Free Weekly Instant Tournament - November 11 - Board 4

Board 4
Both vulnerable

♠ A 9  ♥ Q 10 6 4  ♦ A K J 10 5 3  ♣ K

One spade on my left, pass, pass to me. 

If the one spade bid had come on my right, I would overcall with two diamonds. Some would double for fear of missing a heart fit. But since the opponents have the master suit, missing a heart fit doesn't worry me so much as missing the chance to get my six-card suit into the auction. It is unlikely we can outbid the opponents unless we find a diamond fit.

When the opponents have bailed out at the one-level, however, the situation is different. Now it may well be our hand--possibly for a game--so I'm more concerned about finding a heart fit. That makes doubling more attractive.

I double, LHO passes, and partner bids three clubs. There's not much point in bidding diamonds now. If we don't have a heart fit, our likeliest game is three notrump. Partner probably has some help in spades, so I'm not worried about my single stopper.

I bid three notrump, everyone passes, and LHO leads the king of spades

	NORTH Robot ♠ 10 8 5 4 ♥ A K 9 ♦ 8 7 ♣ Q 8 4 3
	SOUTH Phillip ♠ A 9 ♥ Q 10 6 4 ♦ A K J 10 5 3 ♣ K

West	North	East	South
Robot	Robot	Robot	Phillip
1 ♠	Pass	Pass	Double
Pass	3 ♣	Pass	3 NT
(All pass)

Partner is contributing the expected second stopper in spades, so I have time to set up the diamond suit. If diamonds split, I can take five diamonds, three hearts, and a spade, for nine tricks. The opponents can take at most two spades, a diamond, and the club ace. So I make three notrump. A problem arises only if diamonds are five-zero. But, assuming I attack diamonds by taking a finesse against the queen, I can still take five diamond tricks. If I carelessly cash the diamond ace first, however, I'm in trouble.

At IMPs, taking the diamond finesse is clear, since it guarantees the contract. But at matchpoints could it be right to cash the ace-king, trying to drop a doubleton queen offside? West is known to have the preponderance of high cards after all. But he is also known to have at least five of the seven spades.

This is a fairly common problem. When one opponent is known to have most of the high cards but the other opponent rates to be longer in a given suit, how does that change the odds on how to play that suit? Is the high card disparity or the the length disparity more important?

If West has three or more diamonds, cashing the ace-king doesn't help, so we might as well assume that's not the case.What if we knew West had a doubleton diamond? Would it be right to play for the drop then? 

If we knew nothing about the location of high cards, knowing West had a doubleton diamond would make the finesse a three-to-two favorite. If we assume West has at least 11 HCP, then East is restricted to at most three. (Yes, giving West 11 HCP is a simplification. If he is 5332, he probably has at least 12. If he is shapely, he could have nine or ten. But to keep things simple, we'll assume he has at least 11.)

Crediting East with at most three HCP means that if East has three small diamonds, he could have the heart jack, the club jack, neither, or both. If East has queen third of diamonds, then he can have the heart jack, the club jack, or neither, but he can't have both. So, roughly speaking, the high card constraints have eliminated from consideration about a quarter of those hands where East has queen third of diamonds. In other words, the finesse has gone from being a three-to-two favorite to being a two-and-a-quarter-to-two favorite. It's still a favorite--just less of one.

If it's right to take the finesse even if we knew West had a doubleton diamond, then it must be right if we aren't sure how diamonds split. If West has a singleton diamond, the finesse is a heavy favorite. And it is a heavier favorite yet if he has a void.

I play low from dummy at trick one, RHO plays the spade six and I win with the ace. I play a heart to dummy. West contributes the jack; East, the eight. I play a diamond from dummy and RHO plays the deuce. If I play the ten, West will probably assume I have the jack. If I play the jack, I could easily be missing the ten. So if West has queen-nine fourth of diamonds, the jack may conceal the fact that the suit is running.

I play the jack. West wins with the queen and cashes the spade queen. RHO follows with the seven. West now cashes the spade jack. If he doesn't cash the club ace next, I'll make an overtrick. On this trick, RHO pitches the club seven. Perhaps a diamond pitch will make it appear my diamonds aren't running, so I pitch the diamond three. West continues with another spade, and I claim

	NORTH Robot ♠ 10 8 5 4 ♥ A K 9 ♦ 8 7 ♣ Q 8 4 3
WEST Robot ♠ K Q J 3 2 ♥ J 5 ♦ Q 9 6 4 ♣ A 2		EAST Robot ♠ 7 6 ♥ 8 7 3 2 ♦ 2 ♣ J 10 9 7 6 5
	SOUTH Phillip ♠ A 9 ♥ Q 10 6 4 ♦ A K J 10 5 3 ♣ K

Making four is worth 96%. The overtrick didn't matter much, since the field had difficulty reaching game. Most players balanced with two diamonds and played it there. I find that a little surprising, since I thought the field was fonder of off-shape take-out doubles than I am.

As far as the odds calculation goes, the methodology I described above is fine for an at-the-table approximation. But it assumes all four possible high-card layouts (heart jack, club jack, neither, and both) are all equally likely, which isn't quite the case. For those who care, here is a more accurate calculation:

If we assume East has two spades and three diamonds, then his remaining eight cards are chosen from a population of two jacks and eleven small cards. There are ₁₁C₇ ways he can have one specific jack, ₁₁C₈ ways he can have no jacks, and ₁₁C₆ ways he can have both. Since there are four ways for East to have three small diamonds, the total number of ways for the diamond queen to be offside is 4 * (2 * ₁₁C₇ + ₁₁C₈ + ₁₁C₆). There are six ways for East to hold queen third of diamonds, so the total number of ways for the queen to be onside is 6 * (2 * ₁₁C₇ + ₁₁C₈). That works out to about 51% in favor of the finesse, slightly worse than what we arrived at with our rough approximation.