How do you calculate the probability of a specific finishing order for 3 all in hold em hands
March 24th, 2013 at 3:49:32 PM
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Anyone know how to calculate this?
It's easy with 2 hands all in. However when there is an A,B, and C hold em hand preflop how do you calculate a specific finishing order, such as A in 1st, B 2nd, and C 3rd?
Maybe there is an online calculator that does it but I couldn't find it. Thanks
It's easy with 2 hands all in. However when there is an A,B, and C hold em hand preflop how do you calculate a specific finishing order, such as A in 1st, B 2nd, and C 3rd?
Maybe there is an online calculator that does it but I couldn't find it. Thanks
March 24th, 2013 at 4:31:24 PM
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Quote: clarkacalAnyone know how to calculate this?
It's easy with 2 hands all in. However when there is an A,B, and C hold em hand preflop how do you calculate a specific finishing order, such as A in 1st, B 2nd, and C 3rd?
Maybe there is an online calculator that does it but I couldn't find it. Thanks
Cycle through all 5 board cards, that is 46!/(41!*5!) = 1.4 million showdowns. Sum up all results in the 10 possible categories, and you are done.
March 24th, 2013 at 4:33:19 PM
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http://twodimes.net/poker/ lets you enter as many starting hands as you want -- however it only tells you the overall winner.
March 24th, 2013 at 4:43:31 PM
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Yes that's the problem, I need calculate 1st 2nd and 3rd exactly
March 24th, 2013 at 6:36:06 PM
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Where can I do this?
March 24th, 2013 at 6:42:12 PM
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Quote: clarkacalWhere can I do this?
Poker Odds Calculator
If PA1, PB1, and PC1 are the respective probabilities of each player finishing 1st, then the probability of player A finishing 2nd and 3rd are (ignoring ties):
PA2 = PB1*PA1/(PA1+PC1) + PC1*PA1/(PA1+PB1)
PA3 = 1 - PA1 - PA2
Put this in a spread sheet I suppose if you want a program to do it for you.
March 24th, 2013 at 7:11:38 PM
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Quote: paisiello
This isn't what OP asked I think. He wants specific breakdowns of how often each hand will finish 1st, 2nd, or 3rd. Unfortunately, I am unaware of a program that does this calculation for you.
March 24th, 2013 at 7:20:30 PM
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I'm not real familiar with horse racing but what I'm asking is the probability of a hold em trifecta. 88 coming in 3rd, AK 2nd, and A10 in 1st
March 24th, 2013 at 7:22:25 PM
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deleted
DUHHIIIIIIIII HEARD THAT!
March 24th, 2013 at 7:31:19 PM
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Yes, sorry, I editted the response above.
March 24th, 2013 at 7:31:45 PM
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I think that is just a regular poker calculator, I don't see where it calculates the trifecta
March 24th, 2013 at 7:34:27 PM
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My bad, please see the editted post above.
March 24th, 2013 at 7:43:30 PM
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Quote: paisielloPoker Odds Calculator
If PA1, PB1, and PC1 are the respective probabilities of each player finishing 1st, then the probability of player A finishing 2nd and 3rd are (ignoring ties):
PA2 = PB1*PA1/(PA1+PC1) + PC1*PA1/(PA1+PB1)
PA3 = 1 - PA1 - PA2
Put this in a spread sheet I suppose if you want a program to do it for you.
I'm pretty sure this would be inaccurate. Probability for A cannot be for 2nd and 3rd, it has to be the trifecta for the mid stack to go out in 3rd considering the stack sizes. Or maybe I am misunderstanding
March 24th, 2013 at 8:01:17 PM
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I don't think the stack sizes would have anything to do with which player has the best hand and in which order they finish. Maybe you should give a specific example to clarify your question for me.
Do you agree at least my formula for PA3, the probability of player A finishing 3rd, is accurate?
Do you agree at least my formula for PA3, the probability of player A finishing 3rd, is accurate?
March 24th, 2013 at 8:10:37 PM
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Stack sizes don't have anything to do with the end rankings of the hands but it does affect which order they finish. I didn't include stack sizes in the original post because I just wanted to figure the trifecta probability, but there would be a side pot because of a short stack, medium, and large stack.
As far as the formula I'm afraid it's a bit over my head
As far as the formula I'm afraid it's a bit over my head
March 24th, 2013 at 8:16:52 PM
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OK, it seems you are interpreting the "order of finish" as the amount of chips each player wins?
March 24th, 2013 at 8:25:42 PM
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give me a minute want to try to explain it better
March 24th, 2013 at 8:35:49 PM
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This is the situation:
Player A:300k
Player B:150k
Player C:30k
I am player B so the only way I get eliminated IN THIRD PLACE is to get beat by player A who also either ties or gets beat by player C in this same hand
Player A:300k
Player B:150k
Player C:30k
I am player B so the only way I get eliminated IN THIRD PLACE is to get beat by player A who also either ties or gets beat by player C in this same hand
March 24th, 2013 at 8:39:40 PM
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OK, I think I see now. You want to know the probability of who ends up in what place in the "tournament" assuming they chop immediately after the hand. Correct?
