Tough Game Theory Problem

[David Sklansky]

Forgive me if I have posed this before. Again I believe that the solution, if not already out there, is worthy of a thesis. Especially if it is generalized. I don't know the answer.

Two players ante a dollar. They are both dealt a real number between zero and one. First guy checks or bets a dollar. Second guy calls or folds if bet into. If checked to, he checks or bets a dollar. If first guy checks, he calls or folds if bet into. No raises allowed.

If both players check, or if a bet is called, they are both dealt a second card. Same rules apply for a second round of dollar bets or checks. Highest total wins a showdown.

What is the value of the game to the second guy and what is the optimum strategy?

This question was posed only for game theoriticians who might like something to chew on. No other reason.

[TomCollins]

Not going to post an answer, but game theory is just really interesting.

Do you have any books that you recommend reading on game theory?

[Torgen]

Let's call the two phases of betting "Preflop" and "River" for simplicity.

Simple preflop strategy for both players. (Simple = no bluffing)
(Player 1 abbrev. to P1, likewise P2)

If P1's card is greater than 0.5, he has an equity edge and raises.
If less, he checks.
If P1 raises, his card is greater than 0.5, so its average value is 0.75, meaning P2, getting 3-1 pot odds on a call, must win 25% of the time to call. With P1's average river card being a 0.5, P2 must have a 0.5 or better to call. (Call this case 1 for river strategy discussion.) Otherwise he folds.

If P1 checks, his card is below 0.5, for an average value of 0.25, so if P2's card is greater than 0.25, he should raise. Otherwise he should check behind. (Case 2)

If P2 raises, his card is greater than 0.25, for an average value of 0.625, so P1, getting 3-1 pot odds, should call with a 0.375 or better (case 3) and fold otherwise.

Of course this is not optimal, but can be refined later. (River to follow, unless someone else wants to do it?)

[Grisgra]

David, didn't you ask a similar question before, but limited it to just one round of betting instead of two (i.e., one card instead of two)? If so, the answer there would be a useful starting point.

[laja]

This is the book written by my game theory teacher, he is brilliant. It explains things simply and eloquently, as it was the material used to teach the class.
<a href="http://www.bioinformatics.vg/Books/ga/Game_Theory_and_Strategy_0883856379.html" target="_blank">http://www.bioinformatics.vg/Books/ga/Game_Theory_and_Strategy_0883856379.html[/url]</a>

here is another one, this one goes into prisoner's dilemma and the tournament run by Axelrod. Its implications on cooperation and society are fascinating. http://www.amazon.com/exec/obidos/AS...065225-1851825

[TomCollins]

I don't know much about game theory, but I'm sure bluffing would have to be a part of it. It at least should be considered.

[Torgen]

I know. The intent is that bluffing will be added in a refinement of this straightforward system.

[patrick dicaprio]

bluffing has already been analyzed under game theory, and it has been discussed here and in sklansky's books.

Pat

[gergery]

[ QUOTE ]
Let's call the two phases of betting "Preflop" and "River" for simplicity.

Simple preflop strategy for both players. (Simple = no bluffing)
(Player 1 abbrev. to P1, likewise P2)

If P1's card is greater than 0.5, he has an equity edge and raises.
If less, he checks.
If P1 raises, his card is greater than 0.5, so its average value is 0.75, meaning P2, getting 3-1 pot odds on a call, must win 25% of the time to call. With P1's average river card being a 0.5, P2 must have a 0.5 or better to call. (Call this case 1 for river strategy discussion.) Otherwise he folds.

If P1 checks, his card is below 0.5, for an average value of 0.25, so if P2's card is greater than 0.25, he should raise. Otherwise he should check behind. (Case 2)

If P2 raises, his card is greater than 0.25, for an average value of 0.625, so P1, getting 3-1 pot odds, should call with a 0.375 or better (case 3) and fold otherwise.

Of course this is not optimal, but can be refined later. (River to follow, unless someone else wants to do it?)

[/ QUOTE ]


P1 should raise much more often than just if his card is greater than 0.5 -- (do you see why? whoo-hoo, I’ve always wanted to say that. Now I feel important).

For 2 reasons: 1) there is already $2 in the pot, and his bet risks $1 to win $2. And, 2) P2 could fold and he’ll win outright. I’d guess the combination of those two factors means he should bet if his hand is 0.3 or better. P2 will then recognize this, and since he’s now getting 3:1 on a call, will call with anything better than 0.25 or so.

But then on round two, P1 will need to risk $1 to win the $4 in the pot, and p2 could still fold, so he needs to bet if his hand is better than 0.5 (with the two cards combined, or an average of 0.25 per card). P2 will recognize this and call with anything better than 0.4 (since he risks $1 to win $5).

This is just first order thinking. You’d now need to go back and see what adjustments you should make, given that each player will be betting as above. For example, will P2 really call the first bet with better than 0.25? If he puts P1 on 0.3 or better, then he’d assume P1 has an average of 0.65, and P2 getting 3:1 on his money. And 25% of the time he’ll improve vs. P1 by a margin of .15 or better on the second card. So he is getting correct odds if he has 0.65-0.15, or 0.5. So maybe P2 now starts calling with only 0.5 hands or better.

But then P1 will recognize that, and know that half the time his $1 bet will win, so he will start raising with hands worse than 0.3. So P2 will adjust and call with less than 0.5. And so it goes.

--Greg

[Faro]

What a question. It makes me so angry when I can not grasp a 2X2 probability denisty fx(x) solving for the optimal solution by using eigen values and eigenvectors on two or 4 independent random variables assuming they are independent.

Discrete or continuous the probability density function is unity ( 1 ) integrated with X between 0 and 1 is equal to an expected value E[X] of 0.5. Solve for the variance using a gaussian seems wrong but the discrete case approximation can be:

sigma = sqrt(1/N * sum(0 to N)((xi - E[x]) * (xi - E[x])))

and you have a good location of call or bet on first play but this is to simple.

The probability conditional denisites and the 2X2 matrix needs to be solved for the optimal solution.

Do not see if Bernouli fits in here. Do you see the frustration and the stupidity or no methodology on where to start.