Kleinman Count
michael.a.crane
I'm looking for en explanation of the 'Kleinman Count'. I understand it is for
avoiding leaving blots when rolling large doubles . . . but I could be wrong!
Can anyone help?
Michael
Øystein Johansen
> I'm looking for en explanation of the 'Kleinman Count'. I understand it is for
> avoiding leaving blots when rolling large doubles . . . but I could be wrong!
I believe 'Kleinman Count' is the method of estimating your game winning
probability in a race, based on the pip-count.
K is calculated by this equation:
K = D^2 / S (Eq. 1)
where D = the difference of pip-count of the two players
(Add 4 to compensate for the player on roll additional pips)
S = the sum of each players pip-counts
(Subtract 4 to compensate for the player on roll pips)
If K > 1.0
W = 0.76 + 0.114 * log (K) (Eq. 2)
(log(K) refers to the natural logarithm)
If K < 1.0
W = 0.5 + 0.267 * sqrt(K) (Eq. 3)
Where W = game winning probability in race.
Danny Kleinman did not give us equation 2 and 3, but gives a table of
winning probabilities for the different values of K. Equation 2 is my
own little regression analyses, and equation 3 is the work of Chuck
Bower [1].
Example 1:
I'm on roll got 100 pips in the race, you 113 pips. What's my
probability to win the race?
D = 113 - 100 + 4 = 17
S = 113 + 100 - 4 = 209
K = 17 * 17 / 209 = 1.383
W = 0.76 + 0.114 * log(1.383) = 0.797
So, the Kleinman Count estimates a winning probability in the race of
79.7%
Example 2:
I'm on roll got 100 pips in the race, you 106 pips. What's my
probability to win the race?
D = 106 - 100 + 4 = 10
S = 106 + 100 - 4 = 202
K = 10 * 10 / 202 = 0.495
W = 0.5 + 0.267 * sqrt(0.495) = 0.688
So, the Kleinman Count estimates a winning probability in the race of
68.8%
Now comes the big questions: Can you do this math in your head over the
board?
The answer is no, maybe someone can, but I can't, and I guess most of
the readers can't either. But there is some of this math that can be
done in your head over the board. Calculating the K should be possible
for most of us. You can add and subtract in your head, and you can
usually find a good value for the division to get K.
Remember some key values of K and assign the winning probability for
this values. That's all you need to remember! Forget about the square
roots and logarithms.
K = 0.5 => W = 0.69 (a double)
K = 0.7 => W = 0.72 (a redouble)
K = 1.0 => W = 0.76 (a take)
K = 1.2 => W = 0.78 (borderline take/drop)
K = 1.4 => W = 0.80
K = 2.0 => W = 0.84
Chuck Bower has a more complete table in his posting [1].
I have also implemented these equations as C code:
http://subversions.gnu.org/cgi-bin/cvsweb/~checkout~/gnubg/kleinman.c?rev=1.1&content-type=text/plain
Best regards,
Øystein Johansen 8^)
Chuck Bowers article:
[1] http://www.bkgm.com/rgb/rgb.cgi?view+500
Douglas Zare
This is different from the other Kleinman count for estimating racing chances.
The context is that you have closed out your opponent (or close to it) and want a
quick way to ensure that you are safe from an embarrassing anti-joker (fast
enough that it does not annoy your opponent).
IIRC, you count the number of 6's needed to bear in all of your checkers and bear
off all of your checkers then on your 6 point. Even numbers are supposed to be
safer than odd numbers, but I'm skeptical about the validity of this for a count
of, say, 7 versus 6. It depends on how you can play, for example, 6-5, 5-5, and
6-1, too.
If your 6 point is made and your Kleinman count is 5, then you will leave a blot
on your 6 if you roll 6-6. If your Kleinman count is 3, and you do not have a
(potential) spare on your 5, then you will usually leave a shot with 6-6, 6-5,
and 5-5. High numbers usually cause your Kleinman count to go down by an even
number.
Of course, I read about this in one of Kleinman's books that I bought from CJC,
but I don't remember which one.
