Re: A Poker Problem (Heads Up)

On 27 May 2005 10:17:35 -0700, "David M Einstein" <Deinst@xxxxxxxxx>
wrote:

>
>
>David C. Ullrich wrote:
>> On 27 May 2005 06:07:36 -0700, "Álvaro Begué" <alvaro.begue@xxxxxxxxx>
>> wrote:
>>
>> >David, I am perfectly aware of things like "some series don't converge"
>> >or "some random variables don't have means". I don't think such
>> >considerations are very relevant in this context.
>>
>> You're allowed to think what you want.
>>
>> >I see what your
>> >objections are, although somehow I think the focus on the details is
>> >not letting you see the big picture.
>>
>> But you're starting to sound like what's-his-name here - when you
>> don't have anything to say about the mathematics you ignore the
>> question and say things about me instead. At least you're not
>> being quite so insulting, but I'll give you a hint: Comments
>> like the above are not going to convince anyone of anything.
>>
>> >I've been working as an engineer
>> >for too long, I guess. :)
>> >
>> >Let's make the game finite, so we can avoid all these problems. You are
>> >only allowed to raise N times.
>>
>> That's entirely different - nothing I've said in any of this
>> has any relevance to that game, yes there's certainly a
>> well-defined optimal strategy.
>>
>
>Is this obvious?

Yes.

A strategy is a complete specification of what to do in any
situation. Here a typical "situation" is something like
"you hold a 3, and there have been five raises so far",
and "what to do" is not "fold", "call" or "raise", but
rather a probability measure on the set {fold, call,
raise} specifying that in that situation you should
fold a certain percent of the time, call a certain
percent of the time, and raise a certain percent of
the time, at random.

Say T is the simplex consisting of all probability
measures on the set {fold, call, raise} and S is the
set of all situations; then the set of all stratgies
is X = T^S. Note that X is compact, and also note that
S is _finite_.

Now suppose for a second that you follow strategy
x and the other guy follows strategy y. Then there
is a well-defined value of the game E(x,y), the
expected value to you given those two strategies.
Since everything is finite E(x,y) is just a polynomial
in the components of x and y (I think it's clear that
it's bilinear in x and y.) And in particular E(x,y)
depends continuously on (x, y) in X^2.

Since E is continuous and X is compact we can let
g(x) = min_y E(x,y) and then g is continuous.
now g(x) is the worst that can happen if you follow
strategy x; the optimal strategy is the x which
maximizes g(x). This exists because g is continuous.

(And yes, I think it's fair to say that all this
is obvious even though it took a few paragraphs to
explain - there's nothing the least bit non-trivial
or clever in those paragraphs.)

The problem in the case where an unlimited number
of raises is allowed is that E(x,y) is no longer
continuous. In fact it's not even clear to me that
it's well-defined. E(x,y) is formally still a
bilinear function of x and y, but it's an infinite
sum, and it's not at all clear to me that the
sum converges for every x, y (in fact I think
it's clear that it doesn't converge for some
x, y.) In any case it's no longer clear that
E(x,y) depends continuously on x and y: we
can still give X the product topology, so
X is still compact, but the topology is the
topology of pointwise convergence, and the
sum of a limit doesn't have to be the limit
of the sum if we're talking about infinite
sums.


************************

David C. Ullrich
.

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