Looks simple

It's the first hand of the evening at one of last week's STAC games (only we are vulnerable). After partner opens 3♦ in second seat, we find ourselves in 3NT with no opposition bidding. The lead is the ♥2 (fourth best) and this is what we see:

Dummy
♠	9 4 2
♥	10 7
♦	A Q 10 9 8 6 3
♣	6

My hand
♠	A K Q 8
♥	Q J 3
♦	5 2
♣	K Q 10 3

The opponents clear the hearts, we lead a small diamond, LHO follows with the four and it's decision time. Well, the first thing you notice about the dummy is that it's totally entryless outside diamonds (it's a great preempt, isn't it?).  The defenders have taken two tricks already and have a heart and a club to cash. So, if we're playing IMPs, we need to bring in the entire suit, so must finesse the Q to give us the best chance (about 26%) of making. At first glance, you might think that the chance of making is 50% but in practice, if we finesse the Q and it wins, we must also have a 2-2 split or see the J drop on our right.  Though even then, there's the chance of a dastardly defender deliberately throwing us off holding KJ.

In a diamond contract, there are at least 9 tricks so in effect we aren't competing with the declarers in diamonds. In any case, of 17 tables, only two played in diamonds. So, how do we maximize the number of tricks we will take, keeping in mind that the number may well be less than nine? Finesse the 10 (8 or 9) on the first round, in keeping with the Principle of Least Commitment. This play either breaks even or loses a trick to the Q play when the honors are split. But it enjoys a big win in the case where KJx is on our left and x is on our right (since we have no outside entries to dummy this layout will kill the suit stone dead if we finesse the Q). The 10 play also has a minor win in the unfortunate case where KJxx are all on our left, although this will not be a big comfort at the time. Nothing works of course when KJ are guarded on the right. See the table below for the details.

Layout	Cases	Probability	Q tricks	T tricks	Q expect	T expect
KJxx -	1	4.78%	2	3	0.096	0.143
KJx – x	2	12.44%	2	7	0.249	0.871
Kxx – J	1	6.22%	7	6	0.435	0.373
Kx – Jx	2	13.56%	7	6	0.949	0.814
Jxx – K	1	6.22%	6	6	0.373	0.373
Jx – Kx	2	13.56%	6	6	0.814	0.814
xx – KJ	1	6.78%	6	6	0.407	0.407
x – KJx	2	12.44%	1	1	0.124	0.124
Total	12	76.00%	53	61	4.536	5.157
Approx					4.42	5.08

When the "books" take a look at a suit like this, they assume that entries are available wherever needed.  However, with the excellent SuitPlay program by Jeroen Warmerdam, you can explicitly tell it how many entries accompany the suit, etc.

Since everything must be decided on the two leads towards dummy, we effectively have four realistic lines of play: Q or 10 on the first lead, A or the remaining honor on the second.  My analysis assumes that we will always guess right on the second lead but this will be easier sometimes than at other times.  In particular, if the Q won on the first trick, the Kx-Jx layout will be picked up easily.  But if the Q loses to the K, and LHO plays small to the second trick, we won't know whether he started with xx or Jxx.  Against most defenders, the 10 play may make the second guess easier because players tend to win as economically as possible. So, in a way, playing that 10 first is something of a safety play.

The best matchpoint line, starting with a finesse of the 10 and the one which I happened to choose at the table, did not give me the best chance of making the contract.  But it did give me the maximum expectation of diamond tricks.  As it happens, I lost the first trick to the J.  I was fortunate in that the defender who won the last heart trick didn't have the club ace and chose to lead a spade.  I therefore made my contract exactly.  This was worth only 5.5 out of 16 (approx 33%), however.  I shouldn't really be surprised.  At a club pairs, players perceive a premium on making contracts (and don't typically worry about minimizing the set). Perhaps in the Blue Ribbons I'd have had more company.

The table above has two columns each for the Q and the T plays.  The first is the number of tricks yielded.  The second is the expectation of tricks (the number times the probability).  By summing the expectations, we can compare the overall expectations of the two lines.  However, we poor humans cannot do these kinds of calculations at the table in practice (Chthonic would have no problem of course).  There's another, simplified, method of comparing the lines that Eric Rodwell describes in The Rodwell Files.  He suggests assuming that all possible layouts are equally likely.  This isn't quite accurate of course because, due to considerations of vacant places, KJ74 in one hand is not as likely as, say, KJ opposite 74.  That's because once the K and J have been "placed" in one hand, there are two fewer vacant places in that hand to take the 7, and if the 7 does go with the KJ, there are now three fewer vacant places to accommodate the 4 in the same hand.  However, to a first approximation, we can assume equal likelihood of each layout.  When there are n outstanding cards, there are 2^n possible layouts.  In this case, four missing cards so there are 16 possible layouts.

Once LHO plays a small card (the 7 or 4 in this case) to the first diamond lead, we can immediately eliminate four of the 16 cases. In the table above, the bottom row, labeled Approx, shows the expectations when we use this simplified method of calculation.  The two approximate values are 53/12 and 61/12.  As you can see, the numbers are very close to the accurate values.

However, for a complex situation like this one, even this amount of calculation is too much.  That's why I find myself frequently falling back on the principle of least commitment.