> X doubles. Is this a take in a money game ?
The answer isn't immediately obvious if you've never seen this position
before, and if you know a little about doubling and taking and odds your
thinking might go like this:
"If X's roll ended the game, this would barely be take, because he wins
right away unless he rolls one ace. But this positon is even worse,
because if X misses O has to roll and can still lose, so O should pass."
Well, actually O should take *even* though O only will win about 20% of
the time. And the reason is interesting enough to go into in some detail.
If any of this is new to you, think it over -- not to memorize it, but
because you will win more if you know some basic facts about how to
calculate odds, when to take or drop, and the concept of cube ownership.
O has two choices: drop or take. If O drops, O loses 1 point (I'm
assuming the cube is still centered). What happens if O takes? We don't
know yet -- O could lose 2 points, or win 2 points (or even lose or win 4
points -- more on that later).
O must ask: "I know I'll lose 1 point if I drop; will I lose more or less
than one point, on average, if I take?"
You know that every time you roll the dice, there are 36 possible
combinations? There are, and it's often useful to think of your chances
of winning as some portion of 36, because if your winning chances are less
than 9/36, you drop -- more than 9/36, you take -- and exactly 9/36 --
doesn't matter (*if* there's no chance of gammon, and no possibility of a
redouble). 9/36, or 25%, is the magic number.
Here's a position where O's chances are exactly 9/36:
This is the last roll of the game. X, on roll, will win exactly 27 times
out of 36 (count all 36 rolls and see).
Suppose X doubles and O drops. In 36 games, O will lose 36 points.
Suppose X doubles and O takes. In 36 games, O will lose 2 points 27 times
( = -54) and win 2 points 9 times ( = 18). And 18-54 = -36, so O will
lose the same amount whether O drops or takes.
Suppose X doubles in this position:
Here, O's checkers on both on the 24 point, so this is the last roll of
X is a big favorite but O must take, because O will win this game exactly
10 out of 36 times (27.77%).
Let's go back to the original position now, which is a little more
difficult because we have to look at 2 rolls, not 1, *and* because if
O gets to roll, O can redouble.
We know that X will win immediately 26/36 of the time. For O to win,
first X must miss (10/36), and then O must *not* miss (26/36).
So we can calculate O's winning chances exactly:
10/36 * 26/36 = 260/1296 = 20.06%
But wait -- that's less than 25%, and didn't I say O must take ?
Yes, O must, and the reason is because by taking O gains possession of the
cube. One advantage of owning the cube is that sometimes you get to give
it back! In this position, *if* O gets to roll, he will be a big favorite
and will redouble and X must take.
So let's factor in the cube and see whether O will lose more or less than
1 point on average by taking.
X wins 2 * 26/36 points immediately (that 2 is the cube). X will miss
10/36, after which O will redouble and be a big favorite with the cube at
X wins 26/36 * 2 + 10/36 * 10/36 * 4 = 1.753 , and
O wins 10/36 * 26/36 * 4 = 0.802 , and
in 36 games, O will lose 0.951 points per game on average -- which is just
a little less than the one point O would have lost by dropping,
Isn't that a lot of math to do over the board? Could be,, but many
players will, or have shortcuts to get a close and reliable answer, or be
able to do it off the board and remember the positions it applied to.
I would suggest you remember this position and the proper cube decision;
try thinking of odds as portions of 36; remember the 9/36 (or 25%, or 3/4)
dividing line between taking and dropping; and think about the value of
Daniel Murphy | San Francisco | firstname.lastname@example.org
Monthly tourneys in San Mateo: See www.gammon.com/bgbb/ for details
and some excellently annotated matches. On-line: telnet fibs.com 4321.