On Mar 26, 4:49*am, (Torben Ægidius Mogensen)
wrote:

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- Hide quoted text -

- Show quoted text -

If we agree that owning a 2-cube has the same theoretical meaning at
owning a 4-cube or 16-cube (or whatever level), how do you get 7
possible cube positions? I count only three: centered, owned by me,
or owned by the opponent.

--
Gregg C.