# Games: Backgammon

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Here is a more difficult example of a "Pay me now/ Pay me later" problem. In the diagram position Black has a 5-3 to play. He has already borne off 3 men. He has two choices: (a) 3/Off or (b) 6/3. In both cases he cannot play the 5. Note that the rules of backgammon state that you must play as much of your roll as possible. If you can play only one of two numbers, you must play the larger if you can. Here you cannot play the 5 whatever you do.

Play (a) leaves 11 immediate shots - all the twos. Play (b) leaves 20 shots - all fours and ones. Normally in such a problem the correct play is the one that leaves the least shots. However, in this case it is worth looking at what happens if White misses the first shot.

With play (a) Black will leave a shot only if he rolls 6-6, 5-5, 4-4, 3-3 on his next turn - a total of four rolls. With play (b) he will leave a single shot with 6-1, 6-2, 6-4, 6-6, 5-2, 5-4, 5-5, 4-2, 4-4, 3-1, 3- 3, and 2-2 (19 rolls), and a double shot with 6-5 and 5-1 (four rolls). Even if these shots are missed Black will still have a difficult position in many of the cases, and will often be hit on the third or subsequent roll.

Quite often in backgammon intuition can play a part, and to those experienced in the game the position after play (b) just looks so much better than the position after play (a).

The disjointed nature of the black position after play (a) makes play (b) the right choice in this instance. White will win 50 per cent of the time after play (a) but only 45 per cent of the time after play (b). Note, though, that this type of play, leaving a significant number of extra shots in the original position, is correct only when the alternatives leave a position which will lead to large numbers of additional shots on the very next turn.