Re: A Poker Problem (Heads Up)

>Suppose that you and your opponent both decide to reraise
>until the other folds or calls. How much do you win?
That's not a good strategy if you don't have the nuts. And you can't
both have the nuts.

>Well changing the numbers from 1,000,000 and 999,999 to 5 and 4
>certainly doesn't change whether the problem makes sense.
My point exactly. I was just trying to convince some of you that the
problem makes sense if you think of the 5 card case.

>Do you have a solution for the simplified problem?
Not yet, but I am working on it.

If you want a precise description of the problem, I think I can give it
a try.

A /strategy/ is a mapping that assigns to each possible game situation
a probability distribution of actions (raise, call, fold). A game
situation is given by the history of actions so far and the card in
front of you.

If the strategy S of the opponent is known, one can use Bayes' formula
to deduce as much information as can be deduced about what the opponent
is likely to have, given his actions. Using this information, and
knowing exactly what the opponent is going to do in the future if he
has a particular hand, one can build a strategy that maximizes profit
against a given strategy. Let's call the expected profit
Exploitability(S).

There is probably only one strategy that minimizes the function
Exploitability. Let's call that strategy the perfect strategy.

Now, in the perfect strategy, how many times would you raise if you
have the second best hand? (the answer can be a probability
distribution, not a number)

As for your question of what happens if you raise all the time with the
second best hand, the exploitability of such a strategy is +infinity,
since an arbitrary amount of money can be extracted from it in the
average.

.

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