Game Theory: Unusual Question #3 and #4

[Jerrod Ankenman]

[ QUOTE ]

If A's strategy is:
[0,x] : raise(bluff)
[x,y] : fold
[y,z] : call
[z,1] : raise


[/ QUOTE ]

Btw, it's worth mentioning that while the above strategy for B is co-optimal, it's also dominated by the strategy that raise-bluffs the best folding hands rather than the worst ones. Hopefully everyone sees why.

Jerrod

[well]

[ QUOTE ]
Btw, it's worth mentioning that while the above strategy for B is co-optimal, it's also dominated by the strategy that raise-bluffs the best folding hands rather than the worst ones. Hopefully everyone sees why.

[/ QUOTE ]

If you looked at the assumption made in my previous post

[ QUOTE ]
If A's strategy is:
[0,x] : raise(bluff)
[x,y] : fold
[y,z] : call
[z,1] : raise
And B's strategy is:
[0,a] : fold if raised
[a,1] : call if raised,
where 0<x<y<a<z<1

[/ QUOTE ]

you see that it doesn't matter which hands in [0,y] will be bluffed with,
since z>y anyway!

But if you don't assume A would have thought of that, this would not even
necesserily be B's (co-)optimal strategy I think.

Next Time.

[David Sklansky]

Well, I dunno. The problems that David poses are usually river problems, because those are the ones that he can solve.

Jerrod



I have posed several problems that were not river problems. I have posed several problems that I did not think I knew how to solve without some studying. I chose my problems based on what they could tell us about poker and how likely it seemed to me they would get answered on this forum.

I think that most people realize that he who can find simple logical solutions to problems that journeyman mathmeticians solve with more complex math (without realizing the existence of the simpler solution), would also likely come up with easier and faster solutions to the difficult problems that those same journeyman struggle with (if he ever chose to familiarize themselves with the subject). My father was better at calculus than at least 99.99% of Phd's. But in his Logic course he taught ingeniously simple ways to do these problems without calculus. I'm quite sure that all great mathmeticians and scientists have this capacity. Not something that can be gained merely by learning a bunch of glorified accounting techniques.

[Jerrod Ankenman]

[ QUOTE ]
I think that most people realize that he who can find simple logical solutions to problems that journeyman mathmeticians solve with more complex math (without realizing the existence of the simpler solution), would also likely come up with easier and faster solutions to the difficult problems that those same journeyman struggle with (if he ever chose to familiarize themselves with the subject).

[/ QUOTE ]

You don't have to convince me; we never could have solved the infinite-bet [0,1] game with check-raise by brute force, nor could we have solved the no-limit case without solving these things logically first.

But let's face it, these problems #3/#4 are just do-in-your-head things.

Jerrod

[Jerrod Ankenman]

[ QUOTE ]

If you looked at the assumption made in my previous post

[ QUOTE ]
If A's strategy is:
[0,x] : raise(bluff)
[x,y] : fold
[y,z] : call
[z,1] : raise
And B's strategy is:
[0,a] : fold if raised
[a,1] : call if raised,
where 0<x<y<a<z<1

[/ QUOTE ]

you see that it doesn't matter which hands in [0,y] will be bluffed with,
since z>y anyway!

But if you don't assume A would have thought of that, this would not even
necesserily be B's (co-)optimal strategy I think.

Next Time.

[/ QUOTE ]
Actually, it would. The only requirement that a strategy pair be optimal is that neither player can unilaterally increase his equity by changing strategy. So your solution is in fact optimal.

What's also true is that all strategies where B bluffs the right number of hands in the range below calling are all co-optimal.

The fact that a strategy <A> is dominated doesn't preclude it from being optimal; it simply means that there exists another strategy <A'> which performs better than or equal to <A> against all counterstrategies. The strategy where you bluff [0,z] does worse against some suboptimal strategies from A (like where he calls with some bad hands) and is dominated by the strategy where B raise-bluffs his best folding hands rather than his worst.

