Computing tournament finish place probability

[Bozeman]

Since I have not found any previous mention of this method for computing finish probability, and Sklansky (in TPFAP) says "there is no ironclad way to calculate the chances of coming in second and third based on your chip position", I am reposting my method which I posted about a year ago which got scant notice (perhaps because it was in a reply).

Admittedly this is not ironclad, but given the assumptions that players are of equal ability, blinds and/or antes are small relative to stacks, and therefore that your odds of finishing ahead of any other person are proportional to the ratio of your stacks, the answers are accurate. They therefore provide also a reasonable starting point even if these assumptions are violated.

I compute chance of each place iteratively:

chance of 1st = my_chips/total_chips
chance of 2nd is sum of (chance of each other person taking first*chance you will take 1st among the remaining players)
etc.
As an extension of the assumption that chance of taking first is proportional to stack size.

for example: stacks of 1,1,2 for A,B,C

P(A1)=1/4
P(B1)=1/4
P(C1)=1/2

P(A2)=P(B1)*P(AbeatsC)+P(C1)*P(AbeatsB)
=1/4*1/3+1/2*1/2
=1/12+3/12
=1/3

P(B2)=P(A2)=1/3

P(C2)=2*1/4*2/3=1/3

P(A3)=P(B1)*P(C2givenB1=C/A)+P(C1)*P(B2givenC1)
=1/4*2/3+1/2*1/2
=5/12

P(B3)=P(A3)=5/12

P(C3)=2*P(A1)*P(B2givenA1)
=2*1/4*1/3
=1/6

For the example in HPFAP (3 players with 12k, 6k and 2k):
P(2k1)=1/10
P(2k2)=3/5*1/4+3/10*1/6=12/60=1/5
P(2k3)=1-1/10-1/5=7/10
So 2k has a money expectation (with prizes of $8k, 5k, and 3k) of $3k*7/10+$5k*1/5+$8k*1/10=$3900 instead of the $3950 estimate Sklansky gives.

The numerics begin to get too computionally intensive for the conjunction of more than 10 players and more than ten places (computing probability of placing nth out of N players takes something like N^n computations).

For instances where players are clearly not even, I use this calculation after adjusting the amount of money. For example, suppose a SNG with 9 equal players and one expert who is twice as good, then one could do the calculations with stacks of 2,1,1,1,1,1,1,1,1,1.

Craig

[M.B.E.]

[ QUOTE ]
chance of 1st = my_chips/total_chips
chance of 2nd is sum of (chance of each other person taking first*chance you will take 1st among the remaining players)
etc.
As an extension of the assumption that chance of taking first is proportional to stack size.

[/ QUOTE ]
I'm not sure the second assumption is correct. There's some discussion of this in "Gambling Theory and Other Topics", in essays written not by Mason but by some other guy whose name escapes me. The other guy says (near the end of his last essay) that Mason advocates a method for three-way tournament deals which sort of relies on the second assumption above, but he (the other guy) thinks it's wrong. The essays suggest that you start with the assumption that when someone gets eliminated that player's chips, on average, will be divided equally between the two remaining players. Then you get a system of linear equations to solve, a bit cumbersome with only pen and paper but okay if you have a laptop with you I guess.

Suppose I'm in third with 5,000 and the other two players have 45,000 and 40,000. One problem is that my chance of doing better than third place will depend on the blind structure. If the blinds are high then I've got hardly any chance of making first or second, because my two opponents will just let me be blinded off (assuming there's a decent jump in prize money between third and second). But if the blinds are small, then I do have a fighting chance to claw my way back. So in the first case your algorithm won't work; in the second it might. Also it will depend whether the tournament's limit or no-limit.