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BGonline.org Forums
A comment about the takepoint formula
Posted By: Daniel Murphy In Response To: A comment about the takepoint formula (Phil Simborg)
Date: Saturday, 17 April 2010, at 8:27 a.m.
In the classic position where each player has 1 checker on his 6 point, it is precisely because every redouble will be taken that it is a take. So in this case, I can clearly see the need to understand that aspect and not just whether or not it is an efficient recube.
With one checker each on the six point, the first double is a take, not because the redouble will be taken, but because the redouble will be given exactly 25% of the time, with exactly 75% winning chances. The taker of the first double rates to lose exactly the same number of points whether he drops the first cube, or takes and his redouble is taken, or takes and his redouble is dropped.
And the fact that the take of the redouble is optional means that the redouble is always perfectly efficient. Right?
To the larger point that counting redoubles taken and redoubles passed can be exceedingly difficult in positions more complicated than that one, I concur!
So it seems to me that just knowing, as MCG says, how often you have an efficient recube is all you need to look at.
Oy ... if I could always figure out just that much, I'd be a bgenius.
But let me ask a question: in the problem position, don't (a) the redouble window of ~27-~62% and (b) the fact that redoubles will be racing doubles suggest that redoubling efficiency will be higher than, well, higher than it might have seemed, looking at the one checker trapped behind a 5-prime?
And another: isn't it often enough just to know how many games we win with a redouble, without figuring out how efficient our redoubles will be?
"Bgenius" is not in the glossary. It should be!
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