# Update to prior post dealing with proper play holding Ax opposite Q109xx

There was further discussion and research regarding hand 3 in my prior post.

The issue was the correct way to attack the club suit. One declarer succeeded by leading the ♣Q with the combination below, the other failed when they tried the ♣A, then small.

With inadequate research, I suggested that ♣Q first was correct. Actually, as pointed out by Mark Ralph in a comment to the original blog, ♣A first is correct. It is a long story, so if you just wanted to know the answer, you can quit here.

I (incorrectly) based my assessment of the best lead based on a faulty recollection of a prior article, but my recollection was mostly correct, it is just that the problem to be solved was different. While similar, the problem on this hand and the problem I recalled were actually different.

For this problem, one declarer received a threatening diamond lead knocking out 1 of 2 stoppers. So, he needed to find 2 fast club tricks without losing the lead twice (he only had 1 remaining diamond stopper). So, the first side to win 2 tricks in this suit was going to be the winner – if declarer won 2, he was home with 9 tricks in 3NT. If the defense won 2, they would have 5 tricks and defeat 3NT. This suit always has the power to win 2 tricks by forcing out the ♣K and ♣J, but that isn’t good enough.

The declarer who had both diamond stoppers intact, was merely playing the hand to avoid losing a trick to the danger hand, so he led the ♣Q – best in the context of his situation. But the other declarer, with one diamond stopper dislodged, had to find 2 club tricks without losing 2 club tricks.

I used the odds tables to find the answer.

http://www.automaton.gr/tt/en/OddsTbl.htm

The diagram below shows the analysis. If the ♣K is on the left, leading the ♣A and small to the ♣Q (column 1) always works. If the ♣K is on the right, leading the ♣Q first (column 2) always works. The special cases of singleton ♣J, doubleton ♣Jx and ♣KJ allow for both lines of play to succeed. But, the final special case of singleton ♣K, leading the ♣A first works whether the ♣K is singleton on the left or on the right, where leading the ♣Q first succeeds only if the singleton ♣K is on the right. While this only occurs 1.211 % of the time, it is still the better line of play.

Now back to the problem that caused my confusion. The similar problem was: What is the best line of play to win 4 tricks with this holding:

While similar, the problems are different. There are 4 possible lines of play.

- ♣A, then small to ♣Q (option 1)
- Lead the ♣Q first, letting it ride if not covered (option 2)
- ♣A, then small to ♣10
- Lead the ♣10 first, letting it ride if not covered

With a 3-3 split, each of these have their success depending on Kxx on the left or right, or correspondingly Jxx on the left or right. They offset each other such that each choice of a line of play (out of these 4) has an equal chance of success or failure depending upon the lie of the cards. I’m only going to focus on the first 2 options of how to play the suit, since it can be shown option 3 and 4 are inferior.

With a 4-2 split, ♣KJxx means you can never succeed (remember, the objective is 4 tricks – the defense should not cover). With ♣Jx (or ♣KJ) in either defender’s hand, both option 1 and option 2 succeed. With ♣Kx on the left, no option succeeds. With option 1, you will score your ♣A and ♣Q, but the ♣Jxxx will win a trick making 4 tricks impossible. With ♣Kx on the right, option 1 also fails, but option 2 succeeds! Whether RHO covers the ♣Q or not, the ♣J is the only club trick the defense can score, making 4 tricks for declarer.

With a 6-0 split, no option can score 4 tricks. With a 5-1 split, no option works if the singleton is small (‘x’). Both options work with a singleton ♣J. Option 1 works with a singleton ♣K. Option 2 works only when the singleton ♣K is on the right.

In summary, the option 1 advantage over option 2 only occurs when the ♣K is singleton on the left (1 case). The option 2 advantage over option 1 occurs when ♣Kx is on the right – that means ♣K3, ♣K4, ♣K5, ♣K6 – 4 cases!! So, leading the ♣Q is the superior way to play this suit. The diagram below shows this difference:

Note the percentages changed (compared to the initial problem at the table) because the objective changed – in the first case, win 2 tricks without losing 2; in the second case, win 4 tricks.

So, if the reader got this far…this was interesting to me if no one else. Why did I remember wrong/get this problem wrong? It was because the situation between the problem I remembered was quite similar, but the conditions/objectives of the two problems were sufficiently different that the winning play was different. Can you work this out, real time, at the table, as declarer next time this comes up at the table? I hope that, by going through this tedious analysis, I can. I further hope, if faced with this, that the better play works. It comes with no guarantee, simply a higher percentage of success.

Interesting. It’s a bit like planning the play of any hand… it depends on one’s objective, for example, keeping the danger hand off lead (or from getting the lead too early).