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cfrm sampling theoretical question
http://poker-ai.org/phpbb/viewtopic.php?f=24&t=2661		 Page 1 of 1
Author:  algonoob [ Tue Dec 10, 2013 5:17 am ]
Post subject:  cfrm sampling theoretical question
Hi poker-ai

I tried to modify the monte carlo cfrm algorithm for kuhn poker by not sampling randomly but rather iterating QvK 1M times, followed by QvA 1M times .. and so on for every pair of hands. It didn't converge to the Nash equilibrium. Does this mean the sampling method has to be "random looking" in some way?
I then tried iterating over each holecard pair but perfectly interspersed. i.e. (QK,QA,KQ,KA,AQ,AK)x1000000
and it did converge to the nash equilibrium.

For real poker would it be OK to iterate in this way?
for mycards = 22,23,... AA
for oppcards = 22,23,... AA
for flop = 222,223,... AAA
....

it seems it might not converge as well since the p1 hole card are not interspersed perfectly. though I do not know why it does not converge perfectly in that situation. Anyone have an idea ?

The reason I want to do this is to get perfectly weighted representations for each scenario instead of using a PRNG with no guarantees.

algonoob
Author:  spears [ Tue Dec 10, 2013 7:04 am ]
Post subject:  Re: cfrm sampling theoretical question
This is difficult to put into words. In each iteration of Fictitious Play one player plays his best response to the other's equilibrium response, and then the players change places. The best response strategies are averaged over all time to form the equilibrium response. So if the same cards are played in successive iterations as you describe it's like you have very few iterations so convergence is very slow. CFRM is similar.

Quote:
For real poker would it be OK to iterate in this way?
for mycards = 22,23,... AA
for oppcards = 22,23,... AA
for flop = 222,223,... AAA


I'm 99% sure this would converge slower than a random ordering.

Quote:
The reason I want to do this is to get perfectly weighted representations for each scenario instead of using a PRNG with no guarantees.

I suspect that for a "perfectly weighted representation" you would need the answer before you start. But I think you might be able to bias the sampling to the cards which matter more to the solution. I know no more than that.
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