Problem of the Week #133: January 8


(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.

a) 2/1*(2) 7/5.

I don't want white to anchor. To much gammons lost. And then diversification.

b) 2/1*(2) 7/5.

Dito. And we need only one six to escape. I don't want to duplicate my fives.

In both parts of this week’s problem, I start with the hit, switching points from the 2pt to the 1pt. Black will probably win more games and more gammons by putting a third White checker on the bar. The alternative is to leave White’s blot unmolested on the 1pt, where it becomes an asset for him, by giving him a head start on making an anchor.

Trying to “buy out” of 6s is entirely the wrong concept. In Part (a), for instance, Black could play 7/6 8/5, thus delaying the collapse of his prime. The only 6 Black could play thereafter is when he escapes a rear checker. Other 6s would be blocked. The flaw, of course, is that White would then anchor 30% of the time on the very next throw.

In both problems, I would not touch the back checkers. As they sit, Black can escape with two different numbers. Were he to play 23/22, he would have only one escape number.

After that, it comes down to diversification. In Part (a), Black’s escape numbers are 4 and 5. In Part (b), they are 5 and 6. As he plays the final two aces, Black must avoid duplicating these numbers, while diversifying his spares at the same time.

In Part (a), that’s relatively easy. After moving 7/5 2/1*(2), Black will have 3 and 6 to hit an entering checker on the 2pt, and 4 and 5 to escape.

In Part (b), there is no such neat solution. No matter how he plays, one of his escape numbers will be duplicated. After 7/5, Black will have 3 and 6 aiming at the 2pt. When he plays 8/6 instead, he will have 4 and 5. Either play will work if his next roll is 44. There is no advantage one way or the other with any other doublet, as well. Another alternative is 3/2(2), but it also leaves one of Black’s escape numbers duplicated.

My instinctual preference is for 8/6. I am, however, a bit concerned about this. Given the QF (quiz factor), the fact that I cannot identify some discriminating aspect that differentiates the alternatives means that I have probably missed a major feature of this week’s problem.

My solutions:

Part (a) — 7/5, 2/1*(2)
Part (b) — 8/6, 2/1*(2)

For the Record
I am so often wrong that I like to post my record in these messages. It's kind of a truth-in-advertising thing. Grunch: I have been answering these problems without the use of a bot, and before checking the excellent solutions of others, since Problem 28. My record at this writing is 52%.

7/5,2/1*(2) both. Would rather have 6s duplicated in b than 5s.

(a) I'd go with 2/1*(2) 7/5, so I can escape with a 4 or a 5 and hit back with a 3 or 6 if necessary.

2/1*(2) 7/5


(b) 2/1*(2) 8/6 would leave 5s and 6s to escape and 4s and 5s to hit back (5s duplicated). 3/1*(2) would yield the same result, but with the hole on the 3-pt instead of the 2-pt. It's probably better to have the hole on the 2-pt, so I would go with the former.

2/1*(2) 8/6

I think I agree with higonefive and networth: it seems better to duplicate 6s because there are 1 checker that needs a 6 to escape compared to 2 checkers that need a 5.

I can see the point. You may even be right!

But I can also make this counter-argument. The pieces aren’t frozen. Wherever Black chooses to leave his blots/spares in Part (b), he will be looking to move the one that duplicates his escape number as soon as possible. He needs to get it onto a point that is not duplicated. Since he has several numbers that he can use to shuffle his spares effectively, it is highly probable that he will never face the problem of needing to roll his escape number three times. More precisely, if Black chooses to hit with his escape number, it will more often be the first time he rolls it, than it is the second. By the time he rolls the duplicated number a second time, he will already have moved the duplicated spare.

7/5 2/1*(2) looks right.

Putting him up with switch hit is automatic, devil is in the duplication details.

I agree with the posters who have gone for:

(a) 2/1* (2) 7/5
(b) 2/1* (2) 8/6

...for purposes of minimising duplication.