ª A 9 6

21 The odds change when one defender is known to have extra length in one or two suits

ª A 9 6

¨ 9 7 5 4

§ 9 8 2

ª 10 ª 8 7 4 3

¨ J 10 3 ¨ A K Q 6

§ K Q 10 6 § A 5

ª K Q J 5 2

¨ 8 2

§ J 7 4 3

West North East South

Pass

Pass Pass 1NT 2ª[1]

3© 3ª End

West led §K and East correctly won §A and returned the suit. West won §10 and then §Q and played a fourth round. Since West had shown up with nine cards in hearts and clubs the chance of East holding four spades had increased substantially (see below). Hoping to catch West with a single ª8 or ª7, declarer ruffed with ªA and played ª9.

East showed no flicker so declarer changed tack and won ªK. There was now no way to avoid two down vulnerable and a poor result. East-West can make 4© but hardly anyone will bid it. The position declarer was hoping for was to force East to cover with ª10 and then repeat the finesse. Even if East had held ª10 8 7 4 they can be finessed three times for success. The play would go:

ª9 covered by ª10 and ªJ;

Finesse ©Q;

ª6 covered by ª7 and ªQ;

©A;

ª --

¨ 9 7 5 4

§ --

ª -- ª 8 4

¨ J 10 3 ¨ A K

§ -- § --

ª K 5

¨ 8 2

§ --

Declarer would have been in hand and plays a diamond and the last two tricks would have been made with ªK 5.

But it was not his night. So how good was his play? The chosen line loses when West has ª10 singleton and might lose if it were doubleton. It succeeds if East holds ª10 unless West holds precisely ª8 7 in which case the second finesse would fail. The following table shows these possibilities.

We can assume that East has at least three spades since he opened 1NT with three hearts and two clubs. So only those distributions are relevant. If East has five spades the chosen line is best but it is unlikely as West would probably bid again. So that is eliminated.

Given all that, the probability of a 2-3 spade split is now 54.55% and a 1-4 break is 45.45%. The 2-3 break is still slightly more likely than the 1-4 but less so than if nothing was known about West’s hand. Most of us overestimate the effects of known differences in suit lengths between the opponents. Here nine of West’s cards are known and only five of East’s but the 3-2 break is still favourite. These are the key layouts (ª4 and ª3 are shown as x):

	West	East	Prob	Success or failure	Qualified?
1	ªx	ª10 8 7 x	18.18	Success
2	ªx x	ª10 8 7	5.45	Success
3	ª7 or ª8	ª10 8 x x or ª10 7 x x	18.18	Success
4	ª7 x or ª8 x	ª10 8 x or ª10 7 x	21.82	Failure	But succeeds if East ducks and ª9 is run
5	ª8 7	ª10 x x	5.45	Failure	Ditto
6	ª10	ª8 7 x x	9.09	Failure
7	ª10 x or ª10 8 or ª10 7	ª8 7 x or ª7 x x or ª8 x x	21.82	Failure

That gives a raw total of 41.8% for success and 58.2% for failure. Since there is no 2-3 or 1-4 layout which cannot be played for no losers in the trump suit that clearly makes the “normal” play of bashing out the top spades the preferred route. And it is the way anyone else in the contract will play it which is “playing with the room”, my least favourite excuse for mucking up a hand.

But items 4 and 5 in the table are qualified. Few defenders holding ª10 7 x or ª10 x x will cover ª9. And not many will cover with ª10 8 x. Let’s guess that in practice only 25% of the defenders would (I’d bet on nearer 10%). So these two items with combined probability of 27.27% failure become 20.45% for success and only 6.82% for failure. That swings the odds to 62.27% in favour of the line chosen and only 37.73% against.

At the table declarer, a tall Hampton resident, was irritated to go two down for a 9% result when the simple approach would get a good result (82%). He is now happy that he was unlucky rather than choosing a bad line.

[1] Don’t try this at home!