Trying to understand the doubling cube decision/calculation in xG2

Simple scenario :
7 point game, current score 4-0
cube is @ 2 and I own the cube
current game is 74%/26% chance of winning (and can assume prob G,BG is effectively 0)

Using standard (Kazaross) MET
No double : if I win score it 6-0 => 90.72% MWC, if i lose score is 4-2 => 64.79% MWC so simple weighted value = 84% (pretty close to XG2)
With double : if I win game I win match => 100% MWC, if i lose score is 4-4 => 50% MWC so simple weighted value = 87%

however the NN gives a redouble/take value of 73.7% --> quite a big difference.

only thing I could think of is that the MWC difference (in my basic calculation) is small but in doubling i lose future control of the cube which might have more value than I think. Is there anything else going on?

thanks in advance

If you double to 4 and he takes and you lose, the score won't be 4-4. Your opponent will immediately redouble to 8 so you'll lose the match when you lose.

ok thanks - so its down to the fact you give away control of the cube and he has nothing to lose at 0-4 (losing 0-8 vs 0-12 same result) where as I have significantly more downside

Exactly.
So you were right, cube control had (way) more value than you thought in this specific case.

It would be very rare for you to have a double/take in this situation. Your opponent has an automatic redouble. If he takes and loses he loses the match. That’s still the case if he redoubles. If he manages to win the current game, though, he would win the match if he redoubles, while that isn’t the case if he does not.

You are risking the match on this game by doubling. Therefore you should not double unless your win probability in this game is higher than your current match winning chances. How high would that be? A little algebra is needed. Let’s approximate your winning chance if you don’t double at 91% if you win and 65% if you lose. If p is the probability that you win the current game, then your current match winning chance is 0.91p + 0.65(1-p). We set p equal to this and solve for p.

p = 0.91p + 0.65(1-p) or p = 0.26p + 0.65, 0.74p = 0.65, which gives p=0.878.

Therefore you should not double unless your chance of winning this game is greater than 87.8%. In this case, you’re at 76%, so no double. If you look from your opponent’s point of view, dropping your cube leaves him at 9% MWC. Therefore if he’s got at least a 9% chance in the current game he should take. This indicates that double/take in a situation like this is rare. Your doubling window is 87.8 - 91%, a very narrow one that’s easy to miss. That becomes even more true in a gammonish position.

In such positions, you might even hit a pardoxical too good/take. Consider a hypothetical position where you win single 50%, win gammon 40% and lose 10%. With no gammons, it’s good enough to double. With gammons though your MWC is 40% + .5(91%) + .1(65%) = 92.5%. This is greater than the 90% you have after you double, so no double. You’re too good because making your position worse, namely 0 gammons 90% single wins would make it a double. Of course your opponent still has 9% MWC if he drops, so it’s a take.