Racing cube 6

[Tim Chow]

Back to racing cubes, but now I'm switching to short races.

XGID=-B-AA----------------aa-a-:1:1:1:00:0:0:0:0:10

X:Player 2 O:Player 1
Score is X:0 O:0. Unlimited Game
+13-14-15-16-17-18------19-20-21-22-23-24-+
| | | O O O |
| | | |
| | | |
| | | |
| | | |
| |BAR| |
| | | |
| | | |
| | | | +---+
| | | X | | 2 |
| | | X X X | +---+
+12-11-10--9--8--7-------6--5--4--3--2--1-+
Pip count X: 9 O: 8 X-O: 0-0
Cube: 2, X own cube
X on roll, cube action

---
Tim Chow

[Bradley K. Sherman]

Tim Chow <tchow...@yahoo.com> wrote:
>Score is X:0 O:0. Unlimited Game
> +13-14-15-16-17-18------19-20-21-22-23-24-+
> | | | O O O |
> | | | |
> | | | |
> | | | |
> | | | |
> | |BAR| |
> | | | |
> | | | |
> | | | | +---+
> | | | X | | 2 |
> | | | X X X | +---+
> +12-11-10--9--8--7-------6--5--4--3--2--1-+

> X:9 O:8, X on roll, cube action

Redouble/Take

--bks

[michae...@gmail.com]

This is complicated as it goes to 2 ply.
On first exchange 62,52,42,32,12 (10/36)=27.7% and X is left with 3 checkers whereas the only bad roll for O is 12. Obviously O has a Take.
Redoubling from 2 to 4 can be justified with market losers (20/36 including 66,55,44 that end the game.
Redouble/Take.
If he rolls any of those 10/36, then O would Redouble to 8/pass

[Paul]

I agree with Bradley's meticulously detailed analysis and discussion.
If X doubles, can O take? O is an enormous favourite to bear off in two rolls or fewer. Therefore O cashes if X fails to remove two. Therefore O takes if X has more than 9 numbers that fail to remove two. There are ten such numbers, namely all 2's except 22. We are therefore faced with the question of trying to determine if 10 > 9. This is too hard a question for me, so I consulted some mathematicians at MIT who told me that yes, 10 > 9. O should take if X doubles.

But should X double? X certainly has plenty of market losers. So let's calculate the probability of X winning after doubling. If it's much greater than 50% I will double. (There's a reason my friends call me "Paul Bold Epstein".)

The probability of X failing to win is 10/36 (the 2's problem) + 1/6 * 8/9 * 1/18 (X rolls 41/51/61, O doesn't bear off immediately but X misses) + 1/6 * 1/9 (X rolls as before but O rolls miracle doubles) + 1/12 * 8/9 * 7/36 + 1/12 * 1/9 (as before except with X rolling 11 or 31) + 17/36 * 1/9 (X rolls normally and O rolls miracle doubles). This probability is 135/486 + 4/486 + 9/486 + 7/486 + 4.5/486 + 25.5/486 = 185/486. X's winning probability is therefore 301/486 = approx 62%. 62% is much more than 50% so X should double.

Note that the above calculations imply that O owns the cube. Without the cube, X wins even more.

Redouble/Take.

Paul

[badgolferman]

Double/Take

[peps...@gmail.com]

Above is wrong. The final part of the calculation (17/36 * 1/9) should not include X's doubles that win immediately. So 5/36 * 1/9 needs to be added to X's winning probability. X's winning probability is 308.5/486 which is about 63.5% -- big double.

Paul

[Walt]

I was 4 for 5 using the Trice count in the previous examples. I'm aware
that it is not as reliable in short races, so i may not be as successful
here..

For X, +2 for the spare on the ace and +2 for the extra checker. That
means X is down 13 to 8. How is this a cube? Is X even the favorite?

Another way to think about is that X will be off in two rolls unless he
rolls a non-double 2, while O will be off in two rolls except for
consecutive airballs.

Clear take since O cashes if X rolls a non-double two. I wouldn't ship
it here.

ND/T


--
//Walt

[Walt]

Ok, brain fart here. X does not need to be ahead in the pipcount to
cube. For races less than 62 pips, subtract 5 and divide by seven
rounding down. Often this is zero, and X can cube when a few pips behind.

