[ QUOTE ]
Does anyone know what the probabilities are of finishing last in a tournament where the stack sizes are 1, 2, ..., 100?
[/ QUOTE ]
I still don't, but here are some calculations of the ICM-predicted probability of finishing last in tournaments with stacks of size 1, 2, 3, ..., n.
n=2
1 2/3
2 1/3
n=3
1: 0.5833333333333334
2: 0.2666666666666666
3: 0.15
n=4
1 0.5511904761904762
2 0.24126984126984127
3 0.12976190476190477
4 0.07777777777777778
n=5
1 0.5361360861360861
2 0.22936507936507936
3 0.12031302031302031
4 0.07022977022977023
5 0.04395604395604396
n=10
1: 0.5184164152973668
2: 0.2151557713105984
3: 0.10893008631761322
4: 0.06109750853503585
5: 0.036606485042195436
6: 0.022998223608975937
7: 0.014984758061417186
8: 0.010053105474764727
9: 0.006909801915833682
10: 0.0048478444361987425
n=15
1: 0.5165407213004611
2: 0.21359514003346372
3: 0.10763726163687229
4: 0.06002971007838609
5: 0.03572609821501513
6: 0.02227291325368989
7: 0.014387192991584257
8: 0.009560467771416203
9: 0.006503209213545211
10: 0.0045117668120416956
11: 0.003183781349335634
12: 0.0022802773367725966
13: 0.001654769973221269
14: 0.0012150347599461867
15: 0.0009016552742487154
A brute-force calculation for n=15 (of all probabilities in all places) took only 2 minutes in Mathematica, but that method would run out of space around n=20 on my computer.
<ul type="square">
Mathematica code:
Clear[places]
places[n_, stacks_] := places[n, stacks] = If[Length[stacks] == 1, {1},
(1/Sum[stacks[[j]], {j, Length[stacks]}])
(Table[If[i == 1, stacks[[n]], 0], {i, Length[stacks]}] +
Sum[If[k < n,
stacks[[k]]Prepend[places[n - 1, Delete[stacks, k]], 0],
stacks[[k + 1]]Prepend[places[n, Delete[stacks, k + 1]], 0]],
{k, Length[stacks] - 1}]
)
]
fullplaces[stacks_] := Table[places[i, stacks], {i, Length[stacks]}]
Input: places[1,{2,3,7}]
Output: {1/6, 13/45, 49/90}
Input: fullplaces[{2,3,7}]
Output: {{1/6,13/45,49/90},[1/4,2/5,7/20},{7/12,14/45,19/180}}
[/list]The total number of intermediate distributions calculated is the number of sub-multisets of the opponents' chip counts (when equal stack sizes are entered next to each other), at most 2^n when there are n other players. However, the coefficients are quite complicated.
From the data, I conjecture that as n increases, the probability that the player with the smallest stack finishes last converges to something over 1/2. It's easy to prove some upper and lower bounds, e.g, that the value is greater than the product of (1 - 1/(k choose 2)) from k=3 to infinity, 1/3. The general behavior for the larger stacks may be of more interest.