Probability in tournaments

[act]

Lets say you are in a tournament with 5 players. Everybody is same skill level as yourself, and blinds are small.

You are in third and the chipcounts are:
1) 4000
2) 2500
3) 2000
4) 1500
5) 1000

Now Sklansky would argue that you chance of finishin first is 2000/10000 or 20%. My question is:
how would you estimate your probability of ending second? Or third? Fourth? Fifth (ok I can give you the last one if you give me 1-4 /images/graemlins/grin.gif).

/images/graemlins/confused.gif

[fnord_too]

Here is how I would attack the problem:
20% you finish first.
Of the remaining times...
40% 1 finihes first. Finishing second is effectively the same as finishing first among the other 4, or 2000/6000, for 33% of that 40%
25% 2 fiishes first, and you finish 2nd 2000/7500
etc.

The weighted average for second comes out to be ~25.75%

This attack makes some big assumptions. For third the combinatorics begin to get laborious. I think here you start with
40% 1 wins
of these
--42% 2 finishes second
of these
--57% 3 finishes third.
etc for all combinations of first and second. I don't think you can take any shortcuts ( prob of 1-2 finishing first and second respectively is 4000/10000 * 2500/6000, prob of 1-2 finishing second and first respectively is 2500/10000 * 4000/7500, and these two numbers are not the same.)

[pzhon]

[ 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 &lt; 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 (http://www.math.ucla.edu/~tom/papers/unpublished/gamblersruin.pdf) computes some exact probabilities. The complications suggest that there are no simple formulas for the diffusion model with more players.

[fnord_too]

[ QUOTE ]
Here is how I would attack the problem:
20% you finish first.
Of the remaining times...
40% 1 finihes first. Finishing second is effectively the same as finishing first among the other 4, or 2000/6000, for 33% of that 40%
25% 2 fiishes first, and you finish 2nd 2000/7500
etc.

The weighted average for second comes out to be ~25.75%

This attack makes some big assumptions. For third the combinatorics begin to get laborious. I think here you start with
40% 1 wins
of these
--42% 2 finishes second
of these
--57% 3 finishes third.
etc for all combinations of first and second. I don't think you can take any shortcuts ( prob of 1-2 finishing first and second respectively is 4000/10000 * 2500/6000, prob of 1-2 finishing second and first respectively is 2500/10000 * 4000/7500, and these two numbers are not the same.)

[/ QUOTE ]

Oops, as someone pointed out the chips add up to 11k, add 1k to all of the denominators and disregard all percentages. The methodology is still what I would use, but would not be surprised if someone pointed out a major flaw in my assumptions.