15 positions that minimize waste of pips in bear-off
Stig Eide
On 24 Apr 1996, Dan McCullam wrote:
> Stig Eide <sti...@james.stud.unit.no> wrote:
>
> >The table shows which boards that gives the lowest waste for
> >1-15 pieces left to bear off. The board for 15 pieces is,
> >of course, the most interesting. Because it shows that the optimal
> >bear-in strategy is to get 3 pieces on your 4 point,
> >5 on the 5 point and 7 on the 6 point:
> >
> (table snipped)
>
> Does this mean (as I think it does) that a balanced board isn't such a
> good thing after all?
>
> --
> Dan McCullam dm...@le.ac.uk
> DMc on FIBS
>
>
>
Hi Dan, nice that someone respondes to this! ;)
You're right; A balanced board is not optimal.
I'm not sure what you mean by a balanced board,but if its
5 men on each of your 4,5 and 6 point, its close to optimal.
If you mean equaly distributet at the 6 points like 222333, you'll
waste 2.4 pips more than 'optimal'.
Here's the numbers:
waste= 9.47 board= 222333
waste= 7.27 board= 000555
waste= 7.07 board= 000357
Stig Eide
Stig Eide
On 24 Apr 1996, William C. Bitting wrote:
> Stig Eide (sti...@james.stud.unit.no) wrote:
> : waste= 9.47 board= 222333
> : waste= 7.27 board= 000555
> : waste= 7.07 board= 000357
> : Stig Eide
>
> hmm, it strikes me that the optimal board - 7 5 3 checkers on the
> 6 5 4 points - may be the most efficient in terms of wasting
> fewer pips to bearoff, BUT does this also mean it is the most efficient
> in terms of the number of rolls needed to bear all the checkers off the
> board?
15 men on the acepoint minimize the expected number of rolls needed.
The problem that this article solve is:
Given a number of checkers left, how to place them to waste fewest pips.
That is the same as: Given a pipcount, how to place your checkers
to minimize the number of rolls needed to bear them off. The latter
is how we need to think over the board: No matter how you play
the dice, the pipcount will be the same. The problem is to make the
play that brings you off fastest.
>I would think a 2 2 2 3 3 3 board (6pt to 1 pt) would take fewer
> rolls on average to bearoff all the checkers versus the optimal board for
> pip efficiency. How does one balance the most efficient in terms of
> fewest rolls versus the most efficient in terms of fewest wasted pips?
Given your pipcount, those are just two sides of the same problem.
> A lot of 21's early in the bearoff in the pip efficient board may not
> waste many pips, but it sure will increase the number of rolls need to get
> the checkers off! So, the optimal board in terms of pips may prove of
> little value for many bearoff situations. I've seen a number of 44's
> accomplish so little in boards with lots of men on the 5 and 6 points,
> that when I can, I try to avoid that situation. So, I guess the question
> is when should one try to build a 7 5 3 board - when even or slightly
> behind in the race?
For most positions, the distribution that minimize expected rolls
needed, will also maximize probability of winning.
> The bots seem to do the opposite in many cases -
> building a board with the checkers evenly distributed across all 6
> points. ..wcb on FIBS
I've a feeling that the bots have built their their boards with the
checkers evenly distributed across all 6 points _before_ the game went
into a no-contact game. ;)
Stig
William C. Bitting
: On 24 Apr 1996, Dan McCullam wrote:
: > Stig Eide <sti...@james.stud.unit.no> wrote:
: > >The table shows which boards that gives the lowest waste for
: > >1-15 pieces left to bear off. The board for 15 pieces is,
: > >of course, the most interesting. Because it shows that the optimal
: > >bear-in strategy is to get 3 pieces on your 4 point,
: > >5 on the 5 point and 7 on the 6 point:
: > (table snipped)
: > Does this mean (as I think it does) that a balanced board isn't such a
: > good thing after all?
: > DMc on FIBS
: >
: Hi Dan, nice that someone respondes to this! ;)
: You're right; A balanced board is not optimal.
: I'm not sure what you mean by a balanced board,but if its
: 5 men on each of your 4,5 and 6 point, its close to optimal.
: If you mean equaly distributet at the 6 points like 222333, you'll
: waste 2.4 pips more than 'optimal'.
