Some interesting takepoints
Stig Eide
When your opponent is 3 or 4 away, and doubles you, it is important to
consider cubeownership. Thats because you will, if you take, own the
last chance to redouble for the rest of the game.
I'll take 5 away-3 away as an example. If your opponent doubles you, you
have to think: When can I redouble efficiently?
Lets assume your opponents probability of winning the game is p.
If he drops your redouble, the score is 3 away-3 away.
If he takes and wins, he has won the match,
and if he takes and loses, the score is 1 away-3 away.
This gives your opponent these equitys:
Drop=50%
Take & win=100%
Take & lose=25%
He should take if the expected equity is greater than the equity if he
drops (50%):
50<p*100 + (1-p)*25 solving for p gives
p>1/3
You can redouble your opponent efficiently when your chanses is 2/3.
The problem is now, how often will you be able to redouble your opponent
efficiently?
This can best be shown with the first graph in the history of r.g.b:
|
|
(2/3) |-----------/-------
| /
| _ /~~
(1-p) x/ \ /
| \_/
|
0 +------------------>time/# of rolls
Indicates a possible development of your probability of winning 8-)
If you at a given stage in the game have (1-p) probability of winning,
the probability that you reach the point when you can double efficiently,
before you lose, is (1-p)/(2/3). Hmm, don't think I can prove this...
Since it doesn't matter weather your opponent takes or drop when the
redouble is efficient, we can assume that he drops.
When is it better to take at 5 away-3 away than to drop?
Roy Friedman's equity table in "World class backgammon..." gives:
Drop (5 away-2 away)=28%
Take & lose (5 away-1 away)=17%
Take & redouble eff. (3 away-3 away)=50%
You should take if the expected equity is greater than the equity if you
drop (28%):
28<50*(1-p)/(2/3) + 17*(1-(1-p)/(2/3)) solving for p:
p<0.78
You should take a double at 3 away-5 away if your opponent has less than
78% chances of winning the game. This is far higher than the 67% you get
if you don't add any value for cubeownership.
Of course, its wrong to assume efficient redoubles, so the takepoint
should be somewhere between 67% and 78%, depending on the boardposition.
Here is a table of takepoints -and the redoublepoints. Both seen from
the doublers point of wiev:
3 4 Opponents away
-----------
3 |76-75|82-80|
4 |76-59|82-64|
5 |78-67|78-70|
Yours away 6 |76-75|77-78|
7 |78-74|75-77|
8 |79-70|77-74|
9 |79-70|78-74|
-----------
At (5,3), the table says 78-67, which means that your takepoint, when
you are 5 away and your opponent 3 away, is 78%. And your opponents
takepoint for the redouble is 67%.
General solution for takepoints:
p=(a*(b+d-c)-b*d)/(b*(a-d))
a is your equity if you win 2 points
b is your equity if you win 4 points
c is your equity if you lose 1 point
d is your equity if you lose 2 points
General solution for redoublepoints:
p=a/b
Stig Eide, stig...@avh.unit.no
Kit Woolsey
: When is it better to take at 5 away-3 away than to drop?
: Roy Friedman's equity table in "World class backgammon..." gives:
: Drop (5 away-2 away)=28%
: Take & lose (5 away-1 away)=17%
: Take & redouble eff. (3 away-3 away)=50%
(I cut out most of the analysis above and below this)
Whatever else you do with match equity analysis, please don't use Roy
Friedman's tables. They are just plain wrong. They are based on the
incredible supposition that, when one player needs a gammon and the other
doesn't (such as at Crawford game with trailer having an even number of
points to go and post-Crawford with trailer having at least 3 points to
go) that 35% of the trailer's wins will be gammons! Friedman claims to
have rolled out over 3000 games to come up with this result. This is an
incredible assumption which is far from everybody else's estimate. My
personal estimate of 21 or 22% of the trailer's wins being gammons is
much more accurate, pretty well agreed upon by most good players, and
confirmed by study of over 1000 matches in Hal Heinrich's data base. I
have discussed all this in an article I wrote in Inside Backgammon a
couple of years ago. For the record, below is the match equity table
which I have developed, based upon empirical data from Heinrich's data
base and a computer program I wrote which properly computes the equities
based upon this empirical data. The numbers on top represent the number
of points the leader has to go; the numbers on the side the number of
points the trailer has to go, and the numbers in the matrix represent the
percentage of time the leader can be expected to win. When the leader
has 1 point to go, it is assumed that the Crawford game has not yet been
played.
1 2 3 4 5 6 7
1 50 70 75 83 85 90 91
2 30 50 60 68 75 81 85
3 25 40 50 59 66 71 76
4 17 32 41 50 58 64 70
5 15 25 34 42 50 57 63
6 10 19 29 36 43 50 56
7 9 15 24 30 37 44 50
For example, the leader's advantage at 5 away, 2 away is 75%, not the 72%
that Friedman's tables would have you believe.
Kit Woolsey