Ok, here is the rest of the story. The fellow I was talking about is a very reasonable player, but not as good as justin. He choice the line the justin picked. And i justin is a better player than I am, marginally.
Anyway, i was thinking along the same lines as Rainer Herrmann (above). Let me answer question 2 first. Yes, I play EAST for six spades. The overcall, even playing OBAR bids is not based on much, so five the AK seems unlikely. The raise by WEST and the spot lead confirms the concept that it was six.
Queston 2: The club QUEEN is really bizarre. If west has AK, why not lead one? Why did EAST lead the club Queen missing the JACK (I am looking at it). Two thoughts. East is Qx doubleton, or East is has singleton Queen. I decided with Qx and a come-on signal from WEST, he would continue a club.
So it came down to three lines.
Line one. pull trumps, hook
♦Q with West
Line two, play EAST for
♦Q (several lines work)
Line three, play for double squeeze (described by Justin and others).
The straight up diamond hook of line 3 has to be worse than line three, because you can always fall back on that as a worse case scenario. If the determination that EAST has only one club is right, he is 6-2-4-1, making West 3-1-3-6. So if we are going to do simple math,
♦Q is 4/7th with EAST. I also agree with Rainer Herrmann that with
♠xxx
♥x
♦Qxx
♣AKxxxx west would have bid one more (he misunderstood which E or W showed out on heart). So I am thinking the queen is even more likely with EAST. Thus, I adopted the throwin line, run hearts, cross
♦A, exit
♠Q in the three card ending.
Sure that worked, but I began questioning the logic, thus posting the hand here. The lead a club to set up double squeeze makes a lot of sense. First, they might not find the diamond lead (when you exit a club, and EAST shows out, the DANGER of double squeeze is EXTREMELY CLEAR, however). But the key, is if they do lead a club, you get two chances to make. Chance one, WEST has
♦ Q after all. Chance two, WEST had
♦T9x.
If the hand count was right, and ignoring whatever percentage you use for EAST HAVING to have diamond queen by WEST's simple raise, the odds of EAST having Qxxx (were x can = T or smaller), is 57.1% (4/7); Q with east 42.9% (3/7). But you have to add teh chance WEST has T9x of diamonds. This math is a little more complicated. There are 4 combinations where WEST can hold T9x of hearts (T93, T94, T95, T97), you will have to take my word for it that there are 35 total possible combinations so this is 11.43%.
So the odds -- again with a lot of assumptions -- 57.1% Queen with EAST, 54.3% Queen OR T9 is with WEST. This is very close. So this leaves the two harder things to quantitate. How likely is WEST to bid different with Qxx of diamonds, and how likely is west to find the diamond shift to break up the double squeeze when you thow him in with clubs. If he never finds it, throw him in. If he always finds it, go for the endplay on EAST instead.
There is also the possibility EAST has another club, and WEST has four diamonds. So I came around to thinking the line choosen by the semi-decent player and by justin and almost everyone above is the better line. The T9x of diamonds is what makes the mathematically inferior line better in the long run... after all, the difference is VERY SMALL (3%) and that is based soley on inferred count of stiff club with EAST.
1♥-(ps)-2♥-(2♠)
3♥-(3♠)-4♥-AP
TRICKS
1.) ♠2-(3/5) to JACK-KING
2.) ♣Q-♣T-♣9-♣3 (west thoiught long before the attitude 9
3.) ♥3-K-5-6
4.) ♥J-♣2-♥Q-♥T