Douglas Zare
michael.a.crane
The count I;'m seeking will be low, in the 6 to 9 range give or take a tad. You
assign certain values to checkers in the outer tables and, depending upon your
impending roll you have to keep the 'Kleinman Count' even to stay out of trouble
when bearing in against opposition. Something like being able to take a double
six next roll; I think!
Any ideas now?
Michael
"Řystein Johansen" <ojo...@statoil.com> wrote in message
news:3A9531C7...@statoil.com...
| Řystein Johansen 8^)
michael.a.crane
on it.
Thank you, Douglas,
Michael
"Douglas Zare" <za...@math.columbia.edu> wrote in message
news:3A95647B...@math.columbia.edu...
Mark Driver
moves with large doublets, especially double 6's. Your Kleinman Count is the
sum of the number of crossovers you need to bring all your men home and the
number of men you have on the most retarded points in each quadrant; 6 point
12 point 18 point, and 24 point. A Kleinman Count of 5 represents immeadiate
danger of leaving a shot with double 6's. A Kleinman Count of 7 represents
an immeadiate danger of having to choose between coming down to a Kleinman
Count of 5 and moving a man deep inside your home board. It is generally
desirable to keep your Kleinman Coiunt on even numbers, 8. 6, 4 until all
your men are home."
Quoted from page 130 Vision Laughs At Counting (original edition page
number)
Douglas Zare wrote in message <3A95647B...@math.columbia.edu>...
michael.a.crane
Thanks a lot.
Michael
"Mark Driver" <mdr...@mail.usyd.edu.au> wrote in message
news:973u0p$s80$1...@spacebar.ucc.usyd.edu.au...
Øystein Johansen
>
> This is a great answer . . . but it's not the one I'm looking for ;-(
>
> The count I;'m seeking will be low, in the 6 to 9 range give or take a tad. You
> assign certain values to checkers in the outer tables and, depending upon your
> impending roll you have to keep the 'Kleinman Count' even to stay out of trouble
> when bearing in against opposition. Something like being able to take a double
> six next roll; I think!
>
> Any ideas now?
Ahh-ha!
I don't refer to this count as the Kleinman Count (obviously). The count
you are referring to is described in Jeff Ward's book "Winning is more
fun". I don't have any name on this count, but it may be one of Danny's
inventions.
The count can be applied on closed board situation when you don't want
to leave a shot (i.e. your opponent has a man on the bar, and you have a
closed board).
Since a closed board will use 12 of your checkers the count is all about
the remaining 3 checkers. The whole theory is based on a postulate: "If
6-6 doesn't leave a shot in such situation, no other roll will leave a
shot." I have not found any practical possible positions that violates
this postulate.
Here a short brief:
Divide the board into 5 zones:
Zone 1: 1-point to 5-point (and off)
Zone 2: 6-point to 11-point
Zone 3: 12-point to 17-point
Zone 4: 18-point to 23-point
Zone 5: 24-point (and bar(*))
For each of the 3 checkers, count 1 for each in zone 1, count 2 for each
in zone 2, count 3 for each .... ....
If the total sum is 6 or 4, then 6-6 will leave a shot. The safe values
of the sum is 5, 7 and any number higher than 7.
Remember you don't have to recalculate this count for each roll you're
making, just decrement the sum by one for each time you cross from one
zone to another!
Hope this was the count you where looking for!
Øystein Johansen 8^)
(*) It's theoretical impossible to have a checker on the bar, at the
same time as you have closed out your opponent. The bar is only of
academic interest in this discussion.
Jerri9465
>
>
>
>(*) It's theoretical impossible to have a checker on the bar, at the
>same time as you have closed out your opponent. The bar is only of
>academic interest in this discussion.
No it isn't. I've had it happen and it would be very easy to show situations
where it happens. Say you are playing an ace point game and close your board
at the same time you are forced to run. Then your opponent hits both of your
checkers, and you dance until you hit a lucky shot from the bar with one of
your checkers, He will be closed out, but you will still have one on the bar.
Øystein Johansen
Ooops, I guess you're right. It is possible. I didn't think of the
possibility of two checkers on the bar. Silly me.
The bar should be then be in Zone 5, as stated in the previous posting.
Don't think any player should worry about double sixes when his on the
bar!
-Øystein