Re: poker

David C. Ullrich wrote:
On 7 Feb 2006 12:42:04 -0800, matt271829-news@xxxxxxxxxxx wrote:

David C. Ullrich wrote:
On 7 Feb 2006 10:28:28 -0800, matt271829-news@xxxxxxxxxxx wrote:
[...]

, or
do you not see how proving (1) and (2) shows that he's done,
assuming he actually has proved them?

Correct. This is *exactly* what I don't understand. I'll reply to
Robert's follow-up and explain there why I don't get it, and hopefully
you can check it out there and somewhere between the two help me
understand why it works.

Have you read both (1) and (2) carefully? The result depends
on both of them. If A does this then no matter what B does,
A makes at least 1/10 point per game. If B does that then
no matter what A does, A makes no more than 1/10 per game.
So this and that are optimal for A and B, respectively.

I don't see how you can agree with what you say you agree
with but not agree that these strategies are optimal.
A _is_ making at least 1/10 per game, regardless of what B
does, and there's no strategy for A that makes him more than
1/10 per game (regardless of what B does). What more would
an "optimal strategy" do?


Right. I think I might have finally got my head around this.

1. We exhibit a strategy for B such that A's best counter-strategy
gives A an expectation of 1/10.

2. Therefore A's best strategy cannot give him an expectation greater
than 1/10.

3. We exhibit a strategy for A such that B's best counter-strategy does
indeed give A an expectation of 1/10. This strategy must be optimal for
A, because none can be better (or, strictly, it must be one of possibly
many "equally optimal" strategies for A).

Does that sound about right?

Yes.

(Of course now that you understand the outline of what's being
claimed, how what he said could possibly be right, you have
to go back and verify those proofs - the ones where previously
you said you didn't get past the first line because you didn't
see this or that. Someone needs to keep the guy honest, heh-heh.)

Heh-heh indeed. I already established to my own satisfaction that if A
bets with < 0.1 or > 0.7 and B plays optimally then A's expected payoff
is 0.1, so it just remains to show the second part - if B calls with >
0.4 then A can't do any better than 0.1. I worked this out a different
way actually, dispensing with the probability function which seems
irrelevant (you just need to consider for a given number whether it is
better for A to bet or not bet?), but no matter because I get the same
answer: A's best option is to bet with < 0.1 or > 0.7, which gives the
0.1 payoff. QED.

Thanks for your help.

.

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