[aleatory]

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first trick goes ♠2 - 7 - Q - ruff
trumps split.
bidding aside, what should declarer be thinking here?

[nige1]

aleatory, on 2015-July-12, 11:11, said:

first trick goes ♠2 - 7 - Q - ruff
trumps split.
bidding aside, what should declarer be thinking here?

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OK, I can't see any clever line. You have 11 top tricks and need 2 more. If you ruffed out ♣Q that would only provide one extra trick, so there doesn't seem much point in doing that. Hence: draw trumps, cash ♦A, ♠A, ♣AK. Unless ♣Q drops, run ♦T and repeat the finesse.

[BillPatch]

Without any clues from the bidding or carding, it is even money between nige1's line and playing for the drop in ♦ first else finessing clubs twice against W. But we have the opening lead of the 2. If this is not a false card, and the opponents lead fourth best, the alternative line is better, since W has at least one more free cards than E. (West has 4 spades, E has 5). If opponents lead 3rd,5th, Nige1's line is best.

[BillPatch]

BillPatch, on 2015-July-14, 11:56, said:

Without any clues from the bidding or carding, it is even money between nige1's line and playing for the drop in ♦ first else finessing clubs twice against W. But we have the opening lead of the 2. If this is not a false card, and the opponents lead fourth best, the alternative line is better, since W has at least one more free cards than E. (West has 4 spades, E has 5). If opponents lead 3rd,5th, Nige1's line is best.

Sorry, poor choice of words. "Free cards" describes squeezes. Substitute "places open." If W has 4 spades, W will have 13 minus 4 spades + 2 hearts+ 2 diamonds + one club = 4 places open. E has 13 - 5♠ + 2 ♥ + 2 ♦ + 1 ♣ = 3 places open. West will have the Club Q 4/7 in relation to the seven open spaces or 57% of time, Because of the symmetry of the problem, East will have the ♦queen 3/7 or 43% at the same point if we take Nige1's line. Another possible true card lead from systemic fourth card leaders is third best from three. Since this will give one more additional "space open" for W, or 2 more "spaces open". this will increase the odds of the alternate line.

Note that it was not necessary to actually go through and calculate the number of actual "spaces open" to solve problem given 4th best leads. One excess place open was sufficient. The rest is commentary, for the edification of the reader.

[BillPatch]

Proving the superiority of Nige1's solution required extensive computation, which I doubt I could estimate well at the table. With 3rd-5th leads the lowest card lead indicates 3rd, 5th or 7th best. East play of the Q indicates that W has the J. I enumerated 6 possible 3 card suits with j and 2. 20 possible 5 card suits, and 8 possible 7 card suits. This time we must calculate all the "places open". 3 card W = 2 places open of 7; 5 cards W = 4 of 7; 7 cards W = 6/7
[(31% * 2/7 * 6/84) + (4/7 * 58% * 20/126) + (6/7 * 8/36 * 9)] works - [(31% * 5/7 * 6/84) + (3/7 * 58% * 20/126) + (1/7 * 8/36 * 9%)] fails
Since this is positive Nige1's method is superior.