samsloan <samhsloan@[EMAIL PROTECTED]
> writes:
> On Mar 25, 10:18 am, torb...@[EMAIL PROTECTED]
(Torben Ęgidius Mogensen)
> wrote:
>> samsloan <samhsl...@[EMAIL PROTECTED]
> writes:
>> > One factor to be considered is that the number of possible moves in a
>> > backgammon games is infinite. The players could easily just keeping
>> > hitting each other to infinity.
>>
>> That doesn't matter, as long as the number of possible board positions
>> is finite (which it is).
>> [...]
>> This can be translated into a set of equations that you can solve to
>> find the probability of each possible position being winning or
>> losing. The set of equations is huge, but finite.
>>
>> Torben
>
> Even that is not obvious. There are 21 possible rolls of the dice (6!
> = 21)
6! is 720, actually. But you are right that the number of different
rolls is 21 = 6*7/2. This is still finite, though.
> and three possible positions of the doubling cube
I can't see how this would affect the winning probability.
> plus 24 possible slots for each checker.
You forgot the bar and home, so there are 26 possible positions. But
since the checkers are not distinct, and since black and white pieces
can't coexist (except on the bar and in the home), you get a lot fewer
than the 30^26 different positions you imply.
In any case, my point was that the number of positions is finite (but
huge), so arguing that there are many possible rolls and positions of
doubling cubes and pieces doesn't change that, unless you can show
something is infinute.
The doubling cube is normally limited to 7 possible positions (absent
or 2, 4, ..., 64), but even if you allow unbounded doubling, this
doesn't change the probability of winning.
Torben
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