Over the Board Bearoff Analysis
Dean Gay
+24-23-22-21-20-19-+---+18-17-16-15-14-13-+
13| X X | | |
X| | | |
X| | | |
X| | | |
X| | | |
X| | | |
O| | | |
O| | | |
O| | | |
O| | | |
O| O | | |
12| O O | | |
+-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
O on roll.
Should O double?
This position came up in a game I played today on FIBS. I started to
think it through and quickly became overwhelmed. I know that an exact
calculation can be done on this type of position, and I feel that,
given enough time, and the aid of a calculator, I could *eventually*
come to the correct decision here. But during a game there is a
practical limit to how much can be done. I'm wondering, what is the
thought process that a good player goes through *over the board* to
figure out these type of decisions? Assume whatever score and cube
ownership you like -- or pick a more interesting position if this one
is too easy -- but "think out loud" for me if you will. I'm sure that
your incites would be valuable.
Thanks,
Dean
David Montgomery
In article <330e96f9...@news.pacificnet.net> d...@pacificnet.net (Dean Gay) writes:
> +24-23-22-21-20-19-+---+18-17-16-15-14-13-+
>13| X X | | |
> X| | | |
>12| O O | | |
> +-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
>
>O on roll.
>Should O double?
>I'm wondering, what is the
>thought process that a good player goes through *over the board* to
>figure out these type of decisions?
>
>Dean
Here was my thought process: My pipcount is 12, my opponent's 9,
so I'm down 3 pips. I'll probably waste more than my opponent,
so I really trail by about 4 pips. In a long race, trailing by
about 4 pips and being on roll makes the chances about equal, but
here the race is so short that I think I'm the favorite even if
I'm down 4. Still, I can't be a big favorite, definitely less
than 60% cubeless.
This would be as far as I would go unless doubling were a close
decision with a cpw of more than 50% to less than 60%. For most
scores, it's not enough to double, especially since I can get
many efficient cubes after taking a roll. (For example, after
53 or 63, with two on the deuce point, vs your opponent's one
roll position, you have a very efficient money cube.)
Reality check:
Sconyers CD says O is only a little over 52% here, at the low
end of my rough guestimate. O's effective lead in the race is
a tad over 4 pips.
David Montgomery
monty on FIBS
mo...@cs.umd.edu
Ron Karr
Dean Gay wrote:
>
> +24-23-22-21-20-19-+---+18-17-16-15-14-13-+
> 13| X X | | |
> X| | | |
> X| | | |
> X| | | |
> X| | | |
> O| | | |
> O| | | |
> O| | | |
> O| O | | |
> 12| O O | | |
> +-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
>
> O on roll.
> Should O double?
>
> think it through and quickly became overwhelmed. I know that an exact
> calculation can be done on this type of position, and I feel that,
> given enough time, and the aid of a calculator, I could *eventually*
> come to the correct decision here. But during a game there is a
> thought process that a good player goes through *over the board* to
> ownership you like -- or pick a more interesting position if this one
> is too easy -- but "think out loud" for me if you will. I'm sure that
> your incites would be valuable.
>
There are several ways to look at these positions. What I usually do
(and I'm not a bearoff expert by any means) is quickly run through a few
of them to see if one leads to a decisive result; if none of them do, I
may have to do a more detailed analysis (if I feel like it :-)).
1. Pip count. Often this is too crude in short bearoffs, but sometimes
it can quickly point to the right result. Here, for example, O has 12
pips, X has 9. Being on roll in a race is generally considered to be
worth 4 pips; therefore the pip count alone would indicate that O is a
very slight favorite, unlikely to be strong enough to double.
Then I'd look at factors that would tend to require adjustments to the
pip count. There are formulas, such as Thorp's, that attempt to do this
quantitatively, but in many positions you don't need to get real exact
to know what the correct decision is. Here, for example, O has one
extra checker, which is a minus. And he has 2 checkers on the same
point, which tends to produce less flexibility. Both players have gaps,
but X's are less harmful since all numbers that miss can play
efficiently. So these factors, if anything, would tend to make O's
equity worse, reinforcing the idea that he shouldn't double.