: Here's the numbers:
: waste= 7.27 board= 000555
: waste= 7.07 board= 000357
: Stig Eide
hmm, it strikes me that the optimal board - 7 5 3 checkers on the
6 5 4 points - may be the most efficient in terms of wasting
fewer pips to bearoff, BUT does this also mean it is the most efficient
in terms of the number of rolls needed to bear all the checkers off the
board? I would think a 2 2 2 3 3 3 board (6pt to 1 pt) would take fewer
rolls on average to bearoff all the checkers versus the optimal board for
pip efficiency. How does one balance the most efficient in terms of
fewest rolls versus the most efficient in terms of fewest wasted pips?
A lot of 21's early in the bearoff in the pip efficient board may not
waste many pips, but it sure will increase the number of rolls need to get
the checkers off! So, the optimal board in terms of pips may prove of
little value for many bearoff situations. I've seen a number of 44's
accomplish so little in boards with lots of men on the 5 and 6 points,
that when I can, I try to avoid that situation. So, I guess the question
is when should one try to build a 7 5 3 board - when even or slightly
behind in the race? The bots seem to do the opposite in many cases -
Walter G Trice
>Stig Eide (sti...@james.stud.unit.no) wrote:
>: On 24 Apr 1996, Dan McCullam wrote:
>: > Stig Eide <sti...@james.stud.unit.no> wrote:
>: > >The table shows which boards that gives the lowest waste for
>: > >1-15 pieces left to bear off. ...
>hmm, it strikes me that the optimal board - 7 5 3 checkers on the
>6 5 4 points - may be the most efficient in terms of wasting
>fewer pips to bearoff, BUT does this also mean it is the most efficient
>in terms of the number of rolls needed to bear all the checkers off the
>board? I would think a 2 2 2 3 3 3 board (6pt to 1 pt) would take fewer
>rolls on average to bearoff all the checkers versus the optimal board for
>pip efficiency.
Indeed. Faster still would be a 0 0 0 0 0 15 board, which would take,
on average, a mere 6.979591837 rolls to bear off. However in the
process it would waste, on average, 42.00000000 pips. The point
you are missing is that once contact is broken what happens is
that every time you roll your pip count decreases by exactly
the number of pips that your dice give you -- unless you have begun
to waste pips by bearing off from low points with high numbers.
When wastage happens your pip count goes down by LESS than what
you roll.
In other words, pip count is pretty much a given, and the problem
is to bear off fastest for whatever pip count you have. (Another
way of looking at backgammon is that the object of the game is
to be the first player to reduce his pip count to zero.)
Your 2 2 2 3 3 3 distribution is good, given a 15 checker
position with a pip count of 48. (In fact, subject to these two
constraints, I believe it is optimal, but I'm going on memory.)
However, if you had been bearing in and off properly you would
still have a pip count of 48 (which is not under your control
since the 6 point is still occupied) but you would have borne
off several checkers. Wouldn't you really rather have something
like 3 3 3 1 0 0 ? If you roll 2-1 every time the positions will
both bear off in 16 rolls. But if you roll 6-5 every time
2 2 2 3 3 3 will take 8 rolls and 3 3 3 1 0 0 will take 5 rolls.
>... So, I guess the question
>is when should one try to build a 7 5 3 board - when even or slightly
>behind in the race?
Stig Eide's 'optimal' positions minimize mean rolls to bear off, and
they will be best against the great majority of opposing positions.
(Incidentally, his work is correct -- I did this stuff several years
ago and got the same results.) The particular 7 5 3 board is hard to
compare with alternatives, since most reasonable candidates involve
outfield checkers.
>The bots seem to do the opposite in many cases -
>building a board with the checkers evenly distributed across all 6
>points.
Yes, they do. They make many technically incorrect plays.
-- Walter Trice
David Montgomery
>hmm, it strikes me that the optimal board - 7 5 3 checkers on the
>6 5 4 points - may be the most efficient in terms of wasting
>fewer pips to bearoff, BUT does this also mean it is the most efficient
>in terms of the number of rolls needed to bear all the checkers off the
>board? I would think a 2 2 2 3 3 3 board (6pt to 1 pt) would take fewer
>rolls on average to bearoff all the checkers versus the optimal board for
>fewest rolls versus the most efficient in terms of fewest wasted pips?
You are correct that 333222 will take fewer rolls to bearoff on average
than 000357 -- and 15 0 0 0 0 0 will take even fewer! The problem is
that you don't often get to choose between these bearoff formations.
