by Matthew Kidd

♠KQT6
AT4
AT4
♣876

♠43
KJ653
J987
♣A9

The auction was 1♣ (LHO), Dbl, 1, 3; Pass, 4. The K is led. Plan the play at matchpoints.

Perhaps your 3 call was an overbid but you had reasonable expectations of a nine card fit where ~23 HCP often makes game. Partner seems to think highly of your declarer skill to raise to game on 4-3-3-3 shape. Most pairs will not be in game so you should get a good board just by making it and thus play it as at teams.

You have a spade loser and a diamond loser. If the lead had been the ♣K, you would have a definite club loser too and the play would be simplified to finding the Q. You would place West with the ♣Q and figure East for the KQ or one diamond honor and the ♠A. So West would need the Q to have a respectable opening bid. Sure, East could have the light Q, Q, and ♣J hand but might well have passed on such quacks. Or he might have the ♠A and Q, and no diamond honors but this is unlikely. Playing West for the Q would be the percentage action.

But the K lead introduces a new possibility. If the ♠T finesse succeeds, declarer’s losing club could be pitched on the third round of the suit. Or declarer might finesse twice against the ♠A, though this seem anti-percentage because it plays East for the quack filled hand, at best Q, and ♣KJ.

What can we infer about the defensive honors from the K lead? First, West is less likely to have ♣KQ because with two similarly appealing leads, he would make the other lead about half the time. This is the logic of restricted choice—it’s roughly 2:1 against West holding the ♣KQ. Second, the 18 defensive HCP are probably split 11-7, 12-6, or 13-5. So West rates to have the ♠A along with the known K, either the ♣K or ♣Q and probably at least one of the Q or ♠J. If West holds the ♣KQ and the known K, East rates to have ♠A because otherwise East’s 1 bid was on 5 HCP in quacks. So again West probably has at least one of the Q or ♠J.

What can we infer about the distribution? Spades are 4-3 because with 5+ West would have opened 1♠ or East would have responded 1♠. The K could easily be a singleton because with minimal values and only four diamonds, responder is more likely to have passed or to have tried spades, particularly with ♠A-fourth. This reasoning also means spades are more likely to be 4=3 than 3=4. West almost certainly has 5+ clubs both because responder might prefer a quiet 2♣ raise with four and because a five card or longer suit better justifies opening on many of his marginal HCP hands. It seems that West’s possible shapes leave him with ~2½ hearts on average which doesn’t help in guessing the Q.

Declarer must be careful to preserve a dummy entry. If the A is cashed early, the defense can take their ♠A on the second round and T will not be an entry if either defender started with Qxx. Unless it is a West who ruffs a diamond.

If the ♠J is offside, declarer will go down, losing two spades, a diamond, a club, and possibly a diamond ruff or a heart to Qxx in the East hand. If the ♠J is onside, all is well unless East has the ♠A and can give his partner a diamond ruff when West holds two small trump, leaving East to collect the setting trick with his Qxx. Declarer can guard against this narrow scenario by cashing the A before taking the ♠T finesse but at the cost of losing his entry to the established spade when the Q does not drop doubleton which is too much to pay for the insurance.

So declarer has two lines:

1. Finesse the Q and lose one trick in each of the other suits. This finesse seems close to 50-50 but you have a two way finesse and one direction is probably 55-60%.
2. Play a trump to the K and finesse the ♠T immediately, with the intention of returning to dummy with the A, eschewing the trump finesse in order guarantee dumping the losing club. The spade finesse also seems close to 50-50 but is probably better than average because spades are more likely 4=3 spades than 3=4.

Line B can go down two when it fails, so A might be better in an ordinary matchpoint situation. But since we are in an unusual contract and playing to make, the decision remains close.

Maybe a simulation with Thomas Andrews’ Deal program will help. (See the Two-Six Dream for another simulation example.)

