Odds of finishing second

[Aragorn]

I am trying to figure out the odds of finishing second in a 3-person poker tournament. Suppose the players have $10K, $5K, and $2K in chips. I know their chances of winning are directly proportional to how many chips each has.

But how do you calculate the chances of coming in second? Any ideas would be appreciated.

[Big Country]

[ QUOTE ]
Suppose the players have $10K, $5K, and $2K in chips. I know their chances of winning are directly proportional to how many chips each has.


[/ QUOTE ]

This is only true if you are making the assumption that each player is exactly the same in skill level.

Are you looking for the odds of a particular one of the three players coming in second, or the odds for each ofthem coming in second? (not that I can answer either, just curious)

[burningyen]

Sklansky addresses this in TPFAP. Long story short, there's no easy way to calculate the percentages with more than 2 players remaining.

[schroedy]

On your assumption that chances of winning are directly, and immutably, proportional to stack sizes. I will further assume that your chances of finishing second are directly proportional to stack size if you don't win . . . .

First we calculate that chances of winning.

10,000/17,000 = 58.82%;
5,000/17,000 = 29.41%
2,000/17,000 = 11.76%.

(Totals less than 100% due to rounding.)

Then we look at the chances of finishing second in those cases when you don't win . . .

10,000 stack will not win 41.18% of the time. 29.41% of the time, the big stack will battle the small stack for the second finish, where, on your assumption, it will be a 5:1 favorite. Multiply 29.41% by 83.33% (converting 5:1 to percentages) and you get 24.51%. 11.76% of the time the big stack will face the middle stack for second, and will be a 2:1 favorite. Multiply 11.76% by 66.67% and you get 7.84%. So the big stack has a 32.35% chance of finishing second, and a 91.17% chance of finishing AT LEAST second.

Taking the middle stack next. 58.82% of the time it will be battling the small stack for second where it will be a 5:2 favorite. Multiplying through, the middle stack should finish second in 42.01% of the cases. In the 11.76% of the trials that the middle stack faces the big stack for second, the middle stack is a 2:1 dog, and should win 1/3 times. Multiplying through produces a second place expectation for this leg of 3.92%. Add it all up and the middle stack has a 45.95% chance of finishing second and a 75.36% if finishing at least second.

Finally the small stack. When the big stack wins overall, the small stack is a 5:2 dog to finish second. Multiplying through the small stack gets second 16.81% of the total cases on this leg. When the middle stack wins overall, the small stack is a 5:1 dog to finish second. On this leg the small stacks adds another 4.90% chance of finishing second. Add it up and the small stack has a 21.71% chance of finishing second.

There may be some rounding errors in there, but that is the general approach.

As others have pointed out, your assumptions may be flawed.

[eastbay]

[ QUOTE ]
I am trying to figure out the odds of finishing second in a 3-person poker tournament. Suppose the players have $10K, $5K, and $2K in chips. I know their chances of winning are directly proportional to how many chips each has.

But how do you calculate the chances of coming in second? Any ideas would be appreciated.

[/ QUOTE ]

Search for ICM (independent chip model) for one general solution to this kind of problem.

eastbay

[Aragorn]

Thanks for the response. I did the same calculation and came up with the same answers.

However, I also wrote a program to simulate this by playing out 1 million tournaments and got slightly different answers. I got the big stack finishing second 34.7% instead of 32.5%, the middle stack finishing second 46.9% instead of 45.9%, and the small stack finishing second 18.5%.

The answers are pretty close, but well outside the expected error. I am not sure if there is a bug in my simulation or this method is only an approximation.

Do you know that the calculation method you used is accurate?

Thanks again for the response.

[schroedy]

I simply "logiked" my way through the problem. I am unaware and there may be hidden bugs.

Maybe you should do a series of 100K tournament runs and see how the results fall around expected.