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