Jerrod

[Aisthesis]

Ok, we've seen that B's strategy of calling at above 1/2 and raising at over 3/4 is not optimal. But what is A's best counter to this strategy?

On the one hand, if A ALWAYS calls the raise when he has a hand 3/4 and above, it doesn't change B's EV at all over and above the non-raising scenario: A folds to the raise below 3/4 and loses only the $1 he would have lost anyway. But if both players have hands above 3/4, it's a wash with 0 EV for B. So, the value of the game is just .25 for B, as it is in the non-raising game.

This just seems a little strange to me, because A is only getting 3:1 pot odds on the call. So, it would in a sense seem that if he had the hand .75000000001, he doesn't have the odds to call. B will win his additional dollar 99.99999999% of the time?!

So, if A calls the raise on on the top 2/3 of the hands that B will raise (calls raise at 5/6), what happens? I get that he actually REDUCES B's EV from the original scenario by 1/48 to 1/4-1/48 = 11/48.

Is this correct?

[Bozeman]

I showed above that for (bad) strategies where B calls with > 1/2 and raises with >b, and A uses optimal response, the value is

"1/4-1/8*(1-b)^2"

This leads to 31/128 (=1/4 - 1/128) for b=3/4. The error you made is that A needs to call with 3/4 (not 2/3) as many hands as B raises getting 3:1 odds. If A calls with (less than optimal) top 1/6 of hands, B's EV is 1/4 -1/144. (the correction is 1/3 of what you computed).

Craig

[Aisthesis]

Yep, checked it through, and you're right. Thanks!

It's rather interesting that the ONLY thing making the raise profitable at all is the bluff. If B never bluffs, then A can always find a (tight) calling criterion to make it in fact unprofitable for B to raise (as opposed to the just calling with 1/2 or better and never raising). So, if B never bluffs, A can successfully punish B for raising.

On the other hand, it doesn't take very much bluffing to force A into dramatically looser calls, hence making the raise profitable for B. Moreover, the bluffing criteria and raising criteria for B always force A into an exact call point. It would be kind of interesting to figure out (given B's optimal value-raise threshold) just what the correlation is to A's call treshold if B bluffs more or less than is optimal.

On the other hand, while B's optimal values force A into an exact calling/folding strategy, it doesn't seem to me that this applies the other way around: If A didn't know that B was switching strategies and hence kept calling the raise on the top 1/3 of his hands, instead of bluffing, B could increase his winnings by value-betting only.

I guess what I'm basically saying is: B's optimal strategy forces A into exactly one optimal betting strategy. If B sticks with his optimal strategy and A changes his, A will just lose more money.

But A's optimal strategy by no means forces B into his true optimal strategy. It only does so under the assumption that A will adjust his strategy if B changes his. If A is incapable of adjusting to changes in B's strategy, then B can also capitalize on his inflexibility.

Or yet another way of putting it: A has no way to punish B for inflexibly sticking with the optimal (bluffing) strategy, but B can punish A for inflexibility.

[well]

[ QUOTE ]

Two players, A and B receive one card with a real number between zero and one. High card wins. Player A bets one dollar in the blind. Player B can fold, call, or raise one dollar. If he raises, A can call or fold only. [...]

The second version of this question makes A's blind live. So if B just calls, A can raise if he chooses. Game over.

[/ QUOTE ]

Here's what I found:

Player A
[0,1/12] : raise
[1/12,3/4] : check
[3/4,1] : raise, when called

[0,2/3] : fold
[2/3,1] : call, when raised

Player B
[0,19/36] : raise
[19/36,7/12] : check
[7/12,2/3] : call, fold when raised
[2/3,3/4] : call, call when raised
[3/4,1] : raise

With this, the value of the game for player B is 17/72

Could you please check this solution?

Next Time.

[Bozeman]

Already been checked.

Jerrod and I got the same strategies, and you and I got the same value, Jerrod apparently making an arithmetic mistake.

Craig