Friday, December 4, 2009

More squeeze defense

Jonathan made a good point in comments about defending this end position:

K 10
K 7 6
A 9 3
A 9

West is known to have the spade guard and, if he started with 3 clubs, must have pitched 1 card from an original 6 red cards (the defense started with 7 of each red suit). His goal is to present declarer with a straight 50-50 guess as to whether the suit he pitched from is still guarded or not. When he's 3-3 in the reds, he has no control over that, whichever suit he pitches is unguarded. This happens 40.8% of the time (7c3^2 / 14c6, see below for a faster approximation). So, when he has a long suit, he'll want to pitch from the long suit (still having a guard, 50/59.2 of the time (84.5%)) and from the short suit the rest of the time.

Given that the red lengths are symmetrical, it's easy to work out that with 3-3 in the reds you should discard each suit 50% of the time. With unbalanced red suits, you'd ideally like to do the above arithmetic to figure out the optimum discard frequencies. The problem is, this might take some time, potentially revealing that you aren't easily able to identify the optimum.

Is it therefore ethical with 3-3 to do the same calculation you would want handy if you had been 4-2?

Maybe it's not that hard. Throw out voids and having 1-5 cards in a suit is 1:3:5:5:3, so 3-3 is 25:(15+15+3+3), so with 61 holdings you want to pitch from a long suit 30.5 times and you have a long suit 36 times, or a touch more than 5/6. As evidence that this approximation is effective: 50/59.2 is 84.5%, 30.5/36 is 84.7%.

In case that last paragraph was too fast: having 1 diamond is 7c1 = 7 combos. 2, 3, 4, & 5 diamonds are 7c2, 7c3 etc, which are 21, 35, 35, and 21. There's a factor of 7 in all those numbers, so the odds simplify to 1:3:5:5:3. The ways to hold 1-5 hearts have the same combinations, but 1 diamond implies 5 hearts, 2 diamonds implies 4 hearts, etc, so flip those odds to get 3:5:5:3:1 and pairwise multiply to get the total number of ways to have both 1 diamond and 5 hearts (1x3), 2 diamonds and 4 hearts (3x5), 3d3h 25, 4d2h 15, 5d1h 3. Add them up to get 61 equally likely possibilities, of which 25 are 3-3.

In this weighting, the voids are 1/7 each and the 6 card suits are 1. (30.5 + 1/7) / (36 2/7) would be the precise calculation.

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