Came across a wonderful backgammon position today (on DailyGammon). At first glance, it is totally trival. All the complications are gone. In three rolls the game will be decided. 

Still I managed to squeeze a blunder of nearly 0.2 equity out of it. After 10 minutes of thinking. With a piece of paper. And a calculator.

That’s how mysteriously complicated backgammon can be, even in its most simple positions.

Here we go:

This is a match to 15, score is 2:2.Should white double? Should black take?

I have still not figured out how to solve this one.

What’s your suggestion?

 Spoiler starts here.

The trivial part is: 

• if white rolls 44, 55, 66, white wins (1/12 probability)
• if white rolls something else, and black rolls a double, black wins (11/12*1/6 probability)
• if white rolls something else, black rolls no double, then black has only a single checker on 1 and will win for sure. Unless white can bear off the three stones in two rolls.
  But how likely is this?

I was black, my opponent (correctly) doubled.

I arrived at the conclusion that I need 25% wins to take, the second dot gave me only 15%. So I need to win about 10% of all games in case 3. Since even with a 21 roll, white still has many combinations to win, I concluded that nearly all 2-roll combinations will bring the three checkers home and passed.

Tuns out, the correct cube action is double/take. In this position, black will win almost 30% of games (28,97%, actually; I even did a rollout).

I still have no clue how somebody might solve this OTB. And I’m still amazed that such a simple position is so difficult to solve. But that’s the beauty of backgammon.