2. Another reasonable rule of thumb, which is attributed to Bill
Robertie, is: if I roll an above-average number (not my best) and my
opponent rolls a below-average number (not his worst), will I then
regret not having doubled? Here, it may not be so clear: if O rolls
5-4 or better and X doesn't bear off both checkers, then he'll lose his
market. But 6-3 or 5-3 leave O in a position of having a very efficient
double next time (still a take, X has 28% winning chances), which would
tend to make me think O shouldn't double now.
3. Beyond these, you may need to look at how likely each side is to bear
off in a given number of rolls. For a short position like this, it
shouldn't be too hard. X has 10 numbers to get off in 1 roll
(combinations of 6,5,4, plus 33). And he'll be off in 2 unless he rolls
very small numbers twice. O, on the other hand, has only 2 numbers to
get off in 1, and has a quite reasonable chance of needing 3 rolls.
This would tend to indicate that he should wait a roll.
(Incidentally, Jellyfish Level 7 says O wins 52.3%, which seems quite
consistent with the above reasoning. If it were the last shake, O would
double as the favorite, but not here.)
Match scores make a big difference. If the cube will be dead after you
turn it, it's often correct to double with ANY market losers. Or it may
be correct to wait much longer than for money. It depends on the
doubling window, which varies wildly for different scores. That's a
whole separate discussion.
Ron
Brian Sheppard
Dean Gay <d...@pacificnet.net> wrote in article
<330e96f9...@news.pacificnet.net>...
> +24-23-22-21-20-19-+---+18-17-16-15-14-13-+
> 13| X X | | |
> X| | | |
> 12| O O | | |
> +-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
Certainly not.
The easiest way to see that doubling is bad is to count rolls
that make the opponent wish he had dropped and those that
make you regret doubling.
1) O regrets after X 6-6, 5-5. All other rolls leave X with
at least 10 rolls to win, so O feels fine.
2) X regrets doubling after 1-1, 2-1, 3-1, 4-1, 2-3, 3-4.
Plainly, X should wait.
The other respondents have given good descriptions of ways to
approach such positions. Ron Karr mentioned Thorp's formula,
so I though I'd fill in the details. It usually gives excellent
practical doubling advice when the number of men on the ace point
is not too high and it is especially valuable when the sides
has unequal numbers of men.
1) Compute pip count: O 12, X 9
2) Add 2 for each man left: O 18, X 13
3) Subtract 1 for each point covered: O 16, X 11
4) Add 1 for each man on the ace point: no change
5) If the side to move's count exceeds 30, then add 10% to it (but
not to the other side): no change.
OK, so now double according to the following rule:
if O <= X-2 then O has an initial double
if O <= X-1 then O has a redouble
if X <= O+2 then X has a take.
In this case, O does not have an initial double.
Brian
ag.er...@aaa.dk
In article <330e96f9...@news.pacificnet.net>,
d...@pacificnet.net (Dean Gay) wrote:
>
> +24-23-22-21-20-19-+---+18-17-16-15-14-13-+
> 13| X X | | |
> X| | | |
> X| | | |
> X| | | |
> X| | | |
> O| | | |
> O| | | |
> O| | | |
> O| O | | |
> 12| O O | | |
> +-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
>
> Should O double?
>
> This position came up in a game I played today on FIBS. I started to
> think it through and quickly became overwhelmed. I know that an exact
> calculation can be done on this type of position, and I feel that,
> given enough time, and the aid of a calculator, I could *eventually*
> come to the correct decision here. But during a game there is a
> practical limit to how much can be done. I'm wondering, what is the
> thought process that a good player goes through *over the board* to
> figure out these type of decisions? Assume whatever score and cube
> ownership you like -- or pick a more interesting position if this one
> is too easy -- but "think out loud" for me if you will. I'm sure that
> your incites would be valuable.
>
>
> Dean
I cant claim to be an expert, but I can offer you my initial thoughts
when I saw the position. I assume money game with o holding the cube.
First, realize that x has a really easy take. This is because he can
win either by bearing off in one roll, or by x failing to be off in two
rolls, neither of which is very unlikely.
Second, see that x has many bad rolls, that allow o to redouble, should
x decide to double this turn. (Rolls like 21 31 41 32 43 11)
Third, see that many rolls "break even", ie. allow x to double effiently
next turn, if o fails to bear off. (51 61 42 53 63 22)
Fourth, note that for the double to gain, x needs essentially either to
roll big doubles or to take off _two_ checkers followed by a "miss" by o.
