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Mar 5, 1997, 3:00:00 AM3/5/97

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I think the first posting was garbled so I'll try again

X doubles. Is this a take in a money game ?

+24-23-22-21-20-19+---+18-17-16-15-14-13-+

O

O

X

X

+-1--2--3--4--5--6-+---+7--8--9-10-11-12-+

Mar 5, 1997, 3:00:00 AM3/5/97

to

Yes, after the take X wins 72% at cube 2, then O wins 72% of 28%,

which is about 20%, at cube 4 then X wins the remaining 8% also

at cube 4.

O's equity from taking is

-2*.72 + 4*.20 -4* .8 = -1.44 + .80 -.32 = -.96

--

stein....@fou.telenor.no

... signature funny quote (and more) at http://www.nta.no/brukere/stein

Mar 5, 1997, 3:00:00 AM3/5/97

to

mi...@farnpark.demon.co.uk writes:

>I think the first posting was garbled so I'll try again

>X doubles. Is this a take in a money game ?

>+24-23-22-21-20-19+---+18-17-16-15-14-13-+

> O

> O

> X

> X

>+-1--2--3--4--5--6-+---+7--8--9-10-11-12-+

Yes. If O gets a chance, he will redouble. He has 26 rolls to

bear off, and the equity effect of redoubling is equivalent to

giving him 34 rolls to bear off (this is a useful trick -- you

double the difference from 18, and 26 - 18 = 8, and 18 + 2 * 8

= 34.)

Since X has 10 rolls that miss, O's equity if he takes is the

same as winning 34*10 = 340 times in 1296 games. Since 340 is

greater than 324 (which is 1/4 of 1296), it's a take.

Put another way, if you take your score in 1296 games is:

26*36 2 point losses = -1872

10*26 4 point wins = +1040

10*10 4 point losses = -400

Grand total: = -1232 points, which is not as

bad as the 1296 points you'd lose by dropping.

-- Walter Trice

Mar 6, 1997, 3:00:00 AM3/6/97

to

mi...@farnpark.demon.co.uk writes:

>X doubles. Is this a take in a money game ?

>+24-23-22-21-20-19+-

> O

> O

> X

> X

>+-1--2--3--4--5--6-+-

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:

>+24-23-22-21-20-19+-

> O

> O

>

> X

>+-1--2--3--4--5--6-+-

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:

>+24-23-22-21-20-19+-

> O

> O

> X

> X

>+-1--2--3--4--5--6-+-

Here, O's checkers on both on the 24 point, so this is the last roll of

the game.

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.

>+24-23-22-21-20-19+-

> O

> O

> X

> X

>+-1--2--3--4--5--6-+-

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

4. So:

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

cube ownership.

--

_______________________________________________________

Daniel Murphy | San Francisco | rac...@cityraccoon.com

Monthly tourneys in San Mateo: See www.gammon.com/bgbb/ for details

and some excellently annotated matches. On-line: telnet fibs.com 4321.

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