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RO

Posted By: Tom Keith
Date: Wednesday, 1 May 2013, at 9:26 p.m.

In Response To: White won't resign (Tom Keith)

Mike, Dmitriy, and ah_clem all correctly deduced the right play. Ah_clem, the only one to venture a guess on the difference, overestimated slightly but was still within .05 of the correct answer. ;-) And Tim was right about my purpose in posting the position.

For Blue to lose the game, two things have to happen:

  1. White has to get all his checkers off in only 15 rolls, and
  2. Blue has to use all of 15 rolls to get his own checkers off.

Here's the position again:





White is Player 2

score: 0
pip: 300
1 point match
pip: 53
score: 0

Blue is Player 1
XGID=--AL-o-AA-----------------:0:0:1:45:0:0:0:1:10
Blue to play 45

For (a) to happen, White has to roll 2-2, 3-3, 4-4, 5-5, or 6-6  fifteen times in a row with the added constraint that: for each 4-4 rolled, there must be an offseting 6-6; for each 3-3 rolled, two offseting 6-6's; and for each 2-2 rolled, three offseting 6-6's. (Basically, White has to get 4 crossovers with every roll.) The chance of doing that is 4.873 × 10−16.

For (b) to happen, Blue has to roll fifteen consecutive small numbers, mostly 2-1's. I have a database of bearoff positions that has the probability of a side getting off in exactly n rolls. I looked up n = 15 for the positions corresponding to the two plays and found:

  • After 8/3, 7/3: 0.000000000000000170751647 (about 1.707 × 10−16)
  • After 8/4, 7/2: 0.000000000000000061363873 (about 0.614 × 10−16)

The difference is about 1.093 × 10−16. But that's only the difference in Blue's chances of getting off in 15 rolls. Most of the time it doesn't matter what Blue does because White won't be off in 15. The difference in GWC between the two plays comes from multiplying this difference by the chance that White actually does get off in 15 rolls:

(1.093 × 10−16) × (4.873 × 10−16) = 5.326 × 10−32

So if Blue plays 8/3, 7/3 instead of 8/4, 7/2, he gives up about 0.00000000000000000000000000000005326 in GWC. (It might not sound like much, but make the same error a few decillion times and it really starts to add up!)

This position was inspired by Timothy Chow's post, where he asks: "What is the smallest possible nonzero equity difference between two checker plays in a legal backgammon position? It has to be very small."

I won't be posting a rollout. :-)

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