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%)