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BGonline.org Forums
cube-double anchor
Posted By: Daniel Murphy In Response To: cube-double anchor (ah_clem)
Date: Thursday, 5 July 2012, at 6:57 p.m.
but only 63 actually forces a shot. That's 1/18 to get a ~50% chance of hitting. Not very good odds.
63 does force an immediate shot but 42 does, also. But there is a good chance of future shots as well. From the original position, 2 numbers clear both the 8 and 7 points and 14 numbers clear the 8 point only. But 2 numbers clear the 7 point only, 14 numbers clear neither point and as noted 4 numbers immediately blot.
On Opponent's response, 33 and 22 are bad (unless the 8 or 7 point has been cleared), but Opponent can keep both anchors with 11 21 31 41 51 23 and 24 -- 13 numbers, and simply run from the 20 point with anything else.
Having looked at a few possible continuations, I think it's probably true that unless Player clears both outside points with a doublet, or clears the 8 point with an ace, Opponent rates to get between 6/36 and 9/36 blot-leaving numbers in many future positions, whether or not Opponent still has the 20 point.
I calculate the raw takepoint as 15/42 or about 35%
That's approximately R/(R+G) = 15/(15+27), R being the risk of losing a single point for 4a1aCr equity, G the gain of winning 2 points for 2a3a equity, compared to the equity from dropping and playing from 4a2a.
So far so good. But it can't be right to ignore redoubles. If Opponent adopts a strategy of redoubling immediately, his take point is 33%, which is his 4a2a MWC. But surely he can do better by not redoubling immediately. Most obviously, he won't redouble if he never gets a shot, thereby retaining his chances of winning from 4a1aCr unless he gets gammoned in this game. And if a strategy other than not redoubling immediately improves the initial taker's MWC, then his take point in the original position has to be lower than 33%, not higher.
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