Risky Loser Discard

by Matthew Kidd



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.
  2. Spades are 4-3
  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" }

And the answer is...?

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)
West has 4 spades: 56817

West is a 55% favorite to hold the Q. You’re okay the 1% of the time when it is stiff but doomed the 7% of the time West has Qxxx. Also East has the stiff queen 1% which also works. So line A, finding the Q, works 49% of the time. West has the ♠J 51% of the time, a curiously a bit less than the 57% of the time he holds four spades. The defense to finessing the ♠J where West is able to ruff a diamond with xx is rare, available just 0.3% of the time. The 1% of the time East has the stiff queen doesn’s help line B because it only serves to returns the number of undertricks East has the ♠J. So line B, finessing the ♠J, works just over 50% of the time, edging out line A by less than 2%.

This is really close. The results are only good to about half a percent, varying by that amount when running different 100,000 board simulations. Small changes to the simulation assumptions could easily change the conclusion. I think my assumptions for West, who happened to be Tom Tatham, are reasonable. Kathleen Hennessy was in the East seat. I’m less confident I understand her style, or even the average style, on this particular auction.

This was board 27 (rotated) from the June 22, 2014 La Jolla Unit game.

It’s your lucky day. Either line will work. Too bad I didn’t pick either, choosing to play East for the Qxx, decidedly anti-percentage at 29%.