From the ABT Facebook page, interesting problem late in a money game:



What's our play here and why?

Wow... really tough move. Gotta ask paps for this one. Will get back later!

I play 2/0 (2). Locking in on the backgammon. Even if you get hit and closed out, you would need a lot of bad things to happen to lose the game. A lot.

Stacking the 1 is basically a guaranteed gammon (he needs 3 sufficiently large doubles or to roll 11 and then hit you) and +16. If you rip 2, it breaks down as:

Gammon: 66, 55, 44 (3)
Backgammon: 65, 64, 62, 61, 54, 52, 51, 42, 41, 22 (19)
Hit/cover: 3
Hit: 11

Worst case hit/cover you win a single 92% (ok you get cubed occasionally, but you also enter immediately which should more than balance)

So lower bound of (24*19 + 16*3 + 8*14*.92 - 8*14*.08)/36= 16.61

Looks like you rip 2.

Edit: ****, you backgammon on doubles if you stack the 1, so that needs to be 16*5/6 + 24*1/6 = +17.33 which means you have to calculate all the crap for when you get hit and he doesn't cover. It has to be close.

2 off seems right

Let E(...) stand for expected winnings, and P(...) stand for probability. To simplify things, we assume the cube is at 1. All E( ) values can be multiplied by 8 later.

If we assume that Blue wins neither a gammon nor backgammon after being hit, we have:

E( rip 2 ) = 3*P( White misses and leaves checker in rear quadrant )
+ 2*P( White runs to outer board with 44 55 or 66 )
+ P( White hits )(92% - 8%)

E( rip 2 ) = 3*(19/36) + 2*(3/36) + (14/36)(92% - 8%)

The figure 8% comes from Stick’s chart of winning percentages against an opponent with N checkers off, and 1 checker closed out.

For the next calculation, assume that rolling a set wins the backgammon, but otherwise White loses a gammon.

E( 3 checkers on 1pt ) = 2 * P( White rolls 33 44 55 66 ) + P( White does not roll 33 44 55 66 ) * (3 * 1/6 + 2 * 5/6)
E( 3 checkers on 1pt ) = 2 * (4/36) + (32/36) * (3 * 1/6 + 2 * 5/6)

E( rip 2 ) = 2.08
E( safe ) = 2.15

Mulitply these values by 8, to get the expected winnings when the cube is at 8.

By these calculations, Blue should play 3 checkers down to the 1pt. But these are close enough that my assumptions are probably invalid. In addition to the ones stated above, you can see in my arithmetic that I have also assumed that White can always complete the closeout after hitting. Perhaps Blue wins more than 92% after being hit. As well, some of those victories may be gammons.