Checking those probabilities:
(Note: I will assume the opponents play non-queens at random, which is a safe assumption against good opponents.)
Use Pascal's triangle* to work out the basic numbers:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5
10 10 5 1
32 cases; when LHO follows once and RHO twice, you've down to the bold ones. (I'm showing the cases with more cards on the left on the left side of the table.) But four of the 3=2 (long with LHO) breaks are also gone, because the queen didn't appear on your right, and one 1=4 is gone because it didn't show up on your left, so we're left with:
- 6 3=2, all with the queen on the left.
- 10 2=3, four of which have the queen on the left.
- 4 1=4, all of which have the queen on the right.
In other words my numbers were a bit off above.
- You gain one trick in four cases of Qx xxx.
- You lose a trick half of the time in each of the six cases of xx Qxx, say three cases.
- In six cases of Qxx we don't much care.
- In four cases of x Qxxx, you lose a trick half the time, say two cases.
Net loss one case of the remaining twenty (but five of the nine cases in which it matters): it is about five percent better to finesse, if you assume everyone will bid the slam. If some pairs had not, your play would be right even at matchpoints, because against them the possible overtrick is irrelevant.
* In case it isn't obvious, you generate Pascal's triangle as follows:
- Start with 1 (which actually represents the one way zero cards can break).
- On each line below that, representing one more card, put a 1 on each end; each interior number is the sum of the two above it, offsetting them diagonally:
. . . . . . . . 1 . . (0 cards)
. . . . . . .1....1 . . (1 cards)
. . . . . 1....2.....1 . . (2 cards)
. . . .1....3.....3....1 . . (3 cards)
. . 1....4....6.....4.....1 . . (4 cards)
1.....5...10...10....5.....1 . . (5 cards)
For example, on the last line each 5 comes from adding the 4 and the 1 above it, and the 10s are each the sum of 4 and 6. The result shows the number of ways the cards can be divided, and closely approximates the relative probabilities. It doesn't exactly work for the latter purpose, but it's almost as good as you can do absent other information.