Over the board (or the news reader) the losses seems (to me) to outweight
the gains. I would hold the cube.
Hope this helps,
Lasse
-------------------==== Posted via Deja News ====-----------------------
http://www.dejanews.com/ Search, Read, Post to Usenet
FLMaster39
l look at it this way - when I fail to get off in one roll, my opponent
has 17 rolls to get off in one. So he is 34/36 x 17/36 to win AT LEAST.
How good a cube could this be? The other analyses have been good and
meaningful, but this one is very quick. This isn't close.
Stephen Turner
... and totally spurious. Your opponent has only 10 rolls to get off in one.
--
Stephen Turner sr...@cam.ac.uk http://www.statslab.cam.ac.uk/~sret1/
Stochastic Networks Group, Statistical Laboratory,
16 Mill Lane, Cambridge, CB2 1SB, England Tel.: +44 1223 337955
"Collection of rent is subject to Compulsive Competitive Tendering" Cam. City
Daniel Murphy
flmas...@aol.com (FLMaster39) writes:
>l look at it this way - when I fail to get off in one roll, my opponent
>has 17 rolls to get off in one. So he is 34/36 x 17/36 to win AT LEAST.
>How good a cube could this be? The other analyses have been good and
>meaningful, but this one is very quick. This isn't close.
No, he'll have 10 rolls: 65, 64, 54, 66, 55, 44, 33. That makes the
decision a little closer.
.
--
_______________________
Daniel Murphy
San Francisco
rac...@cityraccoon.com
Daniel Murphy
d...@pacificnet.net (Dean Gay) writes:
> +24-23-22-21-20-19-+--+
> | X X | |
> | | |
> | | |
> | O | |
> | O O | |
Over the board analysis: Where's the cube? Centered or do I have it?
(often makes a difference).
Assuming I have it:
1. Who's the favorite ? I am.
2. Will X pass ? Not likely. It looks like my rolls fall into four
categories. I can (1) Win now; (2) have a double/pass next roll (if
X hasn't won already) (3) be an underdog next time (if I get one); and
have a double/take next time.
Which rolls fall into which categories?
(1) Only 2 rolls win for me now, and then X wins 10/36 on his roll.
(2) Only 13 rolls bear off two checkers or bear off one and put another on
the acepoint (I need a 44, 33, 22, 65, 64, 62, 54 or 52). I'll lose 10/36
of these games anyway and *more* if I double now).
(3) 7 rolls don't take off *any* checkers or will leave a position where
I'm not a favorite to bear off next roll (43, 31, 21, 11) -- in fact,
if I double now I'll have to *drop* X's redouble.
(4) Which leaves 14 rolls (after which X will win 10/36), after which (if
he misses) I can make X take next time.
3. Conclusion: I can't be a very big favorite, and it looks like I should
wait a roll before doubling.
I might stop here, but since I can see *approximately* how often I win in
each case, I might go further (again, approximately):
If I hold the cube now, I win 11 and lose 4 of the Category (1) and (2)
rolls; win 2 and lose 5 of the category (3) rolls; and lose 3 and then
win
2x7 and lose 2x4 of the category (4) rolls, which all adds up to 7.
If I double now, I win 2x10 and lose 2x5 (not 11 and 4 because I can't
double X out next time), lose 2x7 (because X will double me out), lose
2x3 and then win 2x7 and lose 2x4, which adds up to -4.
Never mind *exactly* what "7" and "-4" mean -- 7 is more than -4 so I wait
a roll before doubling.
Chuck Bower
In article <330e96f9...@news.pacificnet.net>,
Dean Gay <d...@pacificnet.net> wrote:
> +24-23-22-21-20-19-+---+18-17-16-15-14-13-+
>13| X X | | |
> X| | | |
> X| | | |
> X| | | |
> X| | | |
> O| | | |
> O| | | |
> O| | | |
> O| O | | |
>12| O O | | |
> +-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
>
>O on roll.
>Should O double?
>
>This position came up in a game I played today on FIBS. I started to
>think it through and quickly became overwhelmed. I know that an exact
>calculation can be done on this type of position, and I feel that,
>given enough time, and the aid of a calculator, I could *eventually*
>come to the correct decision here. But during a game there is a
>practical limit to how much can be done. I'm wondering, what is the
>thought process that a good player goes through *over the board* to
>figure out these type of decisions? Assume whatever score and cube
>ownership you like -- or pick a more interesting position if this one
>is too easy -- but "think out loud" for me if you will. I'm sure that
>your incites would be valuable.
