Re: Bill Chen Type Pure Math/Poker Problem
 

After pondering for a few minutes, I'm thinking the answer is actually simple. Everybody should bet with a frequencey that prevents a renegade from profiting with lesser hands. So for instance if four people ante one dollar and the bet is three dollars, a bluffer could gamble three to win three if everybody else folds. To prevent him from profiting the others must fold no more often than the cube root of one half. Did I miss something?

Re: Bill Chen Type Pure Math/Poker Problem
 

[ QUOTE ]
After pondering for a few minutes, I'm thinking the answer is actually simple. Everybody should bet with a frequencey that prevents a renegade from profiting with lesser hands. So for instance if four people ante one dollar and the bet is three dollars, a bluffer could gamble three to win three if everybody else folds. To prevent him from profiting the others must fold no more often than the cube root of one half. Did I miss something?

[/ QUOTE ]

I got the same and yes it was fairly simple and works basically for the reason you describe. What makes you think it should have been difficult?

Re: Bill Chen Type Pure Math/Poker Problem
 

Yes, that's a good short-cut: for each player, the other
(N-1) players must play independently, but such that the
betting frequency collectively must be equal to the optimal
game-theoretical calling frequency, i.e., so that "bluffs"
break even in the long run.

Exploitative strategy:

Now, suppose we "know" a specific player, A, is playing
suboptimally by picking slightly smaller numbers than
everyone else. Now, if you're in the game, you can play
somewhat "looser" and take advantage of "loose play". Of
course, we have to be sure that A is doing this consistently
for if A just did some very short-term advertising, we'll
end up being the "sucker"! This certainly explains the
dynamics in some of the games.

Re: Bill Chen Type Pure Math/Poker Problem
 

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Everybody should bet with a frequency that prevents a renegade from profiting with lesser hands.

[/ QUOTE ]

Yes. Another way of looking at it is that if your hand is right on the borderline, all your equity is fold equity, since anyone with a hand above the threshold will beat you. One consequence is that it doesn't cost the renegade to play more hands (his fold equity is just as high with a terrible hand), but he won't gain, either, and his strategy will be exploitable.