On Mar 26, 1:45*am, bob wrote:
On Mar 25, 11:18*am, (Torben Ægidius Mogensen)
wrote:





samsloan 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).

The main difference between Backgammon and, say, Checkers is not the
possibility of infinite play but the fact that Backgammon involves
random elements, so few positions are definitely winning or definitely
losing -- all you can say is the probability of winning with perfect
play (i.e., always picking the move that gives you the best winning
probability after moving).

You can solve Backgammon by for each possible position have edges to
every other position that it is possible to get to in one move, and
label each edge with the dice outcome that allow this move).

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

* Agree for backgammon played without a cube, for backgammon played
with a cap on the cube, or for match play. If you are talking about
money backgammon though then the cube position and value makes the
number of positions infinite. Now one might say that its position is
all that matters since if you know the correct theoretical play
holding a 2 cube then you also know the correct theoretical play
holding a 4 or any higher value cube. There is a problem though in
that the equations might not have a solution. As a simple example of
how this might come about suppose we are betting on the flip of a
coin. The first person to toss a head wins. Before each flip a
doubling cube may be used as in backgammon. Solving the equations
gives that every turn is a double and take but this leads to undefined
equities. How to prove there is no situation like that possible in
backgammon?

Bob Koca- Hide quoted text -

- Show quoted text -

Interesting example but I'm surprised by "How to prove there is no
situation like that possible in
backgammon?"

Almost surely, there _are_ situations like that in backgammon. Can't
you mimic the above scenario in backgammon by postulating 18 monster
rolls for each side?
But, suppose there are such situations in backgammon, why does it then
follow that the value of a cube, as well as its position, can affect
theoretical money play? This seems to be a hole in your argument. So
let us assume that your scenario is exactly replicated in backgammon
where there's a stalemate but each side has 18 monster winning rolls,
and no gammons are possible. Please explain why this scenario leads
to the conclusion that the scenario with a 2 cube is essentially
different than the scenario with a 1024 cube.

Paul Epstein