Hi There, Here are two suit combinations from Kelsey's "Matchpoint Bridge" (p. 48-49): [tt] 1) A Q T 4 2) A J T 4 7 6 7 6 [/tt] Assume that in both cases, we begin the suit by finessing the ten, losing to the J/Q respectively. Kelsey now makes 3 claims: At the outset, the odds of making more than one trick is 76 percent for both combinations, After loosing the first finesse in combination 1), the second has a 52 percent chance, After lossing the first finesse in combination 2), the second has a 68 percent chance. This is the explanation provided: "Since the overall chances of making more than one trick by taking two finesses are the same (76 per cent) for each combination, it seems strange that the second finesse should have so much better a chance on [combination 2] rather than [combination 1]. The basic reason is that with [combination 1] you used up a larger slice of your chances on the first round. The losing finesse of the ten used up 50 of the overall 76 per cent, leaving 26 out of the remaining 50, i.e., 52 per cent, for the second Finesse. With [combination 2] only 24 of the overall 76 per cent were used on the first round (the chance of West having both honours). That leaves 52 out of the remaining 76, i.e., 68.4 per cent, for the second finesse." This is impenetrable to me. It seems to me that when I count the placement of outstanding honors, in both cases I find that from the outset, I win 2+ tricks whenever east do not have both honors (3 in 4 cases), and that after loosing to easts honor, only one of the 3 winning cases are left (exactly one of the two "splitted honors" cases). If someone would care to clarify Kelseys argument and demonstrate the error of mine, I'd be most grateful. Also, why are Kelseys numbers 52, 76 etc. rather than 50, 75 etc.? I believe the context for the problem is immaterial, but here it is anyway: [tt] E-W/S J 7 6 4 3 K 6 5 A Q T 4 (A J T 4) J A K T 8 5 2 A 7 2 7 6 8 5 1s - p - 4s - a.p. Lead: CK CK J 7 5 HJ K 9 2 x S3 x A x T J D7 4 A 5 H3 ? [/tt]