>
Nice problem, Dean. Actually, this is NOT at all easy, as you
have probably surmised by the lengths of the responses. I hope this
doesn't sound like scolding but PLEASE IN THE FUTURE (and
this request goes to ALL posters of problems to r.g.bg) INCLUDE THE
MATCH SCORE (OR SPECIFY "MONEY PLAY") AND THE LOCATION OF THE CUBE.
A couple of BTW's: I thought this was a double. (Looks like
I was WRONG!) I'm rather impressed with the mental powers of SOME
of the resonders. To be able to do all this stuff *over the board*
and not make a significant error (or false assumption)...
Of the many responses to this problem, everyone seems to have
assumed money play. Fine, but no one (that I can recall) has
mentioned "beaver". So, I make the following challenge to all of
you dilligent r.g.bg analyzers:
Assume:
1) O doubles to 2 (rightly or wrongly);
2) Both players will make correct cube decisions and checker
plays from now on.
Answer the following questions:
A) Should X beaver?
B) How high can the cube get?
C) What is O's equity (in points) after X's cube decision but
before O rolls?
Obviously I'm twisting Dean's problem a bit. This in no longer
an *over the board* situation.
Alexander Nitschke
A) Yes, X should beaver. The cubeless winning percentage is about 52%
and
X has some redouble equity.
B) The cube can get quite high. Assume the following sequence:
O doubles to 2. X beavers to 4.
O rolls 32. O moves 5/0.
X doubles to 8. O takes. (equity X: no double:0.226,
double/take:0.336)
X rolls 21. X moves 5/2.
O doubles to 16. X takes. (equity O: no double:0.397,
double/take:0.531)
O rolls 51. O moves 5/0 2/1.
X doubles to 32. O takes. (equity X: no double:0.278,
double/take:0.556)
So the cube can reach 32, if both sides roll bad enough :-)
C) Results of analysis:
Cubeless play.
Equity: 0.046167
Cube in centre.
Equity (no double): 0.054989
Equity (double accepted): -0.079247
Equity (double declined): 1.000000
Cube with player 1.
Equity (no double): 0.139322
Equity (double accepted): -0.079247
Equity (double declined): 1.000000
Cube with player 2. (Player 1 cannot double.)
Equity: -0.039623
The equity for O with cube on the side of X (and cube on 1) is -0.04,
so
after double/beaver O's equity is -0.04*4=-0.16.
The analysis is from the program race3 of which the source code can be
found somewhere in the web. This program performs an exhaustive search
and thus delivers correct equities.
I enhanced this program with cubeless play and with match scores. Match
scores strongly influence the doubling decisions. For example, the above
position (which is a money game beaver) is a correct redouble to 4/take
when trailing 1-2 in five point match.
Ciao,
Alexander (acey_deucey@FIBS)
Chuck Bower
In article <3313F8...@mailszrz.zrz.tu-berlin.de>,
Alexander Nitschke <nits...@mailszrz.zrz.tu-berlin.de> wrote:
Does anyone agree with me that these computers are taking all of
the fun out of the game?? (Only kidding, I think....) This race3
thing sorta blew my "homework assignment" out of the water. For
what it's worth (not much these days) I got -0.17 for O's equity if
X beavers, and basically the same scenario for the cube getting to 32.
Most of my work was with pencil, paper, and calculator, with some help
from Larry Strommen's Backgammon Position Analyzer and Edward Thorp's
chapter on 2 vs. 2 endgame cubes in his book "The Mathematics of
Gambling" (previously published in "Gambling Times Mag." around 1980?).
I made a couple of small simplifications which probably explain why
we disagree by 0.01. I'm not sure who should feel better about our
accord, me or the robot.... (And don't try to tell me it doesn't
have feelings! It's probably out there cloning itself like those
sheep.)
