Game Theory: Unusual Question #3 and #4

[Aisthesis]

I'm a little shaky on method for the bluffing solution, but your equations got me going for a non-bluffing solution anyway. Actually, I think I'll refrain from even going through all the individual cases of fold, call or raise.

I'm pretty sure my equations agree with yours, although I couldn't follow all the way. But the great thing is that they simplify to an EV (with R as the raise threshold and .5 as the obvious call threshold) of 11/8*R - R~2.

Hope I didn't mess up anywhere along the road in adding the decisive scenarios, but in any case the solution from there is simple and does yield the best non-bluffing EV seen yet.

Differentiating over R we get:
11/8 - 2R = 0 (for local maximum)
Hence R = 11/16 as the raise threshold

That should make the EV for B 121/256 or a hair over .47

I think the best non-bluffing solution we had up to now was .46

[Aisthesis]

If anyone could explain to me (even, or perhaps more accurately, PREFERABLY with the simplest possible version of this type of problem) how to set up the equations for a bluffing solution, I'd be most appreciative!!

Having absolutely no background in game-theory, I'm having to re-invent it just to get to the non-bluffing solution.

Ankenman's results are too pretty to be wrong on the more complex case, but I'd love to be able to understand how he got there.

[Jerrod Ankenman]

[link to primer material] (groups.google.com)

[Aisthesis]

Thanks! Will take me a little bit to make it through this stuff, but actually this [0,1] game is most interesting...

The first thing that strikes me is the ability to use the continuity here to come back to the discreet "values" of poker hands (which definitely have a lot of mathematical strangenesses over and above that).

[Aisthesis]

I've just made it through lessons 1-3, but that helped immensely--although I think I'll need to get further into it to come up with an algebraic solution even on the simple scenario (still unclear to me how to deal with bluff or even finite pot).

But that brings up my question: Once you actually know what you're doing, is the algebraic approach usually easier than analysis (differentiating for maxima and minima)? I still feel irresistably attracted to analytic solutions and am actually, after looking at the way you set it up, thinking I may be able to get a clean bluffing solution (for the "easy" problem) by differentiating for EV (I'll post it if it works out). I'm actually guessing that the indifference equations turn out to be equivalents of the various equations you get when you differentiate... (??)

[fnord_too]

I have been thinking about the bluffing part. If B's strategy is known then you compute at what point A's call has a 25% chance of winning (same as with no bluffing). I'm going to work on the algebra later today if I have time. If B always raised A calls with any hand over .25, and has a slight edge in the game (it is worth .03 to him I think).

[Gandor]

[ QUOTE ]
[ QUOTE ]
Justin, B's optimal strategy will be to raise with some of his best hand but also to bluff raise with some bad hands.

[/ QUOTE ]

This is incorrect. Let's say he raises all hands .6 (the number is arbitrary) and above, and then bluffs with 10% of his hands. Wouldn't it make more sense to just change that to raise .5 and above instead? Instead of raising 10% of bad hands, just raise the next 10% of hands after the cutoff point.

[/ QUOTE ]

If you select just the next 10% of hands, player A can adjust his strategy knowing he needs a little less to call. Using the bottom 10% to bluff is much more effective. When you raise, player A must decide if you are bluffing or not. If you always raise with the top 40% in your example, then 40% of the time you have a strong hand and player A must be even stronger to call. But player A also knows that 10% of the time you raise, he has you beat. Then to call becomes a guessing game for him.

[well]

I am doing something wrong, but I can't find out what.
Suppose we're looking at the bluff/no-fold strategy.

Player A will call a raise whenever he has something greater than z.
Player B will bluff with something smaller than x, and honestly raise with something over y and will call with the rest.

The bluff:
B bluffs with probability x, and when he does, he
wins 1 with probability z, and loses 2 with probability 1-z.
So we have x(3z-2)

The call:
B calls with probability y-z, and when he does, he
wins 1 with probability x+(y-x)/2, and loses 1 with probability (y-x)/2+(1-y).
So we have (y-x)(x+y-1)

The honest raise:
B does this with probability 1-y, and when he does, he
wins 1 with probability z when A folds, and when called (which happens with probability (1-z)) he will win 2 with probability
(y-z)/(1-z)+(1-y)/(1-z)/2 and lose 2 with probability (1-y)/(1-z)/2.
So here we have (1-y)(2y-z)

Summing these three will result in the EV-function for B which now is
EV(x,y,z)=-x^2-y^2-x-y-z+3xz+yz

[ QUOTE ]
I get that B should bet his best 1/6 hands, bluff his worst 1/18, and A should call with his best 1/3. [...] so the value of the game for B is 5/18.

[/ QUOTE ]

For this solution (x,y,z)=(1/18,5/6,2/3), which results in EV=13/162, or about .0802.

This is not even close to .278

Where did it all go wrong?

Thanks for helping.

[David Sklansky]

You don't need fancy math. A must call with the top one third of his hands to prevent B from profiting from bluffing with bad hands. (B is laying 2-1 on those bluffs)

B should bet the top one sixth (3/18) of his hands for value since those hands are favored to win even if called.

Since A is getting 3-1 when bet into, B should bluff 1/18 of his total hands so that his bet to bluff ratio is 3-1.

Since we know that B makes 25 cents in the non raising game, we need only see how much the existence of a possible raise helps. It doesn't help on his bluffs since they break even. It doesn't help when both players have the top sixth of their hands. It helps only when B is in the top sixth and A is in the SECOND sixth. That gains a bet for B 1/36 of the time. His EV moves from 9/36 to 10/36. .278

[Bozeman]

"I am doing something wrong, but I can't find out what.
Suppose we're looking at the bluff/no-fold strategy.

Player A will call a raise whenever he has something greater than z.
Player B will bluff with something smaller than x, and honestly raise with something over y and will call with the rest."

Because the bluff/no-fold strategy is not optimal: you need to add (the point I neglected to mention in my post, because it was so obvious) that B will fold hands below .5 that he doesn't bluff with.

This makes for the possibility of many cooptimal strategies: any strategy where B bluffs 1/18 of the time (all below .5), calls with .5-5/6 and bets 5/6-1. Thus the best strategy, if A is playing very poorly by calling with too many hands (more than 1/2), would be to bluff with 8/18-9/18.

I don't see how all this solving of the bluffless case is of any use: ANY PLAYER THAT NEVER BLUFFS IS GIVING AWAY EV (unless he is playing an absolute calling station). DO you solvers see anything wrong with my bluffless analysis that shows that B should use the same strategy for the (B) raise game as for the call or fold game if he is not allowed to bluff? And fnord, you are working too hard on the hard way to solve this, check out the game theory primer. Aisthesis: the math is much easier algebraically, though with programs to differentiate for you the differential analysis is not too bad, and getting results both ways provides a check.
Craig