Let's first say that, hypothetically, the cube would never be turned again from the posted position.
White' winning chances are 31/36 x 1/6 = 31/216 = 14,35%.
Therefore, Black's winning chances are 85,65%.
Now, if I refer to Woolsey's MET (1994,
http://www.bkgm.com/rgb/rgb.cgi?view+801), here are the leader-trailer chances:
At 1-away 7-away Crawford = 91% - 9%
At 3-away 5-away = 66% - 34%
So, right now, if we would not factor in the cube handling:
White has a 14,35% X 34% + 85,65% x 9% = 12,59% of match winning chances.
Black has a 14,35% x 66% + 85,65% x 91% = 87,41% of match winning chances.
Now, let's say Black doubles.
If White drops, it becomes 6-0 Crawford (1-away 7-away) and White has 9% of Winning the match.
If White takes, he will theorically re-double, which means that when White wins the game, he wins the match as well. Therefore, that comes down to the initial chances of 14,35% for White.
Clearly, in (a), Black should hold the cube and White will take if he's doubled.
Now, what about in (b) ?
I guess we may have to assume that White, being a weak player, will miss the re-double (or even drop Black's cube!)
If that's the case, When White wins the game, the match becomes 4-4 and it's theorically 50-50 (I say theorically, because if White is in fact weak, it's a bit more than 50% for Black). But when he loses the game, he loses the match.
Therefore, White's winning chances would go like this:
14,35% x 50% + 85,65% x 0% = 7,18%
So if Black suspects that White will either drop the cube or forget to re-double, than Black should double.
(a) No double, take
(b) Double (suspecting either drop or take with no re-double)