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Old 01-09-2012, 12:01 PM   #1
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Join Date: May 2004
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Problem of the Week #133: Solution

Problem of the Week #133: Solution


(a) Money game, White owns a 2-cube.




Black to play 1-1.


(b) Money game, White owns a 2-cube.




Black to play 1-1.


Note: All ‘cash game’ problems assume the Jacoby Rule is in effect. That is, you can’t win a gammon unless the cube has been turned.


The two parts of Problem 133 illustrate an idea called diversification, which is the opposite of duplication. Duplication enables us to reduce our opponent’s good shots by moving so that the same number works for him in different parts of the board. Diversification increases our own good shots by giving us different numbers working in separate parts of the board. It’s a quick and useful tool that will generally point to good plays.

Problem 133a shows the idea in its simplest form. With two of his aces, Black will of course play 2/1*(2), knocking out White’s chance at an ace-point game and buying some more time to release his back checkers. Now he has to decide how to arrange his remaining spares. On White’s side of the board, he needs fours and fives to escape, so on his side of the board, he’ll try to give himself different numbers to attack the open 2-point.

Notice that if he plays 8/6 with his last two aces, he’ll need fours and fives to hit on the deuce point (assuming White enters with a checker). Those are the same numbers he needs on the other side of the board, so in effect he’s duplicated his own good numbers – not the right idea.

A better play is 7/5 with his two aces. That gives him sixes and threes to hit on the 2-point, different numbers from those he needs to escape. By making different numbers work on different sides of the board, you maximize the chance that your next roll will accomplish something.

Problem 133b shows a more difficult situation. This time Black needs fives and sixes to escape instead of fours and fives. Now we have to think a little.

After the mandatory 2/1*(2), Black can play 7/5, giving himself sixes and threes aiming at the 2-point, or 8/6, giving himself fives and fours. Since he needs fives and sixes on the other side of the board, either play duplicates a number: 7/5 duplicates his sixes, and 8/6 duplicates his fives. But notice that the duplication isn’t equal: Black needs two fives to escape his back checkers, but only one six. Since Black will always prefer escaping to hitting (because escaping is harder, and you always do what’s harder first), he won’t hit with a five until he’s first thrown two fives to escape. That makes leaving a five to hit less useful than leaving a six to hit, so the right play is once again 7/5.


Solution: (a) 2/1*(2) 7/5
(b) Same
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