Thursday, November 18, 2010

Determine when to accept a double in backgammon

In this blog post I will examine the precise conditions when to accept a double in backgammon.

The question and it's answer whether to accept a double probably is a classical result in backgammon. Guided by mathematics it is primarily concerned with the expected value of a game.

To recap: If you win a game normally you win the amount indicated by the doubling cube. If you win a gammon i.e. without your opponent bearing off a single checker, you win twice the amount on the doubling cube. If the opponent hasn't borne off any checkers and still has checkers in your home board you win a backgammon and gain triple the amount indicated by the doubling cube.

Let's assume the change of winning the game normally is \(p_{n}\), that of winning a gammon is \(p_{g}\) and that of winning a backgammon is \(p_{b}\). In similar fashion define the chances of losing as \(q_{n},\,q_{g}\) and \(q_{b}\).
Furthermore, let\(v\) be the current value of the doubling cube.

If your opponent offers you a double and you refuse you lose \(-v\) points, guaranteed. But if you accept the double the expected value will be

2vp_{n} + 4vp_{g} + 6vp_{b} - 2vq_{n} - 4vq_{g} - 6vq_{b} = 2v((p_{n} - q_{n}) + 2(p_{g} - q_{g}) + 3(p_{b} - q_{b}))

So you should accept a double if the above expression is greater then \(-v\).

Lets further assume that the changes of losing are winning a gammon are backgammon are zero. Then the above equality simplifies (using the relation \(q_{n} = 1-p_{n}\)) to \(2p_{n}-1 \ge -\frac{1}{2}\) or \(p_{n} \ge \frac{1}{4}\). This is a very classic result of backgammon. For example, it is stated, without proof, in 501 essential backgammon problems, considered to be the bible of backgammon.

In order to gain an other insight we assume that the changes of winning or losing a backgammon or winning a gammon are zero, but losing a backgammon is a possibility. (For example your trapped on the bar while your opponent already beared off succesfully at least one checker.)
The above equations reduces to \(p_{n}-q_{n}-2q_{g} \ge -\frac{1}{2}\). Now the relation \(p_{n} + q_{n} + q_{g} = 1\) tells use that \(q_{n} = 1 - p_{n} - q_{g}\). This simplifies our equations to \(p_{n} \ge \frac{1}{4} + \frac{1}{2}q_{g}\). This tells use you have to have a greater change of winning if your are to accept a double when there is a change of losing a gammon.

The above equations and a keen insight in the chances at the backgammon board can help to guide your decision of taking a double or not.
In an other blog I will investigate the conditions of offering a double.

MinuteMath: A bite size math problem a day

I Recently subscribed to the The Mathematical Association of America twitter feed. The most important reason for me is their daily MinuteMath.

The MinuteMath is a daily bite size mathematical problem of varying difficulty. All the problems are multiple choice and hints, solutions and classifications are given.

It is a fun way to distract your mind for a moment and figure out a interesting little problem. If your interested start following @maanow.