Some edits thanks to later posters especially gnasher. Interesting that none of your experts mentioned the third line, cashing the ace first. If the suit is 5-0 you cannot take 4 tricks. If the suit is 4-1, you can take 4 tricks if RHO has stiff K (cash the A, come to hand, lead to the T), or 1 4-1 break. That looks like the only way to take 4 tricks when the suit is 4-1 assuming we don't divine to finesse the T after (starting in dummy) x-K-x-x. If the suit is 3-2, then we get the following. Low to the T wins whenever the J is in LHO's hand, plus whenever the K is doubleton (Jx {3 cases}, Jxx {3 cases}, xxx {1 case}, KJ {3 cases}, KJx {3 cases}, Kx {3 cases}). Thats 14 3-2 breaks, or 47.5%. Low to the Q wins whenever the K in in RHO's hand, plus whenever the J is in LHO's hand. Specifically we pick up RHO holding (Kany {4 cases}, Kanyany {6 cases}, xxx {1 case} and xx {3 cases}). That's 14 3-2 breaks or 47.5%. Cashing the A wins picks up the first 11 3-2 cases listed above for low to the Q, plus Jx {3 cases}, plus the one 4-1 case, totaling just over 50%. I'm getting more appreciation for the work J-M Roudinesco did! I could well have missed a case or two here, I probably need my head examined for trying to work this out. I used to own a copy of Roudinesco but no more. Unfortunately it seems to be out of print. If someone knows where to come by a copy, please let me know. By the way, such preference cycles ( eg A > B > C > A ...) are known as Condorcet cycles in the literature on voting. curt