First consider the simpler situation that Fred described, where the defenders have all the top spades except for the ace. In this situation, if you were playing an infinite number of hands against this perfect computer, with 5152 you must discard 2D & 1S 1/5 of the time. There are a few ways to see this. One way is as follows: the only relevant layout in which LHO might discard 2D & 1S are 5152 and 2452. On the given information, the former is 5 times more likely, because there are 7*1 ways to put one of the remaining seven spades and all of the remaining three hearts in LHO's hand and 35*1 ways to put four of the remaining seven spades and all of the remaining three hearts in LHO's hand. If you discard 2D & 1S more or less frequently than 1/5 of the time with 5152, declarer will have better than a 50-50 guess about which of the two layouts exists. The question is more complicated for the original problem. The play in diamonds is more flexible because the SK has a greater effect on declarer's strategy (and secondarily, because RHO can vary the frequency with which he plays the SK when he has the heart shortness, based on the defenders' diamond strategy). The short answer (if I haven't made a mistake) is that for any possible optimal defensive strategy, LHO must discard 2D & 1S from 5152 between 6/35 to 34/35 of the time. Long answer that most people will probably not want to read: There are 9 relevant holdings for LHO: S H D probability*140 x Jxx xx 6 xxxx -- xx 15 Kxxx -- xx 20 xx Jxx x 30 xxxxx -- x 12 Kxxxx -- x 30 xxx Jxx -- 20 xxxxxx -- -- 1 Kxxxxx -- -- 6 If the defense discards small spimonds randomly and RHO does not play the SK unduly often, declarer can do no better than play RHO for Jxxx, picking up 84/140 = 3/5 of the layouts. If LHO discards diamonds more or less frequently than random, RHO may or may not be able to make up for this with his SK strategy. If LHO discards 2D & 1S from a holding including short H less than 6 times every 140 relevant hands (only on 35 hands is this possible), then declarer will gain by playing LHO for the long H when he sees 2D & 1S. Likewise LHO needs to discard 1D & 2S from short hearts at least 30 times every 140 relevant hands and 0D & 3S from short hearts at least 20 times every 140 relevant hands. These three conditions are also sufficient to ensure that RHO has an optimal SK strategy. In each of the three cases (n = 6, 30, 20), short-hearted RHO just needs to divide out his n "last discards" between SK and non-SK in such a way that LHO is more likely to have short hearts in each case (for example RHO could play SK exactly balancing RHO's forced SK's, and non-SK slightly less than RHO's forced non-SK's). Since LHO needs to discard fewer than 2D (from a short H holding) on at least 50 out of 140 relevant hands, but there are only 49 such hands on which LHO actually has fewer than 2D, he must discard 2D on at most 34 of the 35 relevant hands with 2D and 1H. I assumed RHO discards randomly from small spimonds and that this doesn't affect the result. Maybe that's not true, but this is more than anyone wanted to read anyway :) I was just curious what the answer was.
