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BGonline.org Forums
Daily Quiz 3/30/11
Posted By: Chuck Bower In Response To: Daily Quiz 3/30/11 (Stick)
Date: Thursday, 31 March 2011, at 1:45 a.m.
Problem 1. OK, here goes my attempt to apply the method I discussed in this thread: http://www.bgonline.org/forums/webbbs_config.pl?noframes;read=92761. Assume not getting hit is a win and getting hit is a loss. Look at the next sequence (Beige roll, Brown roll). Beige wins by racing past Brown (any roll with 9 or more pips = 12 rolls total). 8/36 rolls leave beige on the midpoint. Brown wins immediately by hitting (6 rolls) and we reach a repeater otherwise (30 rolls). That translates to (8/36 * 6/ 36) = 1.33/36 losses and five times that many (6.67/36) REPEATERS (see above link). The remaining 16 Beige rolls force a shot which gets hit on average about 31% of the time. Those multiply out to 5/36 losses and 11/36 wins.
So in summary, cubeless, (5 + 1.33)/36 * (-1) is the equity for losses and (12 + 13)/36 * (+1) for the wins. For games decided in this sequence, the equity contribution is (23 - 6.67)/36 = 16.33/36 = 49/108 ~ (12/13)*0.49 = 0.49 - 0.49/13 ~ 0.45. From the link, this represents (1 - repeater/36)*E(o). Thus E(o) [= E(r) = E, the equity of this position] is 0.45/(1 - 6.67/36) = 0.45/[(108 - 20)/108] = 0.45*108/88 ~ (4/9)*(9*12)/(4*22) = 12/22 = 6/11 = 0.55 = the cubeless equity for Beige. That is definitely a redouble and a tough choice on the take/pass, but it sounds like a (close) take.
Note that these 'repeaters' are really pseudo-repeaters and in fact the analysis over-favored Beige a bit, so his equity is less than 0.55, making the take clear, but probably still > 0.5 meaning (with so many ML's) it's a redouble. Let's see how well my method works in practice.
Summary (if you don't like the above two paragraphs): R+T.
Problem 2. OK, I've found a label for these kinds of problems: WDF, not to be confused with its close relative -- WTF. (In German: Warum diese frage? Hope that's not too much of a butchery of that language. Corrections welcome.) I only see 10(2).
Problem 3. Beige has very few bad rolls this turn. I suppose this could be a Take, but Robertie said "one really good thing about 6-primes is that when things go bad they turn into 5-primes". Throw in Brown's inefficient blot on the acepoint. And if it really is a Take, I just made the double look brilliant. D+P.
Problem 4. With Brown's good board Beige doesn't want to get hit, which means being forced to safing the outfield checker before improving the homeboard. Generic 20-24 games are passes after the midpoint is surrenedered, but this looks like an exception for a couple reasons (the one just mentioned plus the situation in Beige's homeboard). D+T.
Problem 5. Very crudely (but that is often enough for an Meq calculation), say Brown wins a gammon on all 3's (which double hit) plus double 5's for a total of 12/36. Further, failing to roll one of those means no gammon. ==> 33% gammon wins for Brown.
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