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Notice I said equity and not chances of winning it is possible to construct
a position where a player has less than 50 % winning chances, but has a
correct double owing to gammons.
<doubl...@my-deja.com> wrote in message
news:7qb8l9$rk8$1...@nnrp1.deja.com...
>So my buddy Jimmy and I like to have a drink or two when we play BG,
>and we tend to get a little loosey-goosey with the old doubling cube at
>times. Beavers, raccoons, ardvarks,... we've seen them all, and then
>some.
>Well, before I let Jimmy take any more of my money, I thought I'd try
>to ascertain the break point for when to beaver. Obviously, it must be
>somewhere between 25% and 50%. Since there is equity in owning the
>cube, I would not think I need to be all the way to 50%, but where is
>the break point?
>And while we're at it, any thoughts about when to raccoon?
>DoubleThis
>
>Sent via Deja.com http://www.deja.com/
>Share what you know. Learn what you don't.
You have to think you have positive equity, which is not necessarily
indicated by percentage of winning games. Gammons have to be taken
into consideration. If you are more likely to lose a gammon than to
win one, you need better than 50% winning games. If the reverse, then
the reverse.
dk
NIHILIST
<doubl...@my-deja.com> wrote in message
news:7qb8l9$rk8$1...@nnrp1.deja.com...
The "continuous model" gives a take-point of 20%, and a corresponding
beaver-point of 40%. Just as you usually need more than 20% to take, you
usually need more than 40% to beaver. In a short race it depends a lot on
what effect the specific position has on your cube efficiency. It is very
commonly correct to beaver with a 44% or greater cubeless chance of winning.
In a long race the beaver point approaches the theoretical 40%.
Of course it's a bit more complex with gammons.
For raccoons, it is usually true that you should raccoon if your opponent
shouldn't have beavered. But there is a class of exceptions -- consisting of
the Kauder Paradox positions. The sequence double/beaver/raccoon may be
correct (rarely!) And, finally, there are positions which you should beaver
if and only if your opponent can't (or won't) raccoon.
--Walter Trice
doubl...@my-deja.com wrote in message <7qb8l9$rk8$1...@nnrp1.deja.com>...
> [...]
>Of course it's a bit more complex with gammons.
>
>For raccoons, it is usually true that you should raccoon if your opponent
>shouldn't have beavered. But there is a class of exceptions -- consisting of
>the Kauder Paradox positions. The sequence double/beaver/raccoon may be
>correct (rarely!) And, finally, there are positions which you should beaver
>if and only if your opponent can't (or won't) raccoon.
Probably you mean that it is possible for the correct (for both sides) cube
action to be double/beaver (Kauder Paradox positions, Jacoby rule, cube
centered). Also there are positions (again, cube must be centered, Jacoby
rule in effect) in which you should double if and only if your opponent can't
or won't beaver.
If it is correct for X to beaver then X's equity owning the cube must be
greater than (or equal to) zero. If O raccooned, then X's equity owning the
cube is doubled (better choice of words: increased by a factor of two). So O
shouldn't raccoon, unless X's equity owning the cube is zero, in which case
both the beaver and raccoon (and otter, etc.) are "optional." I've yet to run
across a position in which X's equity owning the cube is exactly zero. Do you
know of such a position?
Chris
Correct -- got my wires crossed, I guess, misremembering things I haven't
thought about much for several years. Sorry about that. Kauder paradox
positions are NOT exceptions to the rule "raccoon if and only if your
opponent's beaver was wrong."
>
>If it is correct for X to beaver then X's equity owning the cube must be
>greater than (or equal to) zero. If O raccooned, then X's equity owning
the
>cube is doubled (better choice of words: increased by a factor of two). So
O
>shouldn't raccoon, unless X's equity owning the cube is zero, in which case
>both the beaver and raccoon (and otter, etc.) are "optional." I've yet to
run
>across a position in which X's equity owning the cube is exactly zero. Do
you
>know of such a position?
There is a simple bear-off in which X, on shake, has zero equity, but this
wouldn't fit the bill. I don't know of any position that actually would.
--wt
>
>Chris
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