In the fifth session of the Cavendish pairs, you must decide whether or not to risk a 1NT overcall.
Both vul, North deals. As South, you hold:
1NT is a good description of the hand. 15-17 HCP, balanced, spade stopper. If there is a 4-4 heart fit and partner is strong enough to move towards game, he can bid Stayman and locate the heart fit. A takeout double is distorted with the doubleton diamond, and a 2♦ response will leave you guessing. Passing doesn't come close to describing the hand, and will almost certainly leave you with an impossible decision next round, if there even is a next round.
There are 2 things wrong with the 1NT call. One is that there is at most 1 spade stopper. This may be bad for play in 3NT, but in 1NT it won't be a disaster if the opponents run the spade suit. The other is that there are no spots or source of tricks. If 1NT gets doubled, it could be expensive.
Despite the dangers, there really is no choice but to overcall 1NT. Everything else is highly flawed.
You bid 1NT, ending the auction.
West leads the ♦4 (4th best leads)
Where do you win this trick?
Clearly you will attack clubs first. There can be no advantage leading the first round of clubs from dummy. You have plenty of hand entries if needed. Therefore, it must be right to win the first trick in your hand with the ♦K. In addition to saving the ♦A as a dummy entry, you may find out something about the diamond honors since East will be forced to play third hand high.
You play small from dummy, and capture East's ♦Q with your ♦K. How do you tackle the club suit?
East appears to be more likely than not to own the ♣K, since without that card he might not have the values for his opening bid. Thus, if you believe East is more likely to have 3 clubs you can think about the possibility of taking an intra-finesse. This involves leading a club to the ♣9, losing to the ♣10 or ♣J, and then leading the ♣Q to squash West's ♣10x or ♣Jx.
What tricks do you have, and what tricks might the opponents take? You definitely have 2 hearts and 2 diamonds, and since the ♠A is guaranteed to be onside you have 1 spade trick. If you can get 2 club tricks you will make, provided the opponents don't get 7 tricks first.
For the opponents, if the diamonds are 5-4 all they can get is 3 diamonds, 1 spade, and 2 clubs, since you can certainly play the clubs to guarantee 2 club tricks even if the clubs are 4-1. If the diamonds are 6-3, the opponents might be able to take 4 diamond tricks, 1 spade trick, and 2 club tricks before you have your 7 tricks. Therefore, you should assume that the diamonds are 6-3.
What is the likely club distribution? West has no spades and is assumed to have 6 diamonds. He certainly doesn't have a singleton club, since that would mean he is 6-6 and would surely have taken some action in the auction. It is possible that he is 0-5-6-2, but 0-4-6-3 is more likely Not only is a 4-3 heart split more likely than a 5-2 heart split, but West would be more likely to have bid over 1NT with 6-5. It is better to assume that West has 3 clubs.
Given that assumption, there is no reason not to start clubs by leading the ace. This will give you more information if you have to make a guess on the second round of clubs. Furthermore, it is possible that East has a singleton club. Thus, going for the intra-finesse would be a disaster if East has a singleton ♣J or ♣10, since you would then take only 1 club trick and go down even when the diamonds are 5-4.
If you lead the ♣A and another club, two possible things of interest can happen which would give you a decision to make. East could play the ♣10 or ♣J on the first round, while West follows small on both rounds. Or both opponents could follow small on the first round, and West plays the ♣10 or ♣J on the second round. What would you do in both cases?
If East follows with the ♣10 or ♣J and West plays small clubs on the first 2 rounds of clubs, it is easy. You play the ♣9. Not only is East more likely to have the ♣K from the bidding, but from a restricted choice point of view East is more likely to have started with ♣K10 or ♣KJ doubleton than specifically ♣J10 doubleton.
If both opponents follow with small clubs and West plays the ♣10 or the ♣J on the second round, it isn't so easy. Now you have conflicting clues. The bidding indicates that East is more likely to have the ♣K, but looking at the club suit itself playing the ♣Q is better since it wins in more cases. Which is the percentage play?
There are 4 possible ways the club honors can be divided. West can have KJ10. West can have KJ and East 10. West can have K10 and East J. Or West can have J10 and East K. For 3 of these cases, playing the queen is the winner. For only 1 of these cases, playing small is the winner. Thus, looking at the club suit alone, playing the queen is a 3:1 favorite. What must be determined is whether the bidding information overrides these odds.
