"Worst" possible beaverable position (long)
Christopher Yep
The position where both players have one checker remaining on their 6
pts., with the player on roll having access to the cube, is a very special
position: it is a double and an optional take/pass, even though the second
player only has a 3/16 = 18.75% cubeless winning probability. In fact
for gammon-less positions, this represents the theoretical smallest
cubeless winning percentage in which a player may still have a take. The
reason for this is that player 2 wins exactly 25% of the time with the cube
(if player 1 rolls less than 6, than player 2 can double player 1 out of the
game... actually this pass is optional, although the equity remains the
same whether the redouble is accepted or dropped - see below), and
his redouble if player 1 rolls less than 6 is also perfect in the sense
that it is made with the lowest possible winning percentage (player 2 now
has a 75% winning chance) that will force player 1 to pass (optionally
player 1 can take the redouble for the same equity).
Below is a proof that was given to me by bobk.
--
From: ko...@orie.cornell.edu (Robert Koca)
Subject: .8125 proof
Preliminary lemmas
Lemma 1) Suppose playing continuous game (in sense of Zadeh)
and opponent is not allowed to double but you are. Then should
double when attain probability .75 of winning.
Proof. Easy. Equilibrate opponent take/drop equity.
Lemma 2) Suppose are playing backgammon and opponent is not allowed
to double but you are. Suppose Pr(winning cubeless)=p. If given choice
it is at least as good to start a continuous game starting with Prob(winning
cubeless)=p.
Proof. With condition that opponent can never double added, Zadeh's
arguments now make sense.
Proof of theorem. Suppose in a bg game, Pr(A wins cubeless)=.8125 and
there are no gammon or backgammon chances.
Suppose A offers a double, and pledges that he will never reredouble.
Then B would do at least as well in a continuous game starting with
.1875 cubeless chance. Since A is nice he offers to switch to such a
game also.
Then payoff to B to accepting after all these "favors" from A
equals +2 * Pr(B reaches .75 cubelss chance) - 2*Pr( B doesn't reach
.75 cubeless chance).
Pr(B reaches .75 cubeless)= .1875/(.5625+.1875)=.25.
Thus take equity = 2*(.25)-2*(.75)=-1.
If started with A's cubeless chances > .8125, then calculations give B's
take equity < -1 which implies not a take in original (not more favorable)
bg game.
It is interesting to note that 6 away 6 away attains this bound.
Pretty sure this is airtight, Bob Koca
--
I would like to go farther though. In gammon-less positions, the
theoretical lowest (cubeless) winning percentage that a player MAY still
have and still have a correct beaver is 37.5% by similar reasoning as
above. However I am unable to find such a position. Three questions:
1. Does such a position exist?
2. If no such position is known to exist, then what (currently) is known
as the "worst" possible gammon-less position that is still a correct
beaver.
Here "worst" refers to the cubeless winning position of the player
contemplating a beaver.
3. Same as 2, but now also considering positions involving gammons.
Here, "worst" refers to the equity of the player contemplating a beaver,
*IF* the cube were to be taken out of action after the beaver. In these
positions, for consistency, assume that one of the players has just been
doubled to 2, and is contemplating beavering to 4.
A few more things:
1. (Of course) I am referring to money play here.
2. By "gammons," I really mean "gammons AND backgammons"
3. Although the initial double (before the beaver) will in almost all
(ALL for #2) cases be incorrect, assume that after this, both players have
perfect checker and cube play.
4. In #2, if the position is simple enough, the exact cubeless winning
percentage may be determined exactly. However for #3, with positions in
which gammons are possible, it would seem unlikely that the equities could
be determined exactly - in this case a rollout program would probably have
to be used.
Bob Johnson x6152
>The position where both players have one checker remaining on their 6
>pts., with the player on roll having access to the cube, is a very special
>position: it is a double and an optional take/pass, even though the second
Great article! I was confused, so others may be too.
If the 1st player doubles, then the game boils down to 1 roll; if the
1st player fails to roll 6 or better, then the 2nd player "wins" (or the
equivalent of a win in terms of equity). So, the 2nd player has a
25% chance of winning the game. If the cube were not doubled (which
should not happen) then the 2nd player would stand at only 18.75% to win
(75% of 25%).
The other sections are worth reading too. Thank you Bob Koca!
----- Robert D. Johnson rjoh...@cvbnet.cv.com