you will get a 2 on the next deal; I don't think anyone would argue that. To that end, here's what I get:
52 52 1
48 51 0.941176471
44 50 0.88
3 49 0.06122449
3 48 0.0625
3 47 0.063829787
2 46 0.043478261
2 45 0.044444444
2 44 0.045454545
0.000000017768439
56,279,562.50
whereas the probability is 1:0.000000017768439 and the odds are 1:56,279,562.50
Agree?
Quote from IV_Trader:
The probability of 3 players to get a pair is 1.8%
A. There is a [ 52*51*50*49*48*47] / [ 1*2*3*4*5*6] = 20,358,520 SIX cards combos .
B. Pairs combos : (6*13) * (6*12) * (6*11) = 370,656
B/A = 0.018
Calling for all quants to complite the second part
I don't think this is correct either for calculating the odds of 3 pocket pairs, but your idea should work when done correctly. Your approach I think was to divide the number of possible permutations of three consecutive pairs (Part B) into the number of possible pairs (Part A), but I think both your numbers are wrong.
Part B should be (6*2*13)(6*2*12)(6*2*11). You need to use permutations (the entire set), not combinations, to divide the number of possible ways three consecutive pairs can be dealt into the number of possible ways any three hands of two cards can be dealt. The possible number of ways three hands of two cards can be dealt is simply 52*51*50*49*48*47. Again, permutations should be used here so your are dividing into the total set of possible three, two card hands. If you divide these, you'll get the same answer I did, or 1/4943.3, even though I used iterations. Of course they should be the same and are.
I don't agree with the last three
It needs to be 6/46 for the first number because it doesn't matter which player's pair the first card of the flop matches. There are 6 cards that could be drawn to makes someone's three of a kind. After someone makes three of a kind, there are only 4 cards left that will match one of the last two players' pairs. After that, the last player must see one of the two possible cards to make three of a kind.
So by changing 2 2 2 to 6 4 2, we've multiplied by 6, because there are 6 ways in which the three cards of the flop can come down. It doesn't matter in which order they are dealt, e.g. K84, K48, 84K, 8K4, 4K8, 48K, all the same.