Mark Gritter (markgritter) wrote,
Mark Gritter
markgritter

CP2-7 update: 100K experiment

I unfortunately didn't queue up enough iterations of my experiment before leaving for CA, and didn't take a look at it before today. Results so far:

a.1: $0.03785375 (/100000 changed)
b.1: $0.03545646 (/100000 changed)
a.2: $0.01646984 (14092/100000 changed)
b.2: $0.00252623 (16098/100000 changed)
a.3: $0.00738640 (10890/100000 changed)
b.3: $0.00140493 (9600/100000 changed)
a.4: $0.00651721 (9681/100000 changed)
b.4: $0.00069331 (8678/100000 changed)
a.5: $0.00591959 (8828/100000 changed)
b.5: $0.00013653 (8050/100000 changed)
a.6: $0.00550497 (8216/100000 changed)
b.6: $-0.00024108 (7459/100000 changed)
a.7: $0.00501711 (7555/100000 changed)
b.7: $-0.00076080 (6934/100000 changed)
a.8: $0.00471229 (7019/100000 changed)
b.8: $-0.00090690 (6437/100000 changed)
a.9: $0.00439509 (6518/100000 changed)
b.9: $-0.00123131 (6016/100000 changed)


Much closer to arriving at a fixed strategy.

Here are some hands that still cycle:

3d3h9h9sJs 2d3s4s6c7s QcKhAc or 3d3h3s9h9s 2d4s6c7sJs QcKhAc (about -0.8)
2c5c6hTcTh 2s4d5s6s9h 9sJhKc or 2c5c6hTcTh 2s4d5s6sJh 9h9sKc (about -2.5)
2c2h2s8h8s 3c3s4d9sTd KhKsAh or 2c2h2s3c3s 4d8h9sTdAh 8sKhKs (about -0.6)

These show more of the tradeoff between low and high strength that I expected to see.

I'm not sure whether I still believe whether an unexploitable strategy to the full game would require making random choices between alternative arrangements or not. We can certainly tell at this point that an exploitive strategy can only earn a very small amount (less than 1/100th of a point now and probably much less for more hands.)

10e3 hands: about 40% cycles
10e4 hands: about 20% cycles
10e5 hands: < 6% cycles

Extrapolating a 50% reduction in cycles per order of magnitude hand increase, we should expect the full 635 million hands (call it 10e9) to have less than 1% cycles, perhaps less than 0.5%.

On the other hand, at some point the statistical sample should be large enough to give us the same answers as a full strategy. Obviously this is not the case yet because the expected value of the game is biased towards A in the current samples. It should be zero or else symmetric.
Tags: chinese poker, poker, theory
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