Chuck
bo...@bigbang.astro.indiana.edu
c_ray on FIBS
Claes Thornberg
In article <5evfuv$5...@dismay.ucs.indiana.edu> bo...@bigbang.astro.indiana.edu (Chuck Bower) writes:
> +24-23-22-21-20-19-+---+18-17-16-15-14-13-+
>13| X X | | |
> X| | | |
> X| | | |
> X| | | |
> X| | | |
> X| | | |
> O| | | |
> O| | | |
> O| | | |
> O| | | |
> O| O | | |
>12| O O | | |
> +-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
>
Assume:
1) O doubles to 2 (rightly or wrongly);
2) Both players will make correct cube decisions and checker
plays from now on.
Answer the following questions:
A) Should X beaver?
B) How high can the cube get?
C) What is O's equity (in points) after X's cube decision but
before O rolls?
Obviously I'm twisting Dean's problem a bit. This in no longer
an *over the board* situation.
Chuck
bo...@bigbang.astro.indiana.edu
Ok Chuck, here goes.
Position at start:
Player 1 : 3 men, 12 pips : 0 1 0 0 2 0
Player 2 : 2 men, 9 pips : 0 0 0 1 1 0
Results of analysis:
0. Cube in centre.
Equity (no double): 0.054989
Equity (double accepted): -0.079247
Equity (double declined): 1.000000
1. Cube with player 1.
Equity (no double): 0.139322
Equity (double accepted): -0.079247
Equity (double declined): 1.000000
2. Cube with player 2. (Player 1 cannot double.)
Equity: -0.039623
3. Cube is dead. (No player can double.)
Equity: 0.046167
Probability of winning: 0.523083
Player 1 should not double, wherever the cube is.
Answer to question A
********************
O doubles, wrongly, to 2. X correctly beavers to 4. As can be seen above,
O has negative equity if his double is accepted, therefore X beavers
to increase his equity.
Answer to question C
********************
After X has beavered, O's equity is -0.158494.
How high can the cube get?
O rolls 4 1, and correctly takes one piece of 5/4/0.
Now X is on roll, should X double to 8 and should O take?
Position at start:
Player 1 : 2 men, 9 pips : 0 0 0 1 1 0
Player 2 : 2 men, 7 pips : 0 1 0 0 1 0
Results of analysis:
0. Cube in centre.
Equity (no double): 0.174211
Equity (double accepted): 0.335820
Equity (double declined): 1.000000
1. Cube with player 1.
Equity (no double): 0.225995
Equity (double accepted): 0.335820
Equity (double declined): 1.000000
2. Cube with player 2. (Player 1 cannot double.)
Equity: 0.167910
3. Cube is dead. (No player can double.)
Equity: 0.202180
Probability of winning: 0.601090
Player 1 has a redouble, which player 2 should take.
X (player 1 above) should redouble to 8 and O should take.
X rolls 2 2, and moves 4/2/0 5/3/1.
Now O is on roll, should he redouble to 16 and should X take.
Yes, the position 0 1 0 0 1 - 1 is known to be a redouble/take, while
if X had 2 men on his 2 point, it would be incorrect to redouble (but
correct for an initial double). (This is known as the Jacoby
paradox).
Position at start:
Player 1 : 2 men, 7 pips : 0 1 0 0 1 0
Player 2 : 1 men, 1 pips : 1 0 0 0 0 0
Results of analysis:
0. Cube in centre.
Equity (no double): 0.055556
Equity (double accepted): 0.111111
Equity (double declined): 1.000000
1. Cube with player 1.
Equity (no double): 0.055556
Equity (double accepted): 0.111111
Equity (double declined): 1.000000
2. Cube with player 2. (Player 1 cannot double.)
Equity: 0.055556
3. Cube is dead. (No player can double.)
Equity: 0.055556
Probability of winning: 0.527778
Player 1 has a redouble, which player 2 should take.
So O (player 1 above) should redouble to 16 and X should take, after
which the cube won't get any higher. Either O wins on this roll, or X
wins by redouble/drop (or rolling, since all rolls take the last man
off).
Answer to question B
********************
The cube cannot reach higher than 16, and the sequence of rolls above
will take it to that level, assuming correct cube decisions.
Regards,
Claes Thornberg
--
______________________________________________________________________
Claes Thornberg Internet: cla...@it.kth.se
Dept. of Teleinformatics URL: http://www.it.kth.se/~claest
KTH/Electrum 204 Voice: +46 8 752 1377
164 40 Kista Fax: +46 8 751 1793
Sweden