There was an example in the SBU Simultaneous Pairs on Tuesday evening. You are in three notrump and wish to make four club tricks from this combination:
You start with the king of clubs and both follow small. Then you lead the two of clubs towards the North hand and West follows small. Do you fly with the ace or finesse the ten?
Hopefully everyone can see that it does not matter what you do if the clubs are breaking 3-3. Another layout, East holding the remaining clubs, is irrelevant since you can never make four tricks. This leaves two layouts of interest, those where West hold four clubs and those where West holds them all.
But when West holds five clubs, we can never make four tricks in the suit. So that is (largely) irrelevant. So we should focus on the 4-2 breaks.
At this point I think many just give up. It is all mathematics, computing the probability of West holding four clubs including the two unseen honours is just too difficult at the table.
But there is no need to drag out your logarithm tables to do this - it is very simple. You are assuming East has one remaining card - is it an honour or a small card? Well, there are three cards unseen, the queen, the jack and a small card. So it is 2:1 that East holds an honour, so it must be right to play the ace. If you play small and East wins with an honour, then you will lose another trick to West's honour too.
So that is how to play this combination for four tricks without breaking a sweat.
End of lesson?
No, because there is a little more to this. Perhaps it is not essential to make four tricks in this suit. Perhaps you can afford to make three tricks and get others from elsewhere. Does this remain the best line?
Unfortunately the answer is that it is not the best line for three tricks. And here you do need to know the probabilities. But the chances of a 5-1 break are significant enough to make it right to insert the ten on the second round. This is because the combined odds of QJxx and QJ8xx outweigh the odds of Hxxx.
So there is one line for three tricks and a different one for four tricks. Which should you take for the optimum number of tricks? Actually more important is the entire hand and what tricks you need, who you'd prefer to lose the lead to, any factors that might influence the distribution. It is rare for there to be no other information that just the single suit. But, as it happens, the best line at matchpoints, in total isolation, is to play the ten of clubs on the second round.
However change the six of clubs into the eight in the North hand and you should revert back to the original line of playing the ace!
This is why suit combinations are fickle.
I think the odds for the ace to be right in the first case are 4:3, not 2:1. Easiest way to see it is to look at the a priori probabilities. Since there are four small cards and two honours, there are 8 (=4x2) ways for East to be dealt Hx, and 6 (=4 choose 2) ways for East to be dealt xx. So the odds that he has Hx rather than xx are 8:6, i.e. 4:3.
ReplyDeleteThe problem with the calculation the way you gave it is restricted choice. The defenders' plays from their holdings are not uniformly random; for example East will always play the x and never the H from Hx on the first round. So the simple argument of "there are three cards unseen, and they are equally likely to be in any possible arrangement" doesn't hold.
(The 4:3 calculation assumes that the defenders choose randomly among their small cards. If that's not true, then the answer depends on which particular small cards you have seen.)
OK - another question which is the kind of thing I'm naive about. In some situations it's right for West to split honours when holding QJxx. For example, on some hands it's OK to let declarer win four tricks out of five in the suit overall, but dangerous to let him win all the first three. Since West will sometimes split from QJxx, the fact that he plays low reduces the likelihood that he has that holding. So this could push the odds back towards playing the ace even in the case where you need only three tricks.
So, how to approach this kind of thing in practice? In these "suit combination" questions, the convention seems to be that defenders will play double-dummy. This is a very bad approximation when I am defending (and knowing when to split honours is something that causes me particular uncertainty), and still not great even with top-class defenders.
I think you computed this wrong also. You cannot include information from the small cards. if the clubs are 4-2, there are fifteen possible combinations. (6*5/2 = 15 holdings for east), there are 4 Qx holdings 4 Jx holdings 1 QJ holding, there are 6 xx holdings. Thus it is 9/15 to drop an honour when the clubs are 4-2.
ReplyDeleteYou know it's not QJ-tight though, so can exclude that. That leaves 8/14, or 8:6, or 4:3.
ReplyDeleteA good article explaining why you can't include information from small cards is http://rpbridge.net/7z75.htm
ReplyDelete