[bid_em_up]

Under the given constraints, partner's possible distributions are:

Short ♣, 1 case, 4-4-4-1
Short ♦, 1 case, 4-4-1-4
Short ♥, 3 cases, 4-1-4-4, 4-0-4-5, 4-0-5-4
Short ♠, 3 cases, 1-4-4-4, 0-4-4-5, 0-4-5-4

Given that partner will be short in a major suit 6/8 times and in a minor suit only 2/8 times, and our side will always have 4-5 cards when short in a minor suit, but only 3-4 cards when short in a major suit, I think a minor suit lead has a greater chance of being successful.

I'll pick a club, only because we rate to hold more clubs (either 8 or 9) than any other suit, as long as clubs is not partners shortness.

However, I'm sure this is wrong and I am leading into AKQ10xxxx.

EDIT: I also think that RHO may well have only an 8 card suit headed by the AKJ10, rather than the 9 or 10 card suit that was suggested by others, if that makes a difference.

[straube]

PFormaini, on Mar 6 2010, 11:25 PM, said:

cloa513, on Mar 6 2010, 09:45 PM, said:

PFormaini, on Mar 6 2010, 09:03 PM, said:

Really?

I'm not sure why - since AKQ10xxxxx is far MORE likely to run than AKJxxxxxxx - the odds being nearly 22% for a 3-0 break missing 3 cards - as opposed to only about 8.5% for a 4-0 break missing 4.

So an expert would be MORE likely to assume that the 9 card club suit would run (making a club lead LESS attractive against an expert) than the 10 card suit (making one of those a MORE attractive lead), since the expert would assume from the odds that there was less chance of one of the 10-card suits running than the 9 card suit running.

(Just go to this site that gives the odds and compare "Qxx" to "Jxxx".)

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

So the odds would seem to favor the expert having the 9-card club suit rather than one of the 3 10-card suits - making the club lead LESS attractive than any of the others.

Right?

But if he has those clubs then he must have one loser in the other suits. Ax unless partner opened with less than 11 HCP.

That is true.

So we have arguments that tend to negate each other.

If we trust partner not to violate system requirements, then a club lead looks best, since RHO is LESS likely to hold 9 clubs and 4 side suit winners than the 10 card suit and three Aces.

Even though the odds are much higher that the 9 card suit to the AKQ will run than the 10 card suit to the AK will run.......

Which is why the hand is such a......nightmare.

I think Hrothgar and cloa513 solved this already. Why not trust partner's bidding and then the 7N bidder has (AKJTxxxxxx A A A) with a long suit not clubs. It's immaterial whether a AKQTxxxxx is more likely to run or not than AKJTxxxxxx. The only question is whether it's probable that the AKJTxxxxxx runs. With a ten-card suit, declarer knows that RHO must have shortness in his suit (stiff or void) so he is looking at....


me.....dummy... opener
Q23
Q2..........3
Q3..........2
23..........Q
Q2....................3
Q3....................2
23....................Q
Q...........23
2...........Q3
3...........Q2
.............Q23
.............Q2.......3
.............Q3.......2
.............23.......Q

as possibilities.

In only one case in fourteen does he lose and I might be kind enough to lead away from Qxx as an extra chance.

So dbl and lead a club.