[ QUOTE ]
You are in third and the chipcounts are:
1) 4000
2) 2500
3) 2000
4) 1500
5) 1000
how would you estimate your probability of ending second? Or third?
[/ QUOTE ]
Why do the chips add up to 11,000? Anyway, the probabilities I get by the independent chip model are
For the player with 2000 chips:
1st: 18.2% = 2/11
2nd: 20.4% = 25401/124355
3rd: 22.5% = 50452/223839
4th: 22.5% = 251848/1119195
5th: 16.3% = 1196/7315
The independent chip model can be described in a few ways. One is to imagine that the chips are eliminated one by one. The players are knocked out when all of their chips are gone. Another way is to imagine that the first place is determined proportionately, and then among the losers, the next place is decided proportionately, etc.
Here is some ugly but functional Mathematica code:
<font class="small">Code:</font><hr /><pre>
Clear[places];
places[chipc_, i_] :=
places[chipc, i] =
If[Length[chipc] == 1, {1},
Table[If[j == Length[chipc],
chipc[[i]]/Sum[chipc[[k]], {k, Length[chipc]}], 0], {j,
Length[chipc]}] +
Sum[If[t == i, Table[0, {s, Length[chipc]}],
Append[places[Delete[chipc, t], If[t < i, i - 1, i]]chipc[[t]]/
Sum[chipc[[k]], {k, Length[chipc]}], 0]], {t,
Length[chipc]}]] </pre><hr />
N[places[{4000, 2500, 2000, 1500, 1000}, 3]]
{0.1635, 0.225026, 0.225394, 0.204262, 0.181818}
Keep in mind that this is only a model, and that there are other models. In a limit tournament with small blinds, it might be better to model the chip count movement by diffusion. Diffusion is relatively easy to understand in 2-dimensions, 3 players, and the following
paper by Thomas Ferguson computes some exact probabilities. The complications suggest that there are no simple formulas for the diffusion model with more players.