Cube action?
Carsten Muencheberg
Today on GamesGrid I had a though decision to make:
13 14 15 16 17 18 19 20 21 22 23 24
+------------------------------------------+ X: score: 0
| | | X X X |
| | | X X |
| | | X |
| | | |
| | | |
v| |BAR| | 5-point match
| | | |
| | | |
| | | O |
| | | O O |
| | | O O O |
+------------------------------------------+ O: score: 0
12 11 10 9 8 7 6 5 4 3 2 1
BAR: O-0 X-0 OFF: O-9 X-9 Cube: 4 turn: 0
The cube is on 4. Should O double and play for the match?
If I have calculated correctly, it is double/take, but I'm not sure.
Ron Karr
The double should be the easy part. Since the score is tied, the gain
and loss from doubling the stakes are equal. So the "doubling window"
opens at 50%. In other words, if O is a favorite he can consider
doubling. In addition, the cube will be dead, so X gets no value from
owning it. So if he's a favorite and there are any market losers, O
should double. O is a big favorite, and clearly there are market
losers; after big doubles X has no chance.
The big question is: is it a take? If X drops his winning chances are
15% down 4-0, so he should take with 15%, assuming equal players. A
classic 3-roll position (all 6 checkers on the 1 point for both players)
gives X about 22% winning chances. A 2-roll position gives around 14%.
This position is better than a 3-roll position for O, since he's further
advanced than X. But it's probably not as good as a 2-roll position, so
it's likely to be a take.
Jellyfish says: X has 19% chances here, so take.
Ron
Donald Kahn
Carsten Muencheberg <c...@berlin.snafu.de> wrote:
>Today on GamesGrid I had a though decision to make:
> 13 14 15 16 17 18 19 20 21 22 23 24
> +------------------------------------------+ X: score: 0
> | | | X X X |
> | | | X X |
> | | | X |
> | | | |
> | | | |
> v| |BAR| | 5-point match
> | | | |
> | | | |
> | | | O |
> | | | O O |
> | | | O O O |
> +------------------------------------------+ O: score: 0
> 12 11 10 9 8 7 6 5 4 3 2 1
> BAR: O-0 X-0 OFF: O-9 X-9 Cube: 4 turn: 0
>The cube is on 4. Should O double and play for the match?
>If I have calculated correctly, it is double/take, but I'm not sure.
Correct. If X passes, he has 15% chance to win the match. By taking
he has 18.6 %.
O doubles of course. 81.4% is greater than (.814 x .85) + (.186 x
15)
DK
Chuck Bower
In article <32D84F...@berlin.snafu.de>,
Carsten Muencheberg <c...@berlin.snafu.de> wrote:
>Today on GamesGrid I had a though decision to make:
>
> 13 14 15 16 17 18 19 20 21 22 23 24
> +------------------------------------------+ X: score: 0
> | | | X X X |
> | | | X X |
> | | | X |
> | | | |
> | | | |
> v| |BAR| | 5-point match
> | | | |
> | | | |
> | | | O |
> | | | O O |
> | | | O O O |
> +------------------------------------------+ O: score: 0
> 12 11 10 9 8 7 6 5 4 3 2 1
>
> BAR: O-0 X-0 OFF: O-9 X-9 Cube: 4 turn: 0
>
>The cube is on 4. Should O double and play for the match?
>If I have calculated correctly, it is double/take, but I'm not sure.
A nice bearoff problem, and one which Thorp does a good job with.
Start with doubling window: tie score ==> window opens at 50%. If
X drops, s/he has about 15% chance of winning the match (and a take
is for the match--BOTH sides). Thus the doubling window is:
0.50 < W < 0.85 (where "W" is O's winning chances).
Now we apply the Thorp count to find O's winning chances in this
game IF rolled to the end:
1) Pip count difference = 19 - 13 = 6.
2) # points covered (both have 3 points covered) difference = 0
3) # checkers remaining (both sides have 6) difference = 0
4) # checkers on 1-point (O has 3, X has 0) difference = 3
Race lead is GOOD for O ==> positive
Checkers on 1-point is BAD for O ==> negative
So we can now calculate O's winning chances (in percentage):
w = 74 + 2*(6 - 3) = 80
This is in the window, and not far from a drop (W > 85%). The game
is quite near the end (3-4 rolls) and, without going into the 1296
cases that get O to his/her NEXT doubling opportunity, it looks
like there are a lot of market losers. So DOUBLE-TAKE.
Larry Strommen's Backgammon Position Analyzer gives O 81.0%
winning chances, so the Thorp method was quite close and DOUBLE-
TAKE still looks like the correct action for both sides.
Chuck
bo...@bigbang.astro.indiana.edu
c_ray on FIBS
Dan McCullam
On 13 Jan 1997, Chuck Bower wrote:
<position snipped>
> Race lead is GOOD for O ==> positive
> Checkers on 1-point is BAD for O ==> negative
>
Why is it bad to have checkers on the 1-point? Over three rolls you would
expect to roll at least one 1. O cuold then still bear a man off, whereas
X wouldn't be able to in a similar situation.
