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BGonline.org Forums
Re: Position from MBC
Posted By: Chuck Bower In Response To: Position from MBC (Matt Cohn-Geier)
Date: Wednesday, 28 March 2007, at 9:56 p.m.
Here's how I would (try to) attack this problem OtB.
First do the Match Equity calculation. If you pass you trail -7,-2. That's about 15.5%. If you take and recube it's for the match. So, if you take you need to win at least 15.5% of games. Can you win that much?
You either get one chance to roll big doubles or two chances. If opp rolls doubles his first or second try you get only one chance. Opp rolls doubles on either his first or second try 1 - (5/6)*(5/6) = 11/36. Call that 30% of the time. Thus 30% of the time you must roll big doubles (immediately) to win. 70% of the time you must get off in one or two rolls.
Case A) 30% of the time opp rolls doubles. You need 55 or 66 (2/36 = 5.5%). So in case A you win 0.3 * 0.055 = 0.0165 (call this 1.5%).
Since you need to win 15.5% total and you win 1.5% of the time when opp rolls doubles, you must get (15.5% - 1.5% =) 14% wins from the other scenario. You're given a chance at that scenario (as shown above) 70% of the time so you must be able to bear off in one or two rolls 14%/70% = 20% of the time. Can you?
Case B) You'll roll 55 or 66 on your first roll 5.5% of the time and on your second roll 5.5% of the time. Add those two up and you're at 11%. Can you get off just 9% without rolling 5-5 or 6-6?
Case B') Since you get two rolls, that's 1296 different outcomes. 9% of 1296 is about 117. We need to count two roll sequences that add up to 117.
45 or 46 followed by 65 or 33 or 44 = (2+2)*(2+1+1) = 16.
65 followed by 45 or 46 or 65 or 33 or 44 = 2*(8) = 16.
44 followed by 11 or 22 or 33 = 3.
Ugh. That's only 35 and we're trying to get to 117. I don't think we're going to make it.
D+P
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