(I've tried it -- bearing my last checker into my board to produce
000357, I offer to move all my checkers to the ace point, pointing out
how much extra wastage I'll have. No one's gone for it yet. :-> )
The concept of wastage is relevant for comparing positions with
equal pipcounts, because most checker play problems allow you to
choose among plays with equal pipcounts, and usually the best play
among these is the one with the least wastage. The lowest wastage
positions will be the best play, except sometimes when your chances
are very good or very poor (Kleinman's bull-offs and bear-offs) and
in some positions with very few checkers left where the ability to use
the doubling cube effectively can changes things.
By knowing the general patterns in wastage, you can know the type of
bearoff position you are aiming for, and modify your bearin technique
accordingly. We now know that the best bearoff formation (for a given
number of pips) is one which places most of the checkers on the 4, 5
and 6 points, with preferably more on the 5 and the most on the 6.
This is more important than memorizing the wastage of particular
formations.
Wastage can also be extremely useful in making cube decisions.
>I guess the question is when should one try to build a 7 5 3 board
>- when even or slightly behind in the race?
>..wcb on FIBS
If its just a pure race, and you still have the ability to aim for
000357 (you don't have any checkers yet on the 1, 2 or 3 points),
and you're not too far behind (probably if your chances are more than
a few percent and greater than your chance of getting gammoned)
then this formation should be your goal.
There might be some positions in which you are a huge favorite
to win the race in which its correct to not aim for 000357. However,
I'm not sure which way would be an improvement, if any. Perhaps you
should put more checkers on the 4 point to create a lower volatility
race and reduce your opponent's chances. Or perhaps you should
stack more heavily on the 6 and possibly 5 point, hoping to roll sets
and win a lucky gammon. In either case, you won't give up much by
just aiming for 000357.
In fact, with the exception of positions with very few checkers where
cube utility changes things, and positions where the trailer *must*
rely on doubles to catch up, if you make the low wastage play, even when
it is wrong you will give up very little.
David Montgomery
monty on FIBS
William C. Bitting
O: rolls 4 and 5 ...what is mloner's play ?
Pipcounts: O = 112, X = 101
+-----------------------------------------& X match c 54 ml v jf
| X . . . X X | | X X X X X . |
| X | | X X X |
| | | X |
| |BAR| |
| O | | |
| O O | | |
| X O O O | | O O O |
| X . O O O . | | O O O . . . |
+-----------------------------------------& O ml 1a 5a Crawford
O: (4 5)8-3 6-2 !!
I thought it might be interesting to use the position in the above
board and the actual rolls in the game with alternative moves humans
might choose to see how close one would come to the optimal bear-in
position for bearing off 15 checkers (i.e. 7 5 3 checkers on points
6 5 4). While not exact, the position wasn't too far off - 7 4 4.
While the "human" moves resulted in all checkers being home in
4 moves, versus 5 1/2 taken by mloner, the positions after 9 moves
were identical, so the play for rolls 10 through 13(the last roll)
were also identical. However, no claim is made the human plays are
technically correct.
In general would one expect mloner to take one extra move to bear
its men off versus the technically correct play in 5% of its games,
or would 0.5% be closer to the outcome - or 0.005%? Would there be
some number of games where mloner's play would result in fewer rolls
to bear off all its men versus techically correct plays?
wcb on FIBS
note - all rolls and moves below are O's. X is ignored.
actual moves played alternative moves
1. O: (4 5) 8-3 6-2 1. (4 5) 10-6 10-5
2. O: (1 4) 10-6 8-7 2. (1 4) 10-6 8-7
3. O: (2 6) 8-6 7-1 3. (2 6) 10-4 8-6
4. O: (33)10-7 9-6/2 7-4 4. (3 3) 9-6/2 8-3 7-4
5. O: (5 2) 10-5 10-8 all checkers home; 7 4 4
6. O: (5 6) 8-3 6-off checkers on points 6 5 4
all checkers born-in 5. (5 2) 5-off 4-2
in 5 1/2 rolls 6. (5 6) 6-off 5-off
7. O: (5 3) 5-off 3-off 7. (5 3) 5-off 6-3
8. O: (5 3) 5-off 3-off 8. (5 3) 5-off 3-off
9. O: (1 2) 2-off 1-off 9. (1 2) 2-off 6-5
creates indentical positions for each set of plays
10. O: (3 5) 6-3 5-off 11. O: (4 4) 6-2 4-off 4-off 4-off
12. O: (1 2) 6-5 2-off 13. O: (5 5) 6-1 5-off 3-off 1-off
Tom Keith
In a previous post, Eirik Milch Pedersen and Stig Eide set out to
find the distribution of 15 pieces in the bear-off which minimizes
pip wastage. They discover that the following position
- - - 3 5 7
------------------
1 2 3 4 5 6
has the lowest wastage and go on to conclude that optimal bear-in
strategy is to aim for this position.
