5a 4a 43 to play - H? P? R? D?

[Walt]

XGID=---a-CC-B--AdF---babab-b--:0:0:1:43:0:1:0:5:10

X:You O:O
Score is X:0 O:1 5 pt.(s) match.
+13-14-15-16-17-18------19-20-21-22-23-24-+
| X O O | | O O O O |
| X O | | O O O |
| X | | |
| X | | |
| 6 | | |
| |BAR| |
| | | |
| O | | |
| O | | X X |
| O X | | X X |
| O X X | | X X O |
+12-11-10--9--8--7-------6--5--4--3--2--1-+
Pip count X: 138 O: 126 X-O: 0-1/5
Cube: 1
X to play 43

[badgolferman]

I'll play 13/9, 6/3.

[Bradley K. Sherman]

Walt <walt_...@yahoo.com> wrote:
>Score is X:0 O:1 5 pt.(s) match.
> +13-14-15-16-17-18------19-20-21-22-23-24-+
> | X O O | | O O O O |
> | X O | | O O O |
> | X | | |
> | X | | |
> | 6 | | |
> | |BAR| |
> | | | |
> | O | | |
> | O | | X X |
> | O X | | X X |
> | O X X | | X X O |
> +12-11-10--9--8--7-------6--5--4--3--2--1-+

> X:138 O:126, Cube:1, X to play 43

13/6

--bks

[Freeven]

I think the DMP play is 6/3* 13/9. This leaves O only a few hit and
cover numbers and should win the most games. I don't know if that's the
right idea when gammons are in the mix, but I'd try it here as well.

[Tim Chow]

My first inclination was 6/2 5/2 hoping to hit next time. But we're
far enough behind in the race that maybe we should hit now while we
have the chance. 13/9 6/3* duplicates O's 3's to hit and cover. I
expect it to lose more net gammons but it should win more than enough
extra games to compensate. So I'll go with 13/9 6/3*.

---
Tim Chow

[crf]

OTB, I'm sure I'd play 13/6, but if I counted the pips (and then recounted them because it doesn't "look" like I'm so far behind) and then counted how few hit and covers O has (because it "looks" like there'd be a lot more) I'd hit. Even if I get hit, I might pick up another blot. 13/9, 6/3*.

I'm not fast enough at counting shots/covers OTB. Are there shortcuts for counting shots like what cluster counting does for counting pips?

[Walt]

If you chose H (13/9 6/3*) or R (13/6) congratulations! XGR++ says that
these two very different ideas are indistinguishable equity-wise.

D (13/10 13/9) is fairly close.

I played P (6/2 5/2) thinking that a homeboard point is a permanent
asset and that since O had three homeboard points I needed a third too.

But being behind in the race, a runner to shoot at, and blots to hit if
sent back, dumping checkers deep in the homeboard is not the right idea.
All the good plays either hit or keep all 15 checkers in front of the
runner.

I'll do a full rollout of the top two moves on Monday to see if that can
separate the two.

XGID=---a-CC-B--AdF---babab-b--:0:0:1:43:0:1:0:5:10

X:Player 1 O:Player 2

Score is X:0 O:1 5 pt.(s) match.
+13-14-15-16-17-18------19-20-21-22-23-24-+
| X O O | | O O O O |
| X O | | O O O |
| X | | |
| X | | |
| 6 | | |
| |BAR| |
| | | |
| O | | |
| O | | X X |
| O X | | X X |
| O X X | | X X O |
+12-11-10--9--8--7-------6--5--4--3--2--1-+
Pip count X: 138 O: 126 X-O: 0-1/5
Cube: 1
X to play 43

1. XG Roller++ 13/6 eq:-0.136
Player: 46.59% (G:4.09% B:0.09%)
Opponent: 53.41% (G:5.55% B:0.06%)

2. XG Roller++ 13/9 6/3* eq:-0.137 (-0.001)
Player: 49.91% (G:8.12% B:0.28%)
Opponent: 50.09% (G:17.48% B:0.47%)

3. XG Roller++ 13/10 13/9 eq:-0.163 (-0.027)
Player: 48.32% (G:7.50% B:0.26%)
Opponent: 51.68% (G:17.55% B:0.47%)

4. XG Roller++ 13/9 11/8 eq:-0.202 (-0.066)
Player: 45.74% (G:5.94% B:0.20%)
Opponent: 54.26% (G:11.98% B:0.12%)

5. XG Roller++ 13/10 6/2 eq:-0.266 (-0.130)
Player: 43.63% (G:4.71% B:0.15%)
Opponent: 56.37% (G:9.57% B:0.19%)

6. XG Roller++ 6/2 5/2 eq:-0.273 (-0.137)
Player: 42.69% (G:3.89% B:0.09%)
Opponent: 57.31% (G:6.80% B:0.09%)


eXtreme Gammon Version: 2.10, MET: Kazaross XG2

[Walt]

