From Newsgroup: rec.games.backgammon

XGID=-BB----------------a----b-:1:1:1:00:8:0:0:11:10

X:Daniel O:XG Roller+
Score is X:8 O:0 11 pt.(s) match.
+13-14-15-16-17-18------19-20-21-22-23-24-+
| | | O O |
| | | O |
| | | |
| | | |
| | | |
| |BAR| |
| | | |
| | | |
| | | | +---+
| | | X X | | 2 |
| | | X X | +---+
+12-11-10--9--8--7-------6--5--4--3--2--1-+
Pip count X: 6 O: 8 X-O: 8-0/11
Cube: 2, X own cube
X on roll, cube action

Obviously the match score and cube value makes me highly reluctant
to turn the cube. But the cube isn't dead.
Is my advantage strong enough to double? Should XG take?

Obviously, MET makes this problem completely trivial so please don't use
MET.

Thank you.

Paul
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From Newsgroup: rec.games.backgammon

"peps...@gmail.com" <pepstein5@gmail.com> writes:

XGID=-BB----------------a----b-:1:1:1:00:8:0:0:11:10

X:Daniel O:XG Roller+
Score is X:8 O:0 11 pt.(s) match.
+13-14-15-16-17-18------19-20-21-22-23-24-+
| | | O O |
| | | O |
| | | |
| | | |
| | | |
| |BAR| |
| | | |
| | | |
| | | | +---+
| | | X X | | 2 |
| | | X X | +---+
+12-11-10--9--8--7-------6--5--4--3--2--1-+
Pip count X: 6 O: 8 X-O: 8-0/11
Cube: 2, X own cube
X on roll, cube action

Nice one. No need here for any race formulae, but I am surprised by how
good Kleinman works here, in a very short and extreme position. But of
course the game winning chances can be figured out analytically, even
over the board with a clock.

But this is only half of the task ...

Is my advantage strong enough to double? Should XG take?

Over the board this is a clear double according to Woolseys law: Your
oppenent must be either semi-mad or know about METs to at all consider
taking. (-;

Obviously, MET makes this problem completely trivial so please don't
use MET.

Err, what else? I cannot see another reasonable approach to sort this
out at this match score and cube level. (I assume Neil's numbers and
Turner's formula count as using MET.)

Best regards

Axel
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From Newsgroup: rec.games.backgammon

On Monday, December 11, 2023 at 8:36:44rC>PM UTC, Axel Reichert wrote:

"peps...@gmail.com" <peps...@gmail.com> writes:

XGID=-BB----------------a----b-:1:1:1:00:8:0:0:11:10

X:Daniel O:XG Roller+
Score is X:8 O:0 11 pt.(s) match. +13-14-15-16-17-18------19-20-21-22-23-24-+
| | | O O |
| | | O |
| | | |
| | | |
| | | |
| |BAR| |
| | | |
| | | |
| | | | +---+
| | | X X | | 2 |
| | | X X | +---+
+12-11-10--9--8--7-------6--5--4--3--2--1-+
Pip count X: 6 O: 8 X-O: 8-0/11
Cube: 2, X own cube
X on roll, cube action

Nice one. No need here for any race formulae, but I am surprised by how
good Kleinman works here, in a very short and extreme position. But of course the game winning chances can be figured out analytically, even
over the board with a clock.

But this is only half of the task ...

Is my advantage strong enough to double? Should XG take?

Over the board this is a clear double according to Woolseys law: Your oppenent must be either semi-mad or know about METs to at all consider taking. (-;

Obviously, MET makes this problem completely trivial so please don't
use MET.

Err, what else? I cannot see another reasonable approach to sort this
out at this match score and cube level. (I assume Neil's numbers and Turner's formula count as using MET.)