A couple of other posters have noted that this seems contrary to
backgammon intuition. But then Walter Trice appears to defend
the conclusion, which surprises me because I don't think the
conclusion is right.
I don't doubt that "000357" is the position with lowest wastage
(of the positions considered). But comparing positions of
different pip counts is like comparing apples and oranges.
A player's choice at each roll is between positions of *equal*
pip count. Knowing the lowest wastage of many positions of
different pip counts isn't particularly useful.
Further, by looking at only "bear off" positions, Pedersen
and Eide fail to consider that there might be some "not quite
bear-off positions" with the same pip count which have lower
wastage.
It is tempting to think that in preparing to bear off it must be
better to have all of your checkers in your home table than to
have some of them on your 7 or 8 points. But this is not true.
Paul Magriel has a simple exercise (on page 283 his Book) "for
the skeptic who thinks 'you can't get 'em off 'til you get 'em
in'.":
Set up the pieces [with 15 of X's checkers on X's 6 point and
3 of O's checkers on each of O's 4, 5, 6, 7, and 8 points].
We see that each side has the same number of pips (90). Now
roll the dice and play the same numbers for each side
*simultaneously*. Of course X begins bearing off first, but
you will almost invariably find that at the end, either both X
and O bear the last man off simultaneously, or else O wins.
What would be interesting to know is this: For each pip count,
what position has the lowest expected number of rolls. I think
you would find the positions on this list would all have fairly
flat distributions. E.g.,
X X X
X X X X X X
X X X X X X
------------------
1 2 3 4 5 6
Tom Keith
Walter G Trice
Tom Keith <tak...@io.org> writes:
>In a previous post, Eirik Milch Pedersen and Stig Eide set out to
>find the distribution of 15 pieces in the bear-off which minimizes
>pip wastage. They discover that the following position
> - - - 3 5 7
> ------------------
> 1 2 3 4 5 6
>has the lowest wastage and go on to conclude that optimal bear-in
>strategy is to aim for this position.
>A couple of other posters have noted that this seems contrary to
>backgammon intuition. But then Walter Trice appears to defend
>the conclusion, which surprises me because I don't think the
>conclusion is right.
[much deleted]
>What would be interesting to know is this: For each pip count,
>what position has the lowest expected number of rolls. I think
>you would find the positions on this list would all have fairly
>flat distributions. E.g.,
> X X X
> X X X X X X
> X X X X X X
> ------------------
> 1 2 3 4 5 6
*** But as it happens, you're wrong. The distribution you show has
a pip count of 54. Its expected number of rolls to bear off is
7.883196. Its wastage is 10.38 pips. In contrast, the following
position, also with pip count 54:
X
X X X
X X X
X X X X
------------------
1 2 3 4 5 6
bears off in 7.460891 rolls, on average. Its wastage is 6.93
pips.
You pointed out (in the part of your post that I deleted) that
positions involving outfield checkers may be better than the
ones we have considered. This is true, but it is not relevant
to most bear-off checker play decisions. For example, I am
sure that a better 54 pip distribution than the one I show here
would be 1 checker each on the 17, 18, and 19 points, with 12
already borne off. But this position is difficult to reach.
Better still would be one checker on the 54 point, with 14
borne off. Unfortunately, there is no 54 point.
Incidentally, it can be shown that wastage for one checker on
the N point approaches 4 + 5/7 pips exactly, as N heads for
infinity.
-- Walter Trice
Stein Kulseth
In article <31807D...@io.org>, Tom Keith <tak...@io.org> writes:
|> Further, by looking at only "bear off" positions, Pedersen
|> and Eide fail to consider that there might be some "not quite
|> bear-off positions" with the same pip count which have lower
|> wastage.
Not likely, until you get all checkers in there is no wastage at all.