Here are the assumptions that we will put into the simulation:

1. West has 10-13 HCP and East has 5-8 HCP.
3. East has 4 or 5 diamonds
4. Neither defender has all 5 hearts (otherwise would bid them).
5. West will only bid 1♣ on 10 HCP with 7+ clubs and at least two of the ♣KQJT.
6. 70% of the time East will pass with 5 HCP, no ♠A, and only four diamonds.
7. East would bid 1♠ instead of 1 half the time with ♠Axxx and Qxxx.
8. East would bid 1 instead of 1; half the time with Qxxx and Qxxx.
9. East would raise quietly to 2♣ with either 5 clubs or four of each minor.

To enforce items #6, #7, and #8, we use Tcl’s random number generation function, rand(), to reject a fraction of the relevant cases.

# Fix the visible cards. south is "43 KJ653 J987 A9" north is "KQT6 AT4 AT4 876" west gets KD # Initialize various counters. set nhands 0 set qwest 0 set qwest1 0 set qwest4 0 set qeast1 0 set qeast3 0 set jwest 0 set rufflow 0 set spades43 0 main { set hpw [hcp west] set hpe [hcp east] # Minimum HCP for 1C opener and 1D responder. reject if { \$hpw < 10 || \$hpe < 5 }; set sw [spades west] set se [spades east] set de [diamonds east] # Enforce 4-3 spades. reject unless { \$sw == 3 || \$sw == 4 } # Give East a semblance of a diamond bid. reject if { \$de < 4 }; # Weed out rare case of West having 1H opener with all 5 hearts. Constraints # so far leave West with a 1C opener (never enough HCP for 1NT). reject if { [hearts west] == 5 } # Only allow West to open 1C on a 10 count with 7+ clubs and 2+ of K,Q,J,T. reject if { \$hpw == 10 && ( [clubs west] < 7 || [west has KC QC JC TC] < 2) } # East's behavior is harder to model. Begin by assuming 70% of the time, # East will pass an aceless 5 HCP with only four diamonds. reject if { \$hpe == 5 && \$de == 4 && [west has AS] == 0 && rand() < 0.7 } # Also assume East will prefer S-Axxx to D-Qxxx half the time. reject if { [spades east] == 4 && [west has AS] && \$de == 4 && rand() < 0.5 } # Assume East would bid 1H with all five of them. reject if { [hearts east] == 5 } # Also assume East will prefer H-Qxxx to D-Qxxx half the time. reject if { [hearts east] == 4 && \$de == 4 && rand() < 0.5 } # Assume East will try 2C instead of 1D with 4 clubs and 4 diamonds. reject if { [clubs east] == 5 } reject if { [clubs east] == 4 && \$de == 4 } Increment appropriate counters. incr nhands if { [west has QH] } { incr qwest } if { [west has QH] && [hearts west] == 1 } { incr qwest1 } if { [west has QH] && [hearts west] == 4 } { incr qwest4 } if { [spades west] == 4 } { incr spades43; } if { [west has JS] } { incr jwest if { [east has AS] && [diamonds west] == 1 && \ [east has QH] && [hearts west] == 2 } { incr rufflow } } if { [east has QH] && [hearts east] == 3 } { incr qeast3 } if { [east has QH] && [hearts east] == 1 } { incr qeast1 } accept } deal_finished { puts stderr "Total Hands = \$nhands" puts stderr "West has HQ: \$qwest (stiff: \$qwest1, Qxxx: \$qwest4)" puts stderr "East has HQ-stiff: \$qeast1, Qxx: \$qeast3" puts stderr "West has SJ: \$jwest (ruff low defense works: \$rufflow)" puts stderr "West has 4 spades: \$spades43" }

deal319> deal -i riskyjack.tcl -i format/none 100000

Total Hands = 100000
West has HQ: 54894 (stiff: 1379, Qxxx: 6848)
East has HQ-stiff: 1110, Qxx: 28920
West has SJ: 51022 (ruff low defense works: 285)