Tough Game Theory Problem

[Torgen]

The situation you describe involves bluffing. It was my explicit intent to make a simple strategy based only on the actual strength of respective hands, and then expand it to include both betting a weak hand and checking a strong hand, which would be closer to the optimal strategy (although not necessarily optimal). You can eliminate the iterative process by bluffing a random portion of the time--for example, if it was optimal to bluff 20% as often as your raise is legitimate (I'm not saying that this figure is accurate; it is merely illustrative), you could use the insignificant digits of your 'card' to decide. For example, you would bluff with 0.40~, or 0.41~, or 0.01~, but not with 0.42~. But this would have complicated the first post greatly, and I haven't even gotten around to the simple river strategy yet.

[waffle]

Do you know the answer to the version of this problem with only one street? Is that problem worth exploring as a precursor to this one?

[Cerril]

It's occasionally amusing, as someone who considers himself fairly smart and fairly 'into' mathematics, to look at a message like this one and have absolutely zero clue whether or not it's coherent or gibberish. Heck, even if those words go together to make statements someone up on things would understand, I'm still not sure if they'd go 'uh, no' or 'hmm, that's an interesting point.'

Obviously I'm not insulting your post or your language, it's just funny to realize how out of my depth I am.

[CK1]

Hi Waffle:

The solution for the one street problem is known. I believe the answer is:

The first player should raise when his/her cards are between 0 and 1/8, call when his/her cards are between 1/8 and 5/8, and raise when his/her cards are between 5/8 and 1.

If raised the second player should only call if his/her cards are between 1/4 and 1.

If the first player calls, the second player should raise when his/her cards are between 0 and 11/48, should call when his/her cards are between 11/48 and 5/16, and raise when his/her cards are between 5/16 and 1.

If the second plyer raises, the first player should call 75% of the time
(1/4 to 5/8).

Note that you can get these fractions without writing long equations. Someone should double check my numbers.

CK

[CK1]

I have thought about a more general version of this problem sometime ago. Let us assume that the second card we draw is in between 0 and w. If w=1 this is the same as the problem that David posed. Now two limiting cases of this problem (small and large w) are easy to analyze. For small w (w<<1), we can ignore the second card and I think we can get an exact solution.

The case of large w is more interesting. Now if w is large neither player should fold no matter what in the first round of betting. Thus, the solution in this case for the first round of betting is also simple:
The first player raises if his card is in between 1/2 and 1 and the second player must call independent of what he or she has. There is no bluffing.

If the first player checks, the second player should raise if his/her card is in between 1/4 and 1. Again there is no bluffing. the first player must call independent of what he has.

The second round of betting can be exactly analyzed since the problem reduces to known [0,1] game problems.

I have originally thought about this problem to compare preflop strategies for omaha and texas holdem. Of course w is much larger for Omaha than the w for texas holdem. So one should conclude that if you are playing omaha heads up, you should NEVER bluff preflop and you should play more or less all hands.

Also using this analysis you can show why a good poker player should not play no-limit omaha. because weak players can take away your edge by going all-in preflop. In which case all the complexity of the game disappears.

CK

[jason1990]

I am surprised that the optimal solution of the one-street problem is deterministic. I would have expected it be random. (If your number is X, check with probability f(X), bet with probability g(X), etc.) If there's a proof that doesn't involve long formulas, can you post it? If not, can you provide a link to a proof?

[CK1]

Hi jason:

Similar one-street [0,1] problems were discussed here about 6 months ago. Mainly game theory gives us the optimum bluff fractions. For example, first player must raise one third of the time as a bluff. the second player must call 75% of the time. That is why the second player calls the raise when he/she holds hands that are better than 1/4. So the first player must value bet for 5/8 to 1. Since the first player value bets the top 3/8 of the hands. He or she must bluff the bottom 1/8 of the hands.

you can extend this type of reasoning to all river bets.

CK

[pzhon]

[ QUOTE ]

The case of large w is more interesting. Now if w is large neither player should fold no matter what in the first round of betting. Thus, the solution in this case for the first round of betting is also simple:
The first player raises if his card is in between 1/2 and 1 and the second player must call independent of what he or she has. There is no bluffing.

[/ QUOTE ]
I find the suggestion that the first player should raise half of the time quite surprising. Are you sure? If the position on the second street were decided randomly, I would agree, but the first player has a predictable positional disadvantage.

[KidPokerX]

[ QUOTE ]
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.

[/ QUOTE ]
Hmmm... neither can I

[ QUOTE ]
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.

[/ QUOTE ]

Is there something you wish to prove?

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


[/ QUOTE ]
I'm no mathematician, but is that even a legitimate formula?

Faro, my advice for you is to cut back on the math and balance yourself with some english. Best of luck, my friend.

-KidPokerX

[CK1]

[ QUOTE ]
I find the suggestion that the first player should raise half of the time quite surprising. Are you sure?

[/ QUOTE ]

Hi pzhon:

It is just a value bet. You know the second player will call no matter what, so the first player must raise half of the time. For large w, when you get your second card, your previous bet is almost irrelevant. now you have to reevaluate where you are, and thus, bluff raise, call, or value raise accordingly.

CK