So the lowest wastage possible for a non bear-off position is when you
are certain to reach the lowest-wastage bear-off position, with
equal minimum wastage.
|> Paul Magriel has a simple exercise (on page 283 his Book) "for
|> the skeptic who thinks 'you can't get 'em off 'til you get 'em
|> in'.":
Which is of course not quite the issue here. Note also that Magriel
himself says that you should in general not bother bearing in to
the 1,2 and 3 point in no-contact, the exceptions are when one of
the higher points are empty or have few checkers.
|> What would be interesting to know is this: For each pip count,
|> what position has the lowest expected number of rolls. I think
|> you would find the positions on this list would all have fairly
|> flat distributions. E.g.,
|>
|> X X X
|> X X X X X X
|> X X X X X X
|> ------------------
|> 1 2 3 4 5 6
|>
|> Tom Keith
Compare the maximally flat distribution at pipcount 21 "111111"
with "000112". The latter is much better when rolling high numbers,
and not particularily worse when rolling low (as it then often
transposes into the situation resulting from "111111").
In fact I can't seem to find even one series of rolls where
"111111" is better than "000112".
BTW, Stig would it be possible to repost the original article? I forgot
to save, and it has timed out.
--
stein....@fou.telenor.no
... signature funny quote (and more) at http://www.nta.no/brukere/stein
Tom Keith
In a previous post I questioned whether "000357" really is a good
position to aim for during bear-in. My reasoning was
> ... by looking at only "bear off" positions, Pedersen
> and Eide fail to consider that there might be some "not quite
> bear-off positions" with the same pip count which have lower
> wastage.
But Stein Kulseth points out that such a position is ...
> Not likely, until you get all checkers in there is no wastage at all.
> So the lowest wastage possible for a non bear-off position is when you
> are certain to reach the lowest-wastage bear-off position, with
> equal minimum wastage.
That's a good point. Having given it some thought, I can see that
"000357" very likely is the best 15-checker position with a pip count
of 79 or greater. So, just as Stig Eide and Eirik Milch Pedersen
suggest, it *is* a good position to aim for during bear-in
(if you don't yet have any checkers on your 1, 2, and 3 points).
Thanks for setting me straight.
Tom
Stig Eide
Trice:
> position with a pip count of 48. (In fact, subject to these two
> constraints, I believe it is optimal, but I'm going on memory.)
It's close, but 1 2 3 4 3 2 is the optimal distribution.
As pointed out by several, we can't control the pipcount. And then
our initial posting with minimal pip-waste for 1-15 checkers left
is not of much value, except for the 000357 position which should
be the aim in the bear-in.
As a consequence Eirik made a table of optimal boards for all
possible combinations of pips and checkers left.
Interested can get a postscript-file by emailing him:
th...@stud.unit.no
Here's the optimal board for each pipcount, no matter
how many checkers left. Interestingly, fewest
checkers left is often not optimal:
The 48-pip board 000008 (8 men at 6point)
wastes 1.49 pips more than 000144.
You can always transform pipwaste to expected rolls left:
Just add the pipcount, and divide it with 8.1667.
Enjoy.. 8-)
1:100000(7.17)
2:200000(6.17)
3:110000(5.17)
4:101000(4.62)
5:000010(4.30)
6:000001(4.21)
7:010010(5.05)
8:001010(5.26)
9:001001(5.32)
10:000101(5.11)
11:000011(5.01)
12:000002(5.23)
13:001101(5.40)
14:001011(5.40)
15:000111(5.31)
16:000021(5.45)
17:000012(5.47)
18:001111(5.67)
19:000211(5.79)
20:001012(5.73)
21:000112(5.53)
22:000022(5.65)
23:000013(5.85)
24:001112(6.06)
25:000212(6.05)
26:000122(5.89)
27:000113(5.97)
28:000023(6.05)
29:001122(6.21)
30:000222(6.25)
31:000132(6.23)
32:000123(6.15)
33:000114(6.33)
34:000024(6.38)
35:001123(6.37)
36:000223(6.38)
37:000133(6.36)
38:000124(6.39)
39:000034(6.55)
40:001133(6.56)
41:000233(6.53)
42:000224(6.55)
43:000134(6.51)
44:000125(6.62)
45:000035(6.74)
46:001134(6.66)
47:000234(6.63)
48:000144(6.69)
49:000135(6.67)
50:001234(6.77)
51:001144(6.81)
52:000244(6.76)
53:000235(6.75)
54:000145(6.78)
55:000136(6.83)
56:001235(6.86)
57:001145(6.87)
58:000245(6.82)
59:000236(6.88)
60:000146(6.89)
61:001245(6.92)
62:000345(6.93)
63:000255(6.94)
64:000246(6.91)
65:000156(6.99)
66:000147(7.00)
67:001246(6.98)
68:000346(6.98)
69:000256(6.98)
70:000247(7.00)
71:001346(7.05)
72:001256(7.04)
73:000356(7.03)
74:000347(7.05)
75:000257(7.04)
76:000248(7.09)-001356(7.09)
77:001347(7.11)
78:001257(7.09)
79:000357(7.07)
80:000267(7.11)
81:000258(7.11)
82:000249(7.18)
83:000159(7.22)
84:00014A(7.35)
85:00005A(7.48)
86:00004B(7.68)
87:00003C(7.99)
88:00002D(8.47)
89:00001E(9.19)
90:00000F(10.17)
Where A,B,C,D,E and F represents 10,11,12,13,14 and 15 checkers.
