Poker Probabilities
March 8th, 2012 at 8:19:00 AM
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So I'd like to think my math skills are in the 90th percentile of the general population, but I'm afraid they might be more like the 20th percentile among others with a strong math background, like physicists, engineers, or casino game math analysts, etc. Since I'm a dealer, I often find myself wondering about certain probabilities of events, and what the best way to play a specific situation.
Things arise frequently on Ultimate Texas Hold 'Em, and I'm just wondering if I'm doing the math right. For instance, I was dealing the other day, and on one particular hand there were four to a straight on the board 10-A with no J (can't remember the fifth comm. card). So what I was wondering, is what is the probability of NOBODY completing the straight, after knowing the one card needed? There were five players, including myself, thus ten cards. This is what I came up with:
All possible ten card combinations for remaining deck: 47 nCr 10
All possible ten card combinations without a Jack: 43 nCr 10
So the probability that nobody has a Jack would be (43 nCr 10)/(47 nCr 10) = 37% (a little higher than I thought)
Similarly, a situation that I've seen frequently, is four to a flush on the board, and nobody completes the flush. So assuming the same situation of 5 players, I would come up with this:
All possible ten card combinations for remaining deck: 47 nCr 10
All possible ten card combinations without the needed suite: 38 nCr 10
Probability nobody completes flush: (38 nCr 10)/(47 nCr 10) = 9% (sounds about right)
On a similar note, if somebody is playing the Trips bet, if there is something that will play the Trips, they have the option of folding against the dealer and just collect the Trips payout. So one particular day I was dealing, there was one man playing. There was a board club flush, something like 2-9. I know for certain the 2 was out. He didn't have a club, so if I had ANY club, I would've beat him. Should he have played the hand? Or just collected the Trips payout. He already knows he won't win any money on the Ante/Blind/Play, he'll push or lose.
After a lot of thinking I came up with this, the odds I don't have a club would be : (38 nCr 2)/(47 nCr 10) = 65.03%
The chance that I would have a club would then be just shy of 35%. So if you play 100 hands, and always fold, you'd lose $1000 ($5 Ante/Blind). But if you always play you would lose 35 * $15 = $525. So it would always be beneficial to play this situation.
Then I made a generalization, that the risk for a straight would be more, but not quite reach $1000 (didn't do calculation) and thus if there's any five card made hand on the board, you should always play, because the chance the dealer beats the board is 35% or less for a flush or better.
Is this thinking right?
Things arise frequently on Ultimate Texas Hold 'Em, and I'm just wondering if I'm doing the math right. For instance, I was dealing the other day, and on one particular hand there were four to a straight on the board 10-A with no J (can't remember the fifth comm. card). So what I was wondering, is what is the probability of NOBODY completing the straight, after knowing the one card needed? There were five players, including myself, thus ten cards. This is what I came up with:
All possible ten card combinations for remaining deck: 47 nCr 10
All possible ten card combinations without a Jack: 43 nCr 10
So the probability that nobody has a Jack would be (43 nCr 10)/(47 nCr 10) = 37% (a little higher than I thought)
Similarly, a situation that I've seen frequently, is four to a flush on the board, and nobody completes the flush. So assuming the same situation of 5 players, I would come up with this:
All possible ten card combinations for remaining deck: 47 nCr 10
All possible ten card combinations without the needed suite: 38 nCr 10
Probability nobody completes flush: (38 nCr 10)/(47 nCr 10) = 9% (sounds about right)
On a similar note, if somebody is playing the Trips bet, if there is something that will play the Trips, they have the option of folding against the dealer and just collect the Trips payout. So one particular day I was dealing, there was one man playing. There was a board club flush, something like 2-9. I know for certain the 2 was out. He didn't have a club, so if I had ANY club, I would've beat him. Should he have played the hand? Or just collected the Trips payout. He already knows he won't win any money on the Ante/Blind/Play, he'll push or lose.
After a lot of thinking I came up with this, the odds I don't have a club would be : (38 nCr 2)/(47 nCr 10) = 65.03%
The chance that I would have a club would then be just shy of 35%. So if you play 100 hands, and always fold, you'd lose $1000 ($5 Ante/Blind). But if you always play you would lose 35 * $15 = $525. So it would always be beneficial to play this situation.
Then I made a generalization, that the risk for a straight would be more, but not quite reach $1000 (didn't do calculation) and thus if there's any five card made hand on the board, you should always play, because the chance the dealer beats the board is 35% or less for a flush or better.
Is this thinking right?
March 8th, 2012 at 12:05:41 PM
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The player knows at least their own two cards, so after the river, there are actually seven known cards, and only 45 unknowns.
Simplicity is the ultimate sophistication - Leonardo da Vinci
March 8th, 2012 at 2:49:03 PM
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Quote: AyecarumbaThe player knows at least their own two cards, so after the river, there are actually seven known cards, and only 45 unknowns.
Yes that's true, I guess I'm thinking more from my perspective as the dealer, I haven't seen ANYBODY's cards, so it would be 47. And I think the numbers I put were opposite, 37% and 9% for NOT completing straight and flush respectively would really be the probability FOR completing them.
