Is this a take?
Dan Scoones
Hi all,
Last night I was playing a "money" session with JellyFish Level 7 and
got to the following position.
+24-23-22-21-20-19-+---+18-17-16-15-14-13-+
| O O O O O | | O O O |
| O O O O | | O |
| O | | |
| | | |
| | | |
| | | |[64]
| | | |
| | | |
| | | X |
| X | | X |
| X X X X | | X X |
| X X X X | | X X O |
+-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
X (rook) on roll. Cube action?
The pip count is 84-96 in X's favour. X's lead is just over 14% of
his count.
In Advanced Backgammon, Bill Robertie states that, in long races, the
trailer has a pass if the leader's racing advantage exceeds 12% of his
pip count. That is the case here.
Therefore I doubled, expecting JF to pass. However, JF scooped up
this cube (and went on to lose.)
Applying the Thorp count yields a different picture. Here, the
trailer's adjusted count minus the leader's adjusted count is exactly
0, which indicates a take.
Both the percentage lead method and the Thorp count method are ways of
estimating the correct cube action. Obviously, if they contradict
each other (as they do here), one of them must be yielding an
incorrect estimate. The question is: which one, and why?
Of course, it is likely that the Thorp count method is correct here.
In any case I would be very grateful if some kind soul would shed some
light on this position by asking JellyFish for a rollout.
Interestingly enough, in the "old days" (i.e., the 1970s) it was held
that the trailer has a take down to a 15% racing deficit...
Thanks,
Dan
rook (FIBS)
To reply via email, remove 1 from my address
Jacques Torrione
Dan Scoones a écrit dans le message <3515669c...@News.IslandNet.com>...
Hi there :-)
interesting position I think...
It surprised me that jf evaluate this as a take. I will not speak about the
Thorp formula here ( I am not familiar with this one). I think you can take
with 15% racing deficit in LONG race (more than 120 pips). That is not the
case here. The 14% race deficit indicates a big pass in this race. However
if you look more carefully at the position you can find some arguments for
the take.
1) X has men on 6 points and O on 9 points. So O is better than X on this
point of view ( in a race it's better to have your checkers well distributed
instead of stacked on some points)
2) X has a lot of checkers on his 7 and 8 point. This is bad for the race
because X will have to play deep in his home board with all 6 and 5 during
the bearing in, and so X will waste some pips for the bear off.
3) X has no men on the 4 point and perhaps will not be able to put some
checkers here and it is well know that this cost a lot in the bear off
always for the reason that X will waste some pips)
Compared to X position, O position is good for the race...
So the structure of both position indicates that O can take with a little
bit more than the usual 12% racing deficit.
But is this sufficient for taking with 14% racing deficit ?
Mmmm ??
Don't know how much costs to X his poor distribution...
So I used the computer. Let see the results.
Snowie 3-ply gives X an eq. of 0.598 so double/pass (let say that the
borderline is 0.570 here)
I made a full 3888 rollout in 1-ply cubeless and it gives X an eq. of 0.591
so d/p.
JellyFish level 7 gives X an eq. of 0.563 so d/take.
But I made a full 3888 rollout in level 5 and it gives X an eq. of 0.610
cubeless and 0.558 if O owns the cube (standard deviation is 0.013). So jf's
rollout indicates a clear pass for O.
So it looks like double/pass is the correct cube action here.
Jacques Torrione
Donald Kahn
On Sun, 22 Mar 1998 19:29:42 GMT, ro...@islandnet.com (Dan Scoones)
wrote:
>Hi all,
>Last night I was playing a "money" session with JellyFish Level 7 and
>got to the following position.
>
> +24-23-22-21-20-19-+---+18-17-16-15-14-13-+
> | O O O O O | | O O O |
> | O O O O | | O |
> | O | | |
> | | | |
> | | | |
> | | | |[64]
> | | | |
> | | | |
> | | | X |
> | X | | X |
> | X X X X | | X X |
> | X X X X | | X X O |
> +-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
>
>X (rook) on roll. Cube action?
>
>The pip count is 84-96 in X's favour. X's lead is just over 14% of
>his count.
>
>In Advanced Backgammon, Bill Robertie states that, in long races, the
>trailer has a pass if the leader's racing advantage exceeds 12% of his
>pip count. That is the case here.
>
>Therefore I doubled, expecting JF to pass. However, JF scooped up
>this cube (and went on to lose.)
>
In the usual case, 14% is too much to take against. But here, becuase
of the hole on the 4 point, which will be very damaging unless X can
fill it, this appears to just get under the wire, according to JF
estimates.
deekay
Chuck Bower
In article <3515669c...@News.IslandNet.com>,
Dan Scoones <ro...@islandnet.com> wrote:
>Last night I was playing a "money" session with JellyFish Level 7 and
>got to the following position.
