Classic Type Game Theory Problem

[Xhad]

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So are you saying that player B should always draw when his hand is between 0.5 and 0.586 and player A stands?

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It seems he should. In fact according to my math, B should never stay pat on a hand less than 2 - sqrt(2) unless he knows that A is playing incorrectly.

The reason is that the only potential "bluffing" situations are those in which b > .5, but a + b > 1 (any time one of the above is not true, A should not stand pat even if B will draw 100% of the time because A has a better chance of winning if he simply draws himself). But for all hands from .5 to 2 - sqrt (2), even if A is bluffing at EVERY SINGLE OPPORTUNITY B still has to draw out of respect for the number of legitimate hands A could be standing pat with.

[Darryl_P]

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even if A is bluffing at EVERY SINGLE OPPORTUNITY B still has to draw out of respect for the number of legitimate hands A could be standing pat with.

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I'm not sure he has to draw with 100% probability, though. It looks to be enough to draw with a high probability (equal to (1-b)/b, say).

It doesn't seem right for B to give up so much EV by drawing every single time if he doesn't have to.

Do you know the overall EV for each player under your assumed optimal strategies? If you say A's is more than .5076 then I'd like to pit my player B against your player A and see what happens in a simulation.

[jason_t]

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Player A and Player B are both dealt a real number from zero to one.

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How?

[Darryl_P]

I ended up calculating the overall EV under your strategy as well and it turns out to be 0.4938 for player B which beats my 0.4924

I couldn't do it analytically so I just made a big excel spreadsheet and did it numerically. At least I was able to double check my EV figure which turns out to be correct, but evidently not optimal as your strategy is better.

I should have known something was up if A couldn't gain anything by bluffing. The 0.4924 just seemed so good that I thought it couldn't be beat, but alas, it can. Live and learn.

[tolbiny]

Player A's bluffing strategy has to include the actual value of B's card- ie the probability of drawing over B's card combined with the probablility of B drawing under A's card.

[gumpzilla]

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How?

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I think it's pretty obvious that he means from a uniform distribution, and if generating real numbers (because of a countable computable/uncountable uncomputable issue) bothers you, then use a uniform distribution over some large (a trillion will probably do nicely) number of evenly spaced rationals over that range.

I think it's kind of interesting that nobody has commented on Jerrod's solution in this thread.

[Darryl_P]

I should add that I did my EV calculation above based on BBB's summary in the other forum. Did you catch the modification to A's strategy for values of B's upcard between 0.5 and 0.586?

[Jerrod Ankenman]

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Player A and Player B are both dealt a real number from zero to one.

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How?

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Through the magic of toy game technology, son.

:P

Jerrod

[Xhad]

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even if A is bluffing at EVERY SINGLE OPPORTUNITY B still has to draw out of respect for the number of legitimate hands A could be standing pat with.

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It doesn't seem right for B to give up so much EV by drawing every single time if he doesn't have to.

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He's not giving up that much EV because the hands in that range aren't that likely to back into losers that often.

Basically what it comes down to is that for a patbluff to work, B not only has to draw when he has the best hand, but he also has to back into a worse hand than he already and have that hand not still be winning. The hands in the .5->.59-ish range I suggest just aren't that far above random to begin with.

Of these hands, the hand for B with the widest patbluff range is 2-sqrt(2), which is about .59. That range is everything down to about .41. Patbluffing anything lower than that is -EV for A even if you fall for it 100% of the time (which is why I said draw "unless you know A to be playing incorrectly"), so that means that if you draw to hands in this range you are falling for a profitable bluff at most about 30% of the time, and a significant percentage of those times you will still be ahead of those hands even if you fall for it.

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Do you know the overall EV for each player under your assumed optimal strategies?

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No I don't because I haven't completely finished the problem for the higher ranges of hands. =X However, I am convinced that I have the optimal strategy up to b = 2 - sqrt(2). The math involved can be found here. Note that the inequality is basically EV(catch bluff) > EV (draw) and that 2 - sqrt(2) was the minimum value of b necessary to make it true if A is bluffing as much as he can and still have it be theoretically profitable.

[jason_t]

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How?

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I think it's pretty obvious that he means from a uniform distribution, and if generating real numbers (because of a countable computable/uncountable uncomputable issue) bothers you, then use a uniform distribution over some large (a trillion will probably do nicely) number of evenly spaced rationals over that range.

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It has nothing to do with being bothered by uncountability or computability. There is no uniform distribution on [0,1].