March 24th, 2013 at 8:51:41 PM
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No, I want to know the probability of player B getting eliminated in 3rd place after that hand.
For that to happen, C has to have the best hand, A the second best, and B the worst hand. If A had the best hand both B and C would be out but B would get 2nd place because of the higher stack, regardless of whether C beat his hand
For that to happen, C has to have the best hand, A the second best, and B the worst hand. If A had the best hand both B and C would be out but B would get 2nd place because of the higher stack, regardless of whether C beat his hand
March 24th, 2013 at 9:34:01 PM
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Well, for your specific question regarding player B then, I think my formula stands as before.
However, your original question had to do with the trifecta (to use your expression): you wanted to know the probability for each player finishing in each place. So you are correct that the generic formula should take the specific stack sizes into account. One could work this out using a spread sheet with not too much work involved.
However, your original question had to do with the trifecta (to use your expression): you wanted to know the probability for each player finishing in each place. So you are correct that the generic formula should take the specific stack sizes into account. One could work this out using a spread sheet with not too much work involved.
March 24th, 2013 at 10:29:51 PM
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I am just not good explaining I think. Forget about places in a tourney. Think of it as a cash game. All 3 are all in preflop with the stack sizes as before. How do you calculate the chances that player B goes broke but player C doesn't?
March 24th, 2013 at 10:48:17 PM
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I finally did this in pql at propokertools.com, but I think you got rid of the specific hands in question in the posts. If it is still AT >= AK > 88, then that will happen about 7.66% of the time. It obviously depends on the actual hands though.
PQL Query:
select count(hiRating(p3, river) >= hiRating(p2, river) and hiRating(p2,river) > hiRating(p1, river))
from game="holdem", p1="88", p2="AK", p3="AT"
Results:
Trials COUNT 1
600000 45945 (7.66%)
PQL Query:
select count(hiRating(p3, river) >= hiRating(p2, river) and hiRating(p2,river) > hiRating(p1, river))
from game="holdem", p1="88", p2="AK", p3="AT"
Results:
Trials COUNT 1
600000 45945 (7.66%)
March 24th, 2013 at 11:00:26 PM
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No, I understand what you mean. It is me not explaining it clearly enough I think.
The answer to your question is to use the same formula as before but B and A swapped:
PB2 = probability player B finishes in 2nd place
= PA1*PB1/(PB1+PC1) + PC1*PB1/(PA1+PB1)
PB3 = probability player B finishes in 3rd place
= 1 - PB1 - PB2
PA1, PB1, PC1 are determined from the link given previously.
Since you have specified the stack sizes in your example then the chances of Player B finishng 3rd is the same as having the 3rd worst hand. If you wanted a more generic formula then you would have to include the actual stack sizes in the formula.
The answer to your question is to use the same formula as before but B and A swapped:
PB2 = probability player B finishes in 2nd place
= PA1*PB1/(PB1+PC1) + PC1*PB1/(PA1+PB1)
PB3 = probability player B finishes in 3rd place
= 1 - PB1 - PB2
PA1, PB1, PC1 are determined from the link given previously.
Since you have specified the stack sizes in your example then the chances of Player B finishng 3rd is the same as having the 3rd worst hand. If you wanted a more generic formula then you would have to include the actual stack sizes in the formula.
March 25th, 2013 at 12:45:11 AM
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Quote: tringlomaneI finally did this in pql at propokertools.com, but I think you got rid of the specific hands in question in the posts. If it is still AT >= AK > 88, then that will happen about 7.66% of the time. It obviously depends on the actual hands though.
PQL Query:
select count(hiRating(p3, river) >= hiRating(p2, river) and hiRating(p2,river) > hiRating(p1, river))
from game="holdem", p1="88", p2="AK", p3="AT"
Results:
Trials COUNT 1
600000 45945 (7.66%)
Perfect! Thanks tringlomane. Actually I didn't say that the A10 had two live suits but I doubt that would have made much difference since there would have to be an A or K on the board also for AK to beat 88.
Paisiello I appreciate your answers also. Your equation might work also, I don't know
March 25th, 2013 at 12:50:24 AM
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I went to propokertools and executed the query with the actual suits and the probability went down. Great tool, thanks!
PQL Query:
select count(hiRating(p3, river) >= hiRating(p2, river) and hiRating(p2,river) > hiRating(p1, river))
from game="holdem", p1="8h8d", p2="AcKc", p3="AhTs"
Results:
Trials COUNT 1
600000 42857 (7.14%)
PQL Query:
select count(hiRating(p3, river) >= hiRating(p2, river) and hiRating(p2,river) > hiRating(p1, river))
from game="holdem", p1="8h8d", p2="AcKc", p3="AhTs"
Results:
Trials COUNT 1
600000 42857 (7.14%)