Is it to balance out the importance of a lead in the race?
-----------------------------------------
Dan McCullam McCu...@Uni-Hohenheim.de
DMc on FIBS
Chuck Bower
In article <Pine.A32.3.94.970114142239.23124A-
100...@rs56.ws1.uni-hohenheim.de>,
Dan McCullam <mccu...@uni-hohenheim.de> wrote:
>On 13 Jan 1997, Chuck Bower wrote:
>
><position snipped>
>
>> Race lead is GOOD for O ==> positive
>> Checkers on 1-point is BAD for O ==> negative
>>
>
>Why is it bad to have checkers on the 1-point? Over three rolls you would
>expect to roll at least one 1. O cuold then still bear a man off, whereas
>X wouldn't be able to in a similar situation.
>
>Is it to balance out the importance of a lead in the race?
>
Dan, you are correct that checkers on the 1-pt are not necessarily
bad. When Thorp derived his now famous (?) Thorp Count, he found that
in general, checkers on the 1-pt tended to be overvalued (that is,
counting 1 for each pip PLUS 2 for each checker wasn't enough, so
he added 1 for each checker on the 1-pt).
The Thorp Count is a simple method which attempts to treat all
non-contact bearoff positions. It does a good job, in general. However
there are exceptions. I have found that if a person makes "subjective"
changes to a method such as Thorp's, then poor results tend to occur
more frequently. For example, suppose you look at a position and
decide "well, these checkers on the 1-pt are really OK, so I won't
penalize for them..." This is what I call "subjective changes".
For more info on the Thorp Count, I recommend reading Robertie's
"Advanced Backgammon".
Brian Sheppard
Dan McCullam <mccu...@uni-hohenheim.de> wrote in article
<Pine.A32.3.94.970114...@rs56.ws1.uni-hohenheim.de>...
> On 13 Jan 1997, Chuck Bower wrote:
> > Race lead is GOOD for O ==> positive
> > Checkers on 1-point is BAD for O ==> negative
>
> Why is it bad to have checkers on the 1-point? Over three rolls you would
> expect to roll at least one 1. O cuold then still bear a man off, whereas
> X wouldn't be able to in a similar situation.
>
> Is it to balance out the importance of a lead in the race?
Exactly: it is to balance out the importance of a lead in the race.
The men on the 1-point aren't bad in themselves; certainly you
would rather have a man on the 1-point than on the 6-point! But,
note that having a man on the 6-point has a higher pip-count than
having a man on the 1-point, so that comparison isn't exactly fair.
The problem with men on the 1-point is that they require a
disproportionately large number of pips to bear off. In practice, men
on the 1-point are often borne off using a 2, 3, 4, 5, or 6, which
means that pips were wasted. To compensate for this, we adjust the
pip count by adding a pip for each man on the 1-point. The extra pip
accounts for the wastage.
Brian
Julian
In article <5bds6t$5...@dismay.ucs.indiana.edu>, Chuck Bower <bower@bigba
ng.astro.indiana.edu> writes
>>
>> 13 14 15 16 17 18 19 20 21 22 23 24
>> +------------------------------------------+ X: score: 0
>> | | | X X X |
>> | | | X X |
>> | | | X |
>> | | | |
>> | | | |
>> v| |BAR| | 5-point match
>> | | | |
>> | | | |
>> | | | O |
>> | | | O O |
>> | | | O O O |
>> +------------------------------------------+ O: score: 0
>> 12 11 10 9 8 7 6 5 4 3 2 1
>>
>> BAR: O-0 X-0 OFF: O-9 X-9 Cube: 4 turn: 0
>>
>
> w = 74 + 2*(6 - 3) = 80
>
>This is in the window, and not far from a drop (W > 85%). The game
>is quite near the end (3-4 rolls) and, without going into the 1296
>cases that get O to his/her NEXT doubling opportunity, it looks
>like there are a lot of market losers. So DOUBLE-TAKE.
I would have expected the Thorp count to be a bit generous to X, with
only a couple of rolls left. Certainly my instinct would be to drop as
X; a bit of ad-hoc reasoning says even 33 doesn't gain him a roll, so if
you cancel off O's flunks with a 2 against X's flunks with a 1 or two
consecutive 2s, X is in about a three roll position with only half his
doubles working; so 11% or so. I'm quite willing to accept this is
inaccurate - anyone got one of those exact bearoff evaluators handy?
