Just a brief note on calculating gammon values.
Blue's gammon value is 1, or twice the normal value.
White's gammon value is also elevated -- it's .7
If you knew for sure that the cube would not be turned for the rest of the game then these values are correct. But what are the chances that the game will end with the cube on 1? Isn't it more likely that one of the players will double?
Here are the calculations with the cube on 1, as given above.
| | Blue's gammon value
(2 away) | = |
| win gammon − win single
|
|
| win single − lose single
| | = |
| = |
| = 1 |
| | White's gammon value
(3 away) | = |
| win gammon − win single
|
|
| win single − lose single
| | = |
| = |
| = .7 |
Now here are the calculations with the cube on 2.
| | Blue's gammon value = |
| win gammon − win single
|
|
| win single − lose single
| | = |
| 1.000 − 1.000
|
| | 1.000 − .322
| | = |
| = 0 |
| | White's gammon value = |
| win gammon − win single
|
|
| win single − lose single
| | = |
| = |
| = .477 |
A big difference! (And if White is the one that gets doubled, he will immediately rewhip to 4 and the gammon values will be zero for both sides.)
Another possibility is that the game will end in double/pass. In that case, it is still the gammon values on a 2-cube that are important. The recipient of a double must assume the cube is on 2 in deciding whether or not to play on. And if the doubler overshoots badly (well beyond receiver's point of last take) he diminishes the gammon value even more because there are no gammons on a dropped cube.
In summary: the gammon values in this position are much lower than they appear.