Phill Skelton
> >What would be interesting to know is this: For each pip count,
> >what position has the lowest expected number of rolls. I think
> >you would find the positions on this list would all have fairly
> >flat distributions. E.g.,
>
> > X X X
> > X X X X X X
> > X X X X X X
> > ------------------
> > 1 2 3 4 5 6
>
> a pip count of 54. Its expected number of rolls to bear off is
> 7.883196. Its wastage is 10.38 pips. In contrast, the following
> position, also with pip count 54:
>
> X X X
> X X X
> X X X X
> ------------------
> 1 2 3 4 5 6
>
> pips.
Is it worth pointing out that there aren't fifteen men left
on the board here? If you are going to stick all fifteen men on in
high positions then it will take you more rolls to get all fifteen
off than if they are evenly spread. Granted that for equal PIP COUNTS
you are better of with men on the higher points, but that is largely
becasue you have less men on the board. And pip counts don't really
count for much at all when you are bearing off. If I had fifteen men in
my home board I'd rather have them all on the 1 rather than the 6 any
day.
Phill
Kevin Cline
I would still like to know the best 15-checker positions with
pip counts greater than 79. For example, is
0-0-0-2-3-5-3-2 better than 0-0-0-0-0-15 ?
--
Kevin Cline
Michael J Zehr
In article <3185D4F1...@sun.leeds.ac.uk> ph...@sun.leeds.ac.uk (Phill Skelton) writes:
>> >What would be interesting to know is this: For each pip count,
>> >what position has the lowest expected number of rolls.
>> [ But for pip count of 54, 0 0 1 3 3 4 is better ]
>
> Is it worth pointing out that there aren't fifteen men left
>on the board here? If you are going to stick all fifteen men on in
>high positions then it will take you more rolls to get all fifteen
>off than if they are evenly spread. Granted that for equal PIP COUNTS
>you are better of with men on the higher points, but that is largely
>becasue you have less men on the board. And pip counts don't really
>count for much at all when you are bearing off. If I had fifteen men in
>my home board I'd rather have them all on the 1 rather than the 6 any
>day.
It's certainly important to keep in mind what are choices are and when
we are making them. No one has a problem in understanding that between:
2 1 2 3 3 3 and
2 2 2 4 3 2
the former is better. They have the same pip count but one has 14
checkers and one has 15. This *is* a choice one can make. (A 2 to play
from 2 2 2 3 3 3.)
While we can't chose between 2 2 3 3 3 2 and 0 0 1 3 3 4 on a *single*
play, we have choices a number of plays earlier that can steer towards
one or the other.
To take a simple case, play a '3' from:
0 0 0 4 5 4 1 0 1
We all know it's better to play 9-6 so that next turn after a 64 we play
7-3 6-off, giving us 0 0 1 4 5 4 (14 checkers, 68 pips) instead of
playing 6-3 to get a nice flat distribution and playing 9-3 7-3 with the
64, yeilding 0 0 3 4 5 3 (15 checkers 68 pips). (Or 9-5 7-1 getting
1 0 1 5 5 3 -- this might be better than 0 0 3 4 5 3, but 0 0 1 4 5 4
ought to be better than both of them.)
The same thing applies if we look at a sequence of many rolls instead of
just two: Bear *in* your checkers first, even if they're to high points,
instead of smoothing distribution within your board. Then you starting
bearing *off* while you have a lot of checkers on your high points and
you do end up with a choice between fewer checkers on the high points or
all your checkers nicely spread out.
(There of course will be exceptions with really horrible distributions,
such as stacks on the 4 and 6 when you want to play 6-5 instead of 7-6.
But it's surprising how few checkers it takes on the 5 pt to make the
best choice back to 7-6. Of course, you need a database that extends
out past the 6pt to examine this kind of position in detail.)
-michael j zehr