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Old 10-13-2003, 07:24 AM
Legato Legato is offline
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Join Date: Jun 2003
Location: Sweden
Posts: 102
Default Re: Tournament Finish Probability (long)

Wow, great work. Unfortunately it would take hours for me to understand it all (if even that would be enough). Recently I played a tourney where we wanted to do a split I always used the McEvoy model (which I believe Pokerstars uses in the big one as well), but was informed about the big problem with large stacks potentially getting more than 1st prize which I never thought about before, and which clearly makes the model inadequate.

You seem to know alot about this and I trust your work enough to start using the Malmuth model. Unfortunately I do not understand how it works from your description. Would you please detail it a bit more.

The formula given is:

P(i,2|j,1)=I/(T-J)

My (incorrect) translation:
Probability of player i ending up 2nd granted that player j ends up 1st is equal to player i:s stack divided by the total number of chips minus player js stack.

And using your example 2/1/1:

P(i,2|j,1)=I/(T-J) =>
P(i,2|j,1)=1/(4-2) = 0.5 which you calculate to 0,33.

Would you care to show how u got the figures you got for 2/1/1 stacks?
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