---------------------------------------------------------------------------
Julian Hayward 'Booles' on FIBS jul...@ratbag.demon.co.uk
+44-1344-640656 http://www.ratbag.demon.co.uk/
---------------------------------------------------------------------------
"Any man can be 62, but it takes a bus to be 62A"
- Spike Milligan
---------------------------------------------------------------------------
Julian
In article <Pine.A32.3.94.970114...@rs56.ws1.uni-
hohenheim.de>, Dan McCullam <mccu...@uni-hohenheim.de> writes
>
>> Race lead is GOOD for O ==> positive
>> Checkers on 1-point is BAD for O ==> negative
>>
>Why is it bad to have checkers on the 1-point? Over three rolls you would
>expect to roll at least one 1. O cuold then still bear a man off, whereas
>X wouldn't be able to in a similar situation.
Three men on the one point has the same pip count as one man on the
three point, but the former can't always be borne off in one roll. Men
on 1 are not bad per se, but the straight pip count gives misleading
results if you have a stack there - things look better than they really
are. The Thorpe count is one way of trying to compensate for this.
Hope that helps!
Robert-Jan Veldhuizen
On 13-jan-97 18:42:53, Chuck Bower wrote:
[bearoff/double problem]
CB> A nice bearoff problem, and one which Thorp does a good job with.
I saw the Thorpe count explained here a while ago, but the terms used in that
explanation were so ambiguous that I didn't understand it. Could someone explain
it to me please ?
CB> Start with doubling window: tie score ==> window opens at 50%. If
CB> X drops, s/he has about 15% chance of winning the match (and a take
CB> is for the match--BOTH sides). Thus the doubling window is:
This I understand :).
CB> 0.50 < W < 0.85 (where "W" is O's winning chances).
CB> Now we apply the Thorp count to find O's winning chances in this
CB> game IF rolled to the end:
CB> 1) Pip count difference = 19 - 13 = 6.
CB> 2) # points covered (both have 3 points covered) difference = 0
CB> 3) # checkers remaining (both sides have 6) difference = 0
CB> 4) # checkers on 1-point (O has 3, X has 0) difference = 3
Yup, now how to use these numbers in the formula ?
CB> Race lead is GOOD for O ==> positive
CB> Checkers on 1-point is BAD for O ==> negative
CB> So we can now calculate O's winning chances (in percentage):
CB> w = 74 + 2*(6 - 3) = 80
For instance, where does the 74 value come from ?
Thanks to anyone helping me out here !
--
<tsb>Zorba
Chuck Bower
In article <1327.6960...@xs4all.nl>,
Robert-Jan Veldhuizen <veld...@xs4all.nl> wrote:
>
>I saw the Thorpe count explained here a while ago, but the terms used in that
>explanation were so ambiguous that I didn't understand it. Could someone
>explain it to me please ?
>
(snip)
>For instance, where does the 74 value come from ?
>
The Thorp Count was popularized in Robertie's "Advanced Backgammon". This
method was developed by Edward O. Thorp who is famous for his pioneering work
"Beat the Dealer" which explained how to become a favorite at blackjack. Note
that the CORRECT spelling is "Thorp" without an "e" at the end (even though
Robertie misspelled it in both the 1st and 2nd editions of his classic).
The method as described there is for money play only. I have since
modified it for match play (and THIS is where the "74 value" comes from):
Here is how I perform the Thorp Count (and basically how it is written up
in "Advanced Backgammon")
1) Start with the roller's pip count.
2) Add 2 for each checker remaining on the board.
3) Add an additional 1 for each checker on the 1 pt.
4) Subtract 1 for each home board point covered.
5) If the result up to now is 30 or greater, multiply by 1.1.
(Call this final result "R" for roller)
Repeat steps 1-4 for the NON-roller, BUT NOT STEP 5.
(Call this result "N" for NON-roller)
Now, subtract R from N ( = T) and compare the result as follows:
N - R .ge. -2, roller has an INITIAL double (centered cube);
N - R .ge. -1, roller has a REdouble;
N - R .le. +2, NON-roller has a take.
(here ".ge." means "greater than or equal to", and ".le."--you figure it out!)
As far as how to convert this to game winning chances, I noticed the
following correlation between the Thorp Count (above) and the percentage
cubeless game winning chances required to double/redouble/take as published
in the classic (but hard to find) article "Optimal Doubling in BG" by
Emmitt B. Keeler and Joel Spencer in OPERATIONS RESEARCH, vol 23 p. 1063(1975).
Assuming Roller's pip count is 70, K & S say roller should do the following:
Roller's cubeless Cube Thorp
game winning chances Action Count
70% Double, Take -2
72% REdouble, Take -1
78% PASS +2
Now if you overlook the "singularity" at 78% (Thorp says "barely a take"
and K & S say "barely a pass") you see that there is a linear relation
between the cubeless game winning chances and the Thorp Count:
W = 74 + 2*T (in percentage).
Finally I assumed that you could "extrapolate" the result outside
the 70-78% range. A study of bearoffs showed that this "Extened Thorp
Method" works rather well (IN GENERAL) for winning chances in the range
55-85%. Happy (Thorp) counting!