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Server Time: 12/2/2008 6:48:18 PM PACIFIC |
 Odds;from two pairs to full house in hold'em, Amix1, 2. May 2003 12:51 | ||
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| I am a little confused about the odds. If one has made two-pair on the flop, there are four cards out that will make a full house possible. Deducting the 4 cards from the 47 unseen ones leaves 43 dividing it by 4 gives me 10.75 to 1 odds. Yet the probability tables for making it to a full house indicates 4.97 to 1. Where is my miscalculation? I would appreciate all comments setting me straight! Even the nasty ones, LOL! | ||
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| Re: Odds;from two pairs to full house in hold'em, 4 POKER, 2. May 2003 13:00 | ||
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| It's real simple, The odds are 50/50 in making the full house... You either will make it or you won't !... 50/50 (I'm sorry but all those numbers make me crazy, I'm sure a really good mathematician will help clarify things better for you.) 4 POKER | ||
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| Odd Odds, flintsword, 2. May 2003 13:06 | ||
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| The mathematician in me laughs. You are, ... of course, ... right. Sort of like the term "Pot Odds" meaning how a person acts odd when smoking too much pot ... | ||
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| Re: Odds;from two pairs to full house in hold'em, flintsword, 2. May 2003 13:14 | ||
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| The 4.97 to one refers to the odds of making a full house OR BETTER ... so think about it a bit. Figuring it out yourself will definately help you in the future. Unless someone beats me to the punch, I'll post the math later (when I get home from work ... lol) flintsword | ||
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| Re: Odds;from two pairs to full house in hold'em, Amix1, 2. May 2003 14:47 | ||
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| Thanks for your response, I am looking forward to your further explanation. Regards, Amix1 | ||
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| Re: Odds;from two pairs to full house in hold'em, 4 POKER, 3. May 2003 01:03 | ||
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| on 2. May 2003 13:14 flintsword wrote: > The 4.97 to one refers to the odds of making a full house OR BETTER ... so think > about it a bit. Figuring it out yourself will definately help you in the future. > Unless someone beats me to the punch, I'll post the math later (when I get home from > work ... lol) Looks like they beat you to it ! oh well. 4 POKER > > flintsword | ||
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| Re: Odds;from two pairs to full house in hold'em, Jav, 2. May 2003 13:15 | ||
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| You have seen 5 cards, and you have two pair. But you get two draws to make your full house, not just one. If you had two pair at the turn, with only one more card to draw, then your odds are 10.5:1. The odds for getting a full house when you have two pair on the flop is a combination of the possibilities of getting the full house on the turn and on the river. | ||
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| Re: Odds;from two pairs to full house in hold'em, Amix1, 3. May 2003 07:06 | ||
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| Thanks, I had suspected that possibility. However taking it from the responses, there are a variety of considerations in the equation arriving to the confusing 4.97 to 1printed in Mike Caro's list of probabilities. Again, thanks for your input. | ||
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| Re: Odds;from two pairs to full house in hold'em, shorn, 2. May 2003 13:24 | ||
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| You are 11 to 1 on the NEXT card to make the FH. you need to calculate it with two shots at it if you plan to go to the river and then add those two chances together to determine the actual 4.97 to 1 so: 4/43= 9.3% + 4/42= 9.52% or 18.82% chance or a little less than 5 to 1 or something close to that at least. | ||
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| Re: Odds;from two pairs to full house in hold'em, Amix1, 3. May 2003 07:14 | ||
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| Thanks, your explanation makes a lot of sense. I take it , that all odds referring to conditions on the flop, are based on two more cards to come. Is that correct? | ||
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| Re: Odds;from two pairs to full house in hold'em, stdioh, 2. May 2003 13:59 | ||
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| This plays exactly like a gutshot - 4 outs. The nice thing is that full houses beat straights, so they have better hold-up potential, and you can usually know by action if somebody is likely to have a better full than you. The odds of hitting your tight on the turn are 1 in 11.75 and the odds of not hitting your tight on the turn, but hitting it on the river are 1 in 12.57. Add these fractions together and you get roughly 1 in 6... and a one in six chance is odds of 5:1 against. ... or 4.97 if you care to round things off just so. Likewise, if you have a set on the flop you have roughly a 1/3 chance of eventually getting there to a tight (but a much higher chance of winning if you don't hit, obviously), but your chance of getting there on the turn is only 1 in 6.71 ... you gain outs on the turn because the river can pair any of the flop cards, but can also pair the flop if it did not pair the turn and that adds three outs. Your chance of hitting the river is 1 in 4.60 and thus your chance of not hitting the turn and then hitting the river is 1 in 5.41 which works out very conveniently to 1 in 2.99446 or 1 in 3 for all those you are not obsessive compulsive. That has 2:1 odds against. Of course the real issue with having 2 pair and drawing to a tight is that anybody who has you beat with a set already will also tighten up, so then it becomes a question of having top 2. And you were wondering why top two was so much better than top and bottom :) | ||
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| Re: Odds;from two pairs to full house in hold'em, Amix1, 3. May 2003 07:17 | ||
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| Thanks, your explanation makes a lot of sense. I take it , that all odds referring to conditions on the flop, are based on two more cards to come. Is that correct? | ||
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| Re: Odds;from two pairs to full house in hold'em, stdioh, 5. May 2003 09:05 | ||
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| Yes. When you have 2 pair on the flop you look at odds of getting a card on the turn and add that to the odds of not getting a card on the turn times getting a card on the river. That dictates your chance of getting there eventually...which is the basis for why you would build a pot for yourself. | ||
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| Re: Odds;from two pairs to full house in hold'em, flintsword, 3. May 2003 20:38 | ||
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| Ok Amix1, I promised you the math on Odds for getting a Full House (boat) from two pairs. I see a herd has beaten me to it, but I want to show you a slightly more accurate way of looking at it, since everyone has dealt with "making a full house" whereas I think we should be thinking in terms of "at least a full house", which more accurately includes quads (rare as they are). Example: Ah Qd Flop is: As Qh 5c To make your 'boat' on the turn, you have 4 'outs' = 4/47=0.085 To make your 'boat' on the river, you have again 4 'outs'=4/46=0.087 So far, so good. To calculate total chances of getting a boat on the turn, or the river, to include quads, let us look at the chances of NOT GETTING AN "A" or a "Q" on the turn OR the river. This is (43/47)x(42/46)=(0.915)x(0.913)=0.835 So you have an 83.5% chance of NOT getting an Ace or Queen, which means that your chances of getting AT LEAST A FULL HOUSE is (1-0.835)=.165=16.5% This method builds in the odd chance that you will get a turn and river of AA or QQ, ... Don't we wish! I hope this helps. I am sure I read this type of example in a number of books, and at the time it threw me for a loop, but once you understand the idea, it becomes a snap. I just hope I haven't made a silly math mistake here! flintsword | ||
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| Re: Odds;from two pairs to full house in hold'em, Amix1, 4. May 2003 03:15 | ||
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| Again, thanks for the exhausting run through and number crunching. It certainly seem like a good exercise for the mind. I am just not certain if I would have the ability to do this quick enough at the tables. Again, thanks it is something to think about for sure. :-) | ||
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