Problem of the Week #147: Solution


Cash game, Black owns a 2-cube.




Black to play 2-1.



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.


White got 10 men off, but then got hit while Black still had a chance to make a good prime. Black is about to close his prime, but White has a second checker exposed in his home board. Black has a few choices with his 2-1. What should he do?


We’re deep in the world of the post-ace-point game, where one side gets some men off but then gets whacked. I dubbed these post-ace-point games years ago in Inside Backgammon, but they can arise from all sorts of contact positions, including deuce-point and three-point games, back games, and positions where one or two checkers got closed out. These games can involve a lot of intricate tactics and non-obvious cube action.

Before we start talking about Problem 147, let’s look at two simpler versions, where the right idea is clearer. We’ll start with 147a:




Problem 147a: Black to play 2-1.

In this position, we’ve moved Black’s checker on his ace-point back to the outfield so his choices are now (1) 4/1, slotting the ace-point; (2) 10/7, trying to build a full prime first; (3) something else, perhaps 10/8 and 12/11.

First, let’s note a couple of facts. If Black closes the ace-point without hitting a second checker, his winning chances are around 36%, depending on exactly where his spares end up. If he hits the second checker and completes a closeout, his winning chances are in the 79% to 84% range, again depending on just where his three spares are.

Clearly, hitting the second checker is so important that Black is entitled to take huge risks to get it. In 147a, the right play is 4/1, starting the process immediately.




Problem 147b: Black to play 2-1.

In 147b, Black can’t start the ace-point yet, so his best plan is 10/7, preparing to build a full prime and getting another hitter if White enters.


Now back to our original problem. We know that leaving the ace-point slotted must be right because we need to hit the second checker. And now we know that slotting the bar-point is a useful idea. So isn’t 10/7 the correct play here?

No. Too see why, let’s look at the rollout results for the two main non-covering plays (10/7 and 10/8 12/11) and see what happens in each of four variations:

(1) White fans (probability 25/36)

(2) White enters with 1-2, 1-3, 1-4, or 1-5 (probability 8/36)

(3) White enters with 1-1 and saves his blot (probability 1/36)

(4) White enters with 1-6, hitting one or two men (probability 2/36)

Here’s what the chart looks like:




As we sift through the four sections of the chart, we can see pretty clearly what’s going on.

If White fans next turn, the two plays are almost equivalent. That’s to be expected, since this variation mostly ends in a closeout of one checker. Whether Black plays 10/7 or 10/8 12/11, he quickly runs out of time to wait for White to enter from the bar.

The variation where White rolls 1-2, 1-3, 1-4, or 1-5 is Black’s best, and 10/7 is the play you want to have made here. Black gets a shot at a second checker, and when he misses, he wants to immediately hit on his ace-point to start the process again before White can pick up with 2/1. Playing 10/7 gives him an extra few rolls to enter and hit. (Notice that the primary function of the checker on the 7-point is as a hitter, not as the start of a full prime. Blacks lacks the time to prime.)

When White rolls 1-1, you again want to have moved 10/7. Here the checker does serve a priming function if your entering roll contains a five.

Now for the big ‘oops’: White rolls a 1-6. It’s a great shot for White, but it’s a really great shot if you slotted the 7-point. White hits two men, and his gammon chances vault to 26%. That’s a huge swing in favor of 10/8 12/11.

Putting these results together with their likelihood makes 10/8 12/11 the winner by 0.02. However, it’s only the presence of the 1-6 roll that makes 10/7 the wrong play. Take that single roll out of the picture, and 10/7 becomes the winning play (also by a small amount.)


This problem might not look too important, but I chose it because it’s a good example of a phenomenon that occurs a fair amount in middle game and endgame play; I call them “single-bullet problems”. A single-bullet problem is a situation where there’s a thematic positional play which is best against all rolls but one, but it’s so bad when that roll happens that the verdict swings back to favoring a less constructive but safer play.

These plays are very hard to spot over the board, because we can see that the positional play is clearly better most of the time, and it’s hard to convince ourselves that a play which is right 34/36 or 35/36 of the time can lose enough equity against that one bad shot to be second-best overall. As a consolation, these plays are usually not big errors. Typically they’re only wrong by something in the 0.01 to 0.03 range. But if you remain aware that a single bad roll can represent a decisive swing, you’ll have a better chance of seeing them.


Solution: 10/8 12/11

By curiosity, why would 10/8 12/11 be better than 12/9?

If White enters back with 6-1, Black would have the same hitting numbers after both plays (21 in each case) and if White fails to enter or enters and stays on the ace point, then it provides Black with the same additional covering/hitting numbers: 6-1 and 6-4 after 10/8 12/11, 6-2, 4-4, 2-2 after 12/9.

Would it be because Black will have a smoother bear-in?

If you play 10/8 12/11 you do not duplicate 2's to hit on both sides of the board so you have 6 numbers that hit BOTH blots if white rolls 61 (21, 61, 41) but if you play 12/9 Duplicating 2's then you only have 5 numbers that hit both blots after White rolls 61 (22, 32, 62).

To win you need to hit both checkers BUT hitting them both IMMEDIATELY is very important as if you only hit one then your opponent enters and moves the second checker then you hit the second checker he could easily anchor on your 1 point (as you now have a checker there) which you want to aviod. If you hit both together then only 11 anchors.

I didn't actually count the numbers after White rolls a 1-6; I just relied on diversification. After 10/8 12/11 and 1-6 hitting, Black needs 2s to hit in White's inner board and 1s and 4s to hit in his outfield. After 12/9 he needs 2s and 3s in his outfield plus 2s in White's inner board.

Makes sense. Thanks Grant and Bill.