Proof of doubling with market losers
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> > Against a perfect opponent, double at the first sign of a market loser.
>
> I've heard that put forward, but without what I would consider a proof.
> Does "the first sign of a market loser" include a 1295 to 1 shot? If
> not, where do you draw the line, and how do you prove that that's where
> the line should be drawn?
>
> If there is a simple rule, I don't think there's a simple proof of the
> rule!
I believe there is a simple proof of a slightly more general proposition:
against an opponent who handles the cube perfectly, the strategy of
[doubling if there is any chance of losing your market] is optimal.
Proof in two stages:
First, if your opponent handles the cube perfectly, then your chance of
winning the 2 point match is no greater than your chance of winning a 1
point match. The reason is that your opponent can double at his first
opportunity, making your match winning probability equal to your chance of
winning a 1 point match.
Second, by doubling before you lose your market you ensure that your match
winning probability is *at least* equal to your chance of winning a 1 point
match.
QED
If there's a flaw in this argument, I'd certainly be interested in knowing
what it is :-)
-- Walter Trice
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Douglas Zare writes:
Actually, I'm not sure I agree with the more general proposition. I'd like
to see a definition of "perfect cube handling" and "any chance of losing
your market" first. Should you double if there is a chance that your
opponent's checker misplay will cause you to lose your market, but only
because you are known to play badly while leading Crawford 2-away?
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Walter Trice writes:
Def'n of perfect cube handling: always makes the double and take decisions
that maximize match winning chance.
Any chance of losing your market: yes, that would include losing your
market as a result of opponent's bad checker play. Yes, even if you could
only lose your market because you would play badly in the Crawford game.
> One implicit assumption is that there is no market-losing sequence of two
> initial rolls. That's not a priori obvious, but no reasonable plays seem
> to come close. What about 3-1 8/5 6/5 followed by 4-4 6/2(4)? I'd pass.
Yes, that is an implicit assumption. Whether the assumption is met in all
practical circumstances is a side issue.
I'm still interested in how to get around the simple argument that by
adopting a strategy of doubling before market loss Player B can limit A's
chance to X, while by adopting the same strategy A can ensure that his
chance is X, X being his chance of winning a 1 point match. It seems to me
that this works even with pathological assumptions about checker play,
relative skill, and match equities at the C/2-away scores.
--walter trice
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