>
> +24-23-22-21-20-19-+---+18-17-16-15-14-13-+
> | O O O O O | | O O O |
> | O O O O | | O |
> | O | | |
> | | | |
> | | | |
> | | | |[64]
> | | | |
> | | | |
> | | | X |
> | X | | X |
> | X X X X | | X X |
> | X X X X | | X X O |
> +-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
>
>X (rook) on roll. Cube action?
>
>The pip count is 84-96 in X's favour. X's lead is just over 14% of
>his count.
>
>In Advanced Backgammon, Bill Robertie states that, in long races, the
>trailer has a pass if the leader's racing advantage exceeds 12% of his
>pip count. That is the case here.
(snip)
>Applying the Thorp count yields a different picture. Here, the
>trailer's adjusted count minus the leader's adjusted count is exactly
>0, which indicates a take.
(snip)
>...One of them must be yielding an incorrect estimate. The question is:
>which one, and why?
Good question. My experience with races has led me to use the
percentage method and the Thorp Count to complement each other. That is,
usually they don't both work for the same position. Thorp is best when
both players have ALL their checkers borne in (and/or off), in other words--
no checkers in the outfield. The percentage method tends to work well
in these longer, smoothly distributed races.
As most of you know, I've converted both of these formulas into
cubeless game winning chances. I repeat those modifications:
W = 0.54 + 2*p (where 'p' is the percentage lead of player on
roll as described by Dan above)
W = 0.74 + 2*T (where 'T' is the Thorp Count)
Here we have p = 1/7 and T = 0, as Dan has already said, giving 82.5%
(I have a hangup rounding 0.5!) and 74% respectively. Clearly a
disagreement (again as Dan observed).
There is another slightly more complicated method which, I believe when
fully implemented, works over the full range of both the percentage and
Thorp Counts. That is the Kleinman Count. I have yet to learn the FULL
Kleinman Count. (Anybody want to loan me some time? I promise to pay
you back in my next lifetime....) In it's simple form (for long races
like this one), K = (D + 4)^2 / (S - 4) where D is the pip count
difference and S is the pip count sum. If you plug the above pip counts
into this formula, you find K = 1.45. So what do you do with THAT??
Danny (Kleinman, that is) created a table based up the standard
normal distribution. If you memorize the table, then you can convert K
to cubeless winning chances. Some people just memorize the "key" money
values of 68% (or 70%), 72%, 78%. In tournament backgammon, though,
you often want to know winning chances over a MUCH wider range. How
do you get THAT??
I've developed an algorithm (which isn't really THAT hard to use,
especially if you're willing to learn a couple chapters of Art Benjamin's
book "Mathemagics"). I "reveal" the algorithm here for the first time.
(A roll of the drums, please!)
if K < 1: W = 0.50 + 0.267*sqrt(K)
This reproduces the Kleinman Table to 0.5% or better over the range
0.50 < W < 0.75. If you don't like 0.267, then 0.27 works even better
except between 73% and 76%, but even there it is off less than 1%.
Now, what about K >= 1? Well, I derived a logarithmic fit there
(which works up to 95%) but even Art doesn't normally do logarithms
in his head! (...though he could if he saw the need, I'm sure.)
I then saw that it's not THAT hard to memorize the table if you see
the following pattern:
(approx.)
W K delta K
0.76 * 1.0 ---
0.77 1.1 0.1
0.78 1.2 0.1
0.79 1.3 0.1
0.80 * 1.4 0.1
0.81 1.55 0.15
0.82 1.7 0.15
0.83 1.85 0.15
0.84 * 2.0 0.15
0.85 2.2 0.2
0.86 2.4 0.2
0.87 2.6 0.2
0.88 2.8 0.2
0.89 * 3.0 0.2
0.90 3.33 0.33
0.91 3.67 0.33
0.92 * 4.0 0.33
0.93 4.33 0.33
0.94 4.67 0.33
0.95 * 5.0 0.33
0.96 * 6.0 1.0
0.97 * 7.0 1.0
0.98 * 8.0 1.0
0.99 (does it matter???)
(Note: * are "key" values of K in my algorithm.) As above, this part
of the algorithm matches Kleinman's table to 0.5% or better.
Notice that in the above problem the Kleinman Count says X will win
just over 80%. I did 180 level-6 cubeless JF3.0 rollouts and it got X
winning 79.7% with an S.D. of 0.2%. Are you impressed with the Kleinman
Count? One problem doesn't mean a whole lot, but I've been amazed
at how accurate it is in general.
So where do you go from here? Well, as I mentioned, this is the
original Kleinman Count. There are advances (which don't involved any
more square roots!) which make adjustments to the Count for distributional
anomolies (similar to the adjustments made in the Thorp Count). That is
the part I still need to learn. Kleinman's many books (which are valuable
to the serious student) cover the extensions. You can get them (and about
everything else on the BG market) for Carol Cole ( c...@flint.org ).
Special thanks to Marc Gray (FlashGammon on FIBS) for urging me to
learn the Kleinman Count.
Chuck
bo...@bigbang.astro.indiana.edu
c_ray on FIBS