You absolutely do need a MET to solve it. But this is a quiz.
Solve it by guessing or knowing the MET, rather than by googling it.
The MET is the hard part -- the game winning chances are trivial.
I lose if I fail to roll 22/33/44/55/66 and if my opponent rolls 33/44/55/66. The probability of me losing is 31/324. So the probability of me winning is 293/324 which is just over 90.4%.
(I did the computation that 293/324 is slightly less than 90.4% mentally, without using a computer or even pen and paper.)
Obviously, the winning chances don't give the full story. It also matters
how often the cube reaches 8.
Paul
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From Newsgroup: rec.games.backgammon

On Monday, December 11, 2023 at 9:42:05rC>PM UTC, peps...@gmail.com wrote:

On Monday, December 11, 2023 at 8:36:44rC>PM UTC, Axel Reichert wrote:

"peps...@gmail.com" <peps...@gmail.com> writes:

XGID=-BB----------------a----b-:1:1:1:00:8:0:0:11:10

X:Daniel O:XG Roller+
Score is X:8 O:0 11 pt.(s) match. +13-14-15-16-17-18------19-20-21-22-23-24-+
| | | O O |
| | | O |
| | | |
| | | |
| | | |
| |BAR| |
| | | |
| | | |
| | | | +---+
| | | X X | | 2 |
| | | X X | +---+
+12-11-10--9--8--7-------6--5--4--3--2--1-+
Pip count X: 6 O: 8 X-O: 8-0/11
Cube: 2, X own cube
X on roll, cube action

Nice one. No need here for any race formulae, but I am surprised by how good Kleinman works here, in a very short and extreme position. But of course the game winning chances can be figured out analytically, even
over the board with a clock.

But this is only half of the task ...

Is my advantage strong enough to double? Should XG take?

Over the board this is a clear double according to Woolseys law: Your oppenent must be either semi-mad or know about METs to at all consider taking. (-;

Obviously, MET makes this problem completely trivial so please don't
use MET.

Err, what else? I cannot see another reasonable approach to sort this
out at this match score and cube level. (I assume Neil's numbers and Turner's formula count as using MET.)

You absolutely do need a MET to solve it. But this is a quiz.
Solve it by guessing or knowing the MET, rather than by googling it.
The MET is the hard part -- the game winning chances are trivial.
I lose if I fail to roll 22/33/44/55/66 and if my opponent rolls 33/44/55/66.
The probability of me losing is 31/324. So the probability of me winning is 293/324 which is just over 90.4%.
(I did the computation that 293/324 is slightly less than 90.4% mentally, without using a computer or even pen and paper.)
Obviously, the winning chances don't give the full story. It also matters how often the cube reaches 8.

Paul

I just computed it using the Rockwell / Kazaross MET. This validates XG's verdict
but doesn't validate XG's assessment of my play. XG must be using a different MET.
In other words, I got this wrong. However, I don't think XG would have flagged my error
with the Rockwell Kazaross MET. I'll google XG's MET. Hmm, I think XG does use it.
At this point, I may has well give away the result and working.
If I hold the cube. Then I have a 90.4% (GWC) of being a 96.6% favourite (1A, 11A)
And I have a 9.6% chance of being an 84% favourite (3A, 9A)
Combined MWC from holding is 90.4% * 96.6% + 9.6% * 84% = 95.39%.
Now suppose I double, and XG drops. Then I have MWC probability 96.6%.
Suppose I double and XG takes. Then I win the match outright with probability 5/36.
Suppose that I get one of my 31 bad rolls. XG then redoubles to 8 and I of course take.
Now for XG to win the match, XG needs to bear off immediately (probability 1/9) and
then win the match from a level score (combined probability of 1/18). So if I roll badly (probability 31/36)
then I win the match with probability 17/18. Combined match winning probability is therefore
5/36 + 31/36 * 17/18 = 617/648 = 95.22% which is worse for me than holding. ND/T.
Paul
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From Newsgroup: rec.games.backgammon

On December 11, 2023 at 3:06:40rC>PM UTC-7, peps...@gmail.com wrote:

Suppose I double and XG takes. Then I win
the match outright with probability 5/36.
Suppose that I get one of my 31 bad rolls.

What is that 36 doing there? How can there
be 31 bad rolls when "there are 21 possible
dice rolls at every turn"??
Your calculations are all wrong. If you don't
trust me, you can ask a "mathgammonician
like Tim or Doug Zare". Although he doesn't
have a PHD but only MS, even Kit Woolsey
may know since, unlike the other two who
aren't, he is a "gamblegammon giant" and
thus he may deserve to be respected as a
"mathgammonician" also... ;)
MK
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