Hi All,

Similar sort of topic to Yogiman's last cube question. These confuse me, but it seems to me that they are one of those parts of backgammon that can be learned with a principle or two.

So here goes:

+20AANC/ZyAAAA
:AxkAAIAAAAAA

Looks like a take when you combine hitting and race chances. And who doesn't like 16 cubes with a chance to redouble to 32?

Yes spot on. Although dropping is not a massive error.

I have two more questions if anyone would be kind enough to shed some light on for me.

1) Is it a take solely because we have a slim chance of hitting him?

2) Don't these numbers mean we have less than a 25% chance of winning and, if so, why are taking? If not, please don't point and laugh, I am still not sure how to read them:

Kamba --

Quick answers to your two questions:

1) The chance of Black hitting White is not so slim. White has 7 immediate shot numbers (52, 43, 41, and 33) so Black hits 2+ of those. There are also 7 numbers that leave indirects (11, 21, 31, 32) and Black hits roughly 1 of those 7, for a total of 3+ hits in 36 games. That's roughly 8-9% right there. Without that chance it would certainly be a pass.

2) You only need a 25% chance of winning on the very last shake, when there is no recube vig. Earlier in the game, you can take with less than 25%, because owning the cube will enable you to win a few games by doubling you would otherwise have lost, or let you win some games at double the stake.

The theoretical limit for taking with a long way to go in the game is about 20%. (In a game with no gammons.)

There's a cute bearoff position where you can take with <19% chances.

In a spoiler in case someone wants to try figuring it out. Which should be possible -- think perfect recube efficiency.

Thanks All, as always very helpful.