What do you know about the other missing honors? East definitely has AQJ of spades. East played the ♦Q at trick 1. Unless this is a falsecard, which is highly unlikely since East can't know your problem and will be more concerned about letting his partner know what is going on, West has the ♦J. So East has 9 HCP in the pointed suits.
If East doesn't have the ♥Q, he almost certainly needs the ♣K to get up to an opening bid. He could be there if he has both pointed jacks and the pair opens light, but it is safe to say that if West has the ♥Q then East has the ♣K.
If East does have the ♥Q, he probably doesn't need the ♣K to get up to his opening bid. He already has 11 HCP and a good suit, which is enough for a lot of pairs to open. Add in one more jack and just about everybody would open. In addition, the ♣K might get East up to a strong notrump, as one more jack hits the 15 mark. A hand such as ♠AQJxx ♥QJx ♦Qxx ♣Kx sure looks like a hand most players would open a strong notrump. Thus, if East has the ♥Q the 1♠ opening bid actually decreases the probability that East has the ♣K.
The above analysis indicates that it must be percentage to go up ♣Q. The ♥Q is about equally likely to be in the East or West hand. Thus, the bidding clue doesn't override the 3:1 odds clue from the club suit itself.
You choose to lead a small club from your hand without first cashing the ace. West follows small. What do you play from dummy, and what will your plan be after that?
It is clearly correct to insert the ♣9. It would be a disaster to play the ♣Q and lose to a singleton ♣K. Also, as we have seen, East is more likely to have the ♣K than West.
Assuming the ♣9 loses to the ♣10 or the ♣J, it is right to cash the ♣A on the second round. As seen, leading the ♣Q from dummy risks a disaster if East has a singleton. Also, East is more likely to have a doubleton club than West.
You put in the ♣9. This forces the ♣K. The clubs are 3-2 and the diamonds are 5-4, so eventually you are able to lead up to your ♠K and make an overtrick. The full hand is:
Obviously West should have split his honors. Had he done so, should you play the ♣Q or play small?
What does West have for the split? As discussed, he doesn't have a singleton. With KJ or K10 doubleton, he likely would have gone up ♣K. With ♣KJx, ♣K10x, ♣Jx, or ♣10x he simply would have played small. With ♣J10xx he probably would have played small. Therefore, his only two likely holdings are ♣J10x and ♣J10 doubleton (♣KJ10 is also possible).
There are two possible J10x holdings, ♣J106 and ♣J103. There is only one possible J10 doubleton holding. Against this, West is more likely to hold the club length, since with 11 red cards he might have bid. These are conflicting clues, and the answer is not clear at all. Note that leading the ace would resolve the issue if West has J10 doubleton, but on the actual hand would have led to losing 2 club tricks if the percentage play were taken.
Some readers may question my math when I say that after cashing the ♣A (both opponents playing small) and leading a club towards the dummy (West playing the ♣J or the ♣10) that on the information about the club suit itself, playing the queen is a 3:1 favorite since there are 4 possible honor layouts. They may say: Suppose West plays the ♣10 on the second round of clubs. That eliminates ♣KJx as a possibility, so the only possible holdings are ♣KJ10x, ♣K10x, and ♣J10x. Therefore, there are only 3 possible holdings.
This is fallacious thinking. The point is that the play of the ♣10 doesn't tell you anything more than the play of the ♣J would tell you, namely that one of the original 4 relevant holdings exists. The fact that West plays the ♣J or ♣10 doesn't give you new information, so the original odds haven't changed.
The relationship between the bidding and the location of a critical card is something which is often misanalyzed. Only if the presence or lack of presence of the card would have changed the player's bid does it figure into the equation. I have often heard a player say: I played him for the ♥K because he overcalled 1♠. If the overcaller is known to have ♠AKJxx that argument is nonsense, since the overcaller would have made the same 1♠ overcall whether he had the ♥K or not. However, if the overcaller is known to have ♠AJxxx and nothing else, then the argument may be valid since he probably wouldn't have overcalled if he didn't have the ♥K.
Plus... it's free!