Math problem from Imagination and technique in bridge

[inquiry]

I love bridge problems and bought the book Imagination and technique in Bridge by Tom Bourke and Martin Hoffman at the NABC in Atlanta. It was in the half price bin, and since I was buying a bunch of other stuff he knocked another four dollars off it. I have been doing one hand a day like an upscale newspaper bridge column. The hands so far have been challenging and entertaining. I "missed one" so far, but I don't think I actually missed the best line. I will not repeat the hand here, but will discuss the main issue in generalities. If you have that book (published 2003) it is hand 9.

 

On the hand, one opponent used Michaels and is assumed to be 5-5 in the reds.

 

Your partnership holds eight spades and seven clubs. There are two lines of play for the contract both of which I considered before picking my solution. One line is 100% if the Michaels hand has two spades and one club, the other line of play works (see exception below) if the Michaels hand has one spade and two club.

 

The solution the authors suggest to the problem is based on mathematics and some logic. The math starts off soundly (I think) in that the Michaels hand is more likely to have a doubleton in the suit in which you hold the least cards (clubs) by a 5 to 4 ratio. Spades was also trumps, so if the Michaels hand had a singleton club, the logical premise presented is he might have lead a club at trick one. These two factors tipped the \the scales towards playing him for 1 spade and 2 clubs (which was the successful solution on the hand held).

 

The 4 out of 9 chances he has two spades is 44%, the 5 out of 9 chance of two clubs is 56% (ignoring 3=0 splits). However, the line of play when the Michaels hand has two clubs requires that he hold xx of clubs while his partner holds Kxxx. There are 15 combination for the 2=4 split, 10 of them with the king "correctly" placed for the contract to make on the suggested line when clubs split 2=4 and five of them when the king is in the short hand and the suggested line would fail

 

So my position is that the odds have now tilted. The line playing for spades to be 2=3 is solid 44% (actually fractionally higher), while the 56% odd on the 2=4 club split has decreased by 1/3 to the high 30% range

 

Did I make a mistake in my analysis or is the fact no club was lead strong enough to still favor playing for 2=4 clubs and just hope the king is in the long hand.

[cherdano]

Hard to say without the auction, but the inference from the lead is quite likely to trump the 44% versus 35% a priori odds. Say opening leader would choose a singleton club 50% of the time, and a doubleton club 20%. Then you have to compare half of 44% with 0.8 of 35%. I would say these are very conservative estimates.

[gszes]

fly in the ointment???? Does it make any mathematical

difference to the lines of play (not given) if the michaels

bidder has 2 small spades and the stiff club K (which they

would not ummmm ususally lead??)? This may be an

irrelevant thought but just in case--------------------------

[BillHiggin]

I looked at this in terms of combinations (and included the 3-0 possibilities which have some significance - at least in terms of percentages).

The total possibilities are proportional to [combinations of 11 things taken 3 at a time] = 165

Zero spades and 3 clubs will happen 20/165 times

One spade and 2 clubs will happen 75/165 times - 50/165 do not include the club K and 25/165 do

Two spades and 1 club will happen 60/165 times - 10/165 will have a singleton K of clubs which presumably will not be considered for a lead.

Three spades and no clubs will happen 10/165 times.

 

36% for spade doubleton and 30% for club doubleton

Playing for the doubleton club is best if we think that a singleton club lead would happen less than 20% of the time - otherwise it is better to play for the doubleton spade.

[rhm]

On a purely