Off-Topic: Backgammon Puzzle

Here's a very simple position in Backgammon to show the correct use of the doubling cube. Let's assume the stake is $100. In this position White doubles. Should Black accept or refuse the double?

In this position there are 26 possible dice rolls for White that will win the game, and 10 that will lose. (The losing rolls are 1-2, 1-3, 1-4, 1-5, 1-6, 2-1, 3-1, 4-1, 5-1, 6-1). Put another way, White has a 72.2% chance of winning, Black has a 27.8% chance of winning. Since the odds are against Black, should he refuse the double?

If Black refuses the double he will lose $100.

If Black accepts the double he has a 27.8% chance of winning $200 and a 72.2% chance of losing $200. Let's put that in a formula.

(27.8/100) * 200 – (72.2/100) * 200

which is 55.6 – 144.4

which is   -88.8

That means that Black will only lose an average of $88.80 by accepting the double, an average profit of $11.20 over refusing it.

When your opponent offers a double you should accept it if you think you have a better than 25% chance of winning. This is easily proved by the following calculations.

26% chance:  (26/100) * 200 – (74/100) * 200   =   52 – 148   =   -96   ($4 profit)
25% chance:  (25/100) * 200 – (75/100) * 200   =   50 – 150   =   -100   (no change)
24% chance:  (24/100) * 200 – (76/100) * 200   =   48 – 152   =   -104   ($4 loss)

In this position it's easy to calculate the odds. In most positions it isn't.