Posted Fri May 26, 2006 5:12 pm GMT by Cricket_Fire
Was reading a gambling site, and came across 1000000 questions like this.
| Quote: | | First of all, if there's a better site on gambling on the web, I sure haven't seen it! It was also cool to put a name to a face when watching the Travel Channel. We always bring this question up at my monthly game, and decided it was time for an answer. In a 5-card draw game of "trips to win", where you must have 3 of a kind or better to win the pot, if I get dealt 2 pair, is it better to keep only one of the pair, and get 3 new cards to try and match the first pair, or should I keep the 2 pair and get one card to try to match either pair? Assume 6 players at the table, no wild cards, that players can draw 3 cards, four with an Ace, and experience shows that any 3 of a kind will probably win the hand, making pulling the full house not that much more advantageous than just the 3 of a kind. Thanks! - Dave A., Cincinnati, Ohio |
The response was:
| Quote: | | Thanks for the kind words. I'm familiar with this game. Let's assume your intial hand was JJQQK and you keep the two jacks. The number of ways to get one jack and two other cards on the draw is 2*combin(45,2) = 1,980. The number of ways to get two jacks on the draw is 45. The number of ways to get a three of a kind on the draw is 10*4+1 = 41. So, the number of ways to improve the hand to a three of a kind or better is 1,980+45+41 = 2,066. The total number of ways to choose 3 card out of the 47 left is combin(47,3) = 16,215. So, the probability of improving the hand to three of a kind or better is 2,066/16,215 = 12.74%. If you kept the two pair, the probability of improving to a full house is 4/47 = 8.51%. So, assuming a three of a kind would probably win, I agree that keeping just one pair (the higher one) is the better play |
Now, in the equation in bold, what does combin mean?
Hope this was clear,
Thank you!
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Posted Fri May 26, 2006 6:00 pm GMT by jbark
that is math shorthand for combinations, in particular the number of two card combinations that could be dealt from a deck of 45 cards. this is also written as 45C2, which is read 45 choose 2.
an example: suppose you want to know the odds of flopping two aces to give you quads (since you got AA in your pocket). the odds of flopping two aces would be:
number of ways to draw two aces is 2C2*48 or the number of ways to choose the two aces from the two aces remaining in the deck times the number of non-ace cards that could be the third card int he flop. so 48 ways to catch a flop with AA
number of flops that DO NOT include an ace is 48C3 which is the number of ways to choose three cards from the remaining 48 (non-ace) cards in the unseen cards. that number is 17,296
so your odds would be 48 to 17,296 or about 1 in 360 with slight rounding
similar is the idea of permutations: a permutation is a combination of cards selected in order. in a combination the order does not matter. in poker i do not believe the order of the cards being dealt matters so using combinations makes sense.
hope that clears it up....
cheers,
jerry
BTW: the above odds could be computed without using combinations, i did it this way to create an interesting, relevant example.
Posted Fri May 26, 2006 6:25 pm GMT by Cricket_Fire
Thanks for the long write up!
I think I've got ya.
| Quote: | | number of ways to draw two aces is 2C2*48 or the number of ways to choose the two aces from the two aces remaining in the deck times the number of non-ace cards that could be the third card int he flop. so 48 ways to catch a flop with AA |
How did you get 48? 2C2*48. How do you figure that out to be 48? Why is there a *2 in this equation, but not the other?
| Quote: | | number of flops that DO NOT include an ace is 48C3 which is the number of ways to choose three cards from the remaining 48 (non-ace) cards in the unseen cards. that number is 17,296 |
So, where did 17 296 come from?
Thank you!
Posted Fri May 26, 2006 7:27 pm GMT by Adam Marshall
A comb(x,y) is x! / (x-y)! y!
So if wanted to figure out the possible number of two card combinations in a 52 card deck (so As - Kd is the same as Kd - As), it'd be...
52! / (52-2)! 2!
The trick is that you end up with a ton of cancelling out. 52! over 50! ends up being just 52x51.
So anyway,(52x51)/2 = 1326
The reason that the response is 2(combin 45,2) is because they're trying to figure out how many three card combinations exist from the remaining cards that would include only one jack. Subtracting the discarded cards and all four jacks, we get the 45. There's two jacks left in the deck, so you multiply by two at the end.
By the way, jbark, I don't think you want the "number of flops that DO NOT include an ace" in your example. You just want the total number of possible flops and divide into that, yes? That'd put the result at around 1 in 408 or so. I think the problem is that you are comparing quads vs no improvement at all and leaving out all the one ace flop situations.
Posted Fri May 26, 2006 7:36 pm GMT by jbark
odds in favor are (ways to succeed)/(ways to not succeed) so flops with two aces and flops without would give the odds of quads.
(flops with two aces)/(all possible flops) would be the probability of flopping quads.
i think its correct.
cheers
jerry
Posted Fri May 26, 2006 7:43 pm GMT by jbark
Cricket,
glad to write this out,
to answer:
How did you get 48? 2C2*48. How do you figure that out to be 48? Why is there a *2 in this equation, but not the other?
i probably should have written that this way: (2C2)*48, 2C2 is the number of ways to choose two aces from two remaining aces in the unseen cards(2C2 = 1 BTW). 48 is the number of unseen cards left to select the third card of the flop from (52 minus two in your pocket minus the two aces)
So, where did 17 296 come from?
48C3 = 17,296 i used an online probability calculator to get the value, since my statistical calculator is at work. i believe Excel has that function built in as well.
cheers
jerry
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