Also, for really short races like this, the spare on the ace shouldn't
matter. That gives a point of last take of zero, and X can cube when
two behind. X is down by three, which would indicate an initial double
but not a redouble.

I still say ND, but it's not as clear as it seemed. Huge take.



--
//Walt

[Paul]

With only two rolls left, why not try calculating the position by hand? (In other words, why doesn't every single other person on this planet do exactly what I do?)

Although my calculations indicate a redouble, I didn't prove it by any means. I used the heuristic -- double if X is over 50% to win. But this isn't fully accurate, of course.

Also, I could be wrong. What conclusion do you get to, when you calculate it by hand? (I admit it's harder than any count).

Paul

[Tim Chow]

XGID=-B-AA----------------aa-a-:1:1:1:00:0:0:0:0:10

X:Player 2 O:Player 1
Score is X:0 O:0. Unlimited Game
+13-14-15-16-17-18------19-20-21-22-23-24-+
| | | O O O |
| | | |
| | | |
| | | |
| | | |
| |BAR| |
| | | |
| | | |
| | | | +---+
| | | X | | 2 |
| | | X X X | +---+
+12-11-10--9--8--7-------6--5--4--3--2--1-+
Pip count X: 9 O: 8 X-O: 0-0
Cube: 2, X own cube
X on roll, cube action

Paul gave an excellent analysis that I won't repeat here. It's important, of course, that X misses with any 2 except 22. Nudge X's checker from the 3pt to the 2pt and O should pass (see variant).

Analyzed in Rollout
No redouble
Player Winning Chances: 67.24% (G:0.00% B:0.00%)
Opponent Winning Chances: 32.76% (G:0.00% B:0.00%)
Redouble/Take
Player Winning Chances: 67.24% (G:0.00% B:0.00%)
Opponent Winning Chances: 32.76% (G:0.00% B:0.00%)

Cubeful Equities:
No redouble: +0.389 (-0.125)
Redouble/Take: +0.514
Redouble/Pass: +1.000 (+0.486)

Best Cube action: Redouble / Take

Rollout:
1296 Games rolled with Variance Reduction.
Dice Seed: 271828
Moves and cube decisions: 4-ply
Confidence No Double: ± 0.000 (+0.389..+0.389)
Confidence Double: ± 0.000 (+0.514..+0.514)

eXtreme Gammon Version: 2.19.208.pre-release

-------
Variant
-------

XGID=-BA-A----------------aa-a-:1:1:1:00:0:0:0:0:10

X:Player 2 O:Player 1
Score is X:0 O:0. Unlimited Game
+13-14-15-16-17-18------19-20-21-22-23-24-+
| | | O O O |
| | | |
| | | |
| | | |
| | | |
| |BAR| |
| | | |
| | | |
| | | | +---+
| | | X | | 2 |
| | | X X X | +---+
+12-11-10--9--8--7-------6--5--4--3--2--1-+

Pip count X: 8 O: 8 X-O: 0-0

Cube: 2, X own cube
X on roll, cube action

Analyzed in Rollout
No redouble
Player Winning Chances: 80.65% (G:0.00% B:0.00%)
Opponent Winning Chances: 19.35% (G:0.00% B:0.00%)
Redouble/Take
Player Winning Chances: 80.65% (G:0.00% B:0.00%)
Opponent Winning Chances: 19.35% (G:0.00% B:0.00%)

Cubeful Equities:
No redouble: +0.635 (-0.365)
Redouble/Take: +1.154 (+0.154)
Redouble/Pass: +1.000

Best Cube action: Redouble / Pass

Rollout:
1296 Games rolled with Variance Reduction.
Dice Seed: 271828
Moves and cube decisions: XG Roller++
Search interval: Gigantic
Confidence No Double: ± 0.000 (+0.635..+0.635)
Confidence Double: ± 0.000 (+1.154..+1.154)

eXtreme Gammon Version: 2.19.208.pre-release

---
Tim Chow

[Bradley K. Sherman]

Paul <peps...@gmail.com> wrote:
> ...

>I agree with Bradley's meticulously detailed analysis and discussion.

> ...

I am using my enteric nervous system to analyze these positions
and, unfortunately, the analysis is too large for the margins
of the USENET.

--bks