On 5/3/2013 4:58 AM, crf wrote:
> OTB, I'm sure I'd play 13/6, but if I counted the pips (and then recounted them because it doesn't "look" like I'm so far behind) and then counted how few hit and covers O has (because it "looks" like there'd be a lot more) I'd hit. Even if I get hit, I might pick up another blot. 13/9, 6/3*.
>
> I'm not fast enough at counting shots/covers OTB. Are there shortcuts for counting shots like what cluster counting does for counting pips?
>

I suck at shot counting, so I don't know that I can tell you much.
Magriel covers it in chapter 3 and there aren't many shortcuts.

Count 11 for each direct shot. 20 for each double direct shot. 27 for
a triple direct shot. Add combination numbers (eg 23 and 41 for a shot
that's 5 away) making sure to not count the ones that are blocked, and
not count any roll twice.

Indirect shots, it helps to remember

7 - 6
8 - 6
9 - 5
10 - 3
11 - 2
12 - 3
15 - 1
16 - 1

[Tim Chow]

On May 3, 4:58 am, crf <friesen.cr...@gmail.com> wrote:
> Are there shortcuts for counting shots like what
> cluster counting does for counting pips?

You might like this little trick:

http://pion.ch/backgammon/birds.php

An equivalent formula that I personally like a bit better is that if
there are X numbers that kill bird 1, Y numbers that kill bird 2, and
Z numbers that kill both birds, then the number of ways to kill both
birds is

2*X*Y - Z^2
---
Tim Chow

[crf]

These work great when the only things in sight are shooters and targets, but when there are intervening blocked points in between, I start getting tripped up -- start with 20 (or 11 or 27), add combinations that work, adjust for whether they're doublets or not, subtract out combinations that don't work, remember if a combination for one number includes another number already accounted for... it doesn't have to be very complicated before I have to revert back to "OK, what does 11 do? 12? 13?..."

Seems like there has to be a better way.

[crf]

> You might like this little trick:

Yes, that's quite nice. Pythagorean shot counting!

> 2*X*Y - Z^2

Also nice, easier to count the intersection than the union.

I'll have to play around with this, thanks!

[Tim Chow]

On May 5, 4:51 am, crf <friesen.cr...@gmail.com> wrote:
> Seems like there has to be a better way.

I doubt it. There are just too many things that can happen, and
sometimes those "little" exceptions are what matter the most in a
particular position. But if you come up with something, it would be
interesting to hear about it.

---
Tim Chow

[Freeven]

On 5/3/2013 4:58 AM, crf wrote:

> I'm not fast enough at counting shots/covers OTB. Are there shortcuts for counting shots like what cluster counting does for counting pips?

This may not quite answer your question, but some might find it helpful.
Here are some simple methods I came up with to help my students figure
direct and indirect shots until they (hopefully) get the numbers memorized.

For single, direct shots:
------------------------

1) Add 10 to the distance from the target.
2) Add 1 for any extra hits that are possible only because doubles can
be played four times.

Example: You hit with any 4. How many ways to hit?

1) 10 + 4 = 14
2) Add 1 for 11: 14 + 1 = 15

So you hit with 15 rolls.


For single, indirect shots:
--------------------------

1) Subtract the distance to the target from 13.
2) Add 1 for any extra hits that are possible only because doubles can
be played four times.

Example: You hit with any 9. How many ways can you hit?

1) 13 - 9 = 4
2) Add 1 for 33: 4 + 1 = 5

So you hit with 5 rolls.


For double, direct shots:
------------------------

1) Add 16 to the sum of distances from the targets.
2) Add 1 for any extra hits that are possible only because doubles can
be played four times.
3) Add 1 if one of the targets is twice as far away as the other.

Example: You hit with any 6 or 3. How many ways can you hit?

1) 16 + 3 + 6 = 25
2) Add 1 each for 11 and 22: 25 + 2 = 27
3) Add 1 because 6 is twice 3: 27 + 1 = 28

So you hit with 28 rolls.


Obviously, you still have to be aware of numbers that are blocked, but
that's normally just a minor adjustment once you have the raw numbers.

[crf]

> This may not quite answer your question, but some might find it helpful.

Actually, this is very much the kind of thing I was looking for, not so much the specific formulas, but the approach. Graphing these out on a 6x6 grid, I see how I was grouping rolls into overlapping 'clusters' more complicated than they need to be.

And, point taken, it shouldn't take long to memorize the 21 raw answers and then the only counting you do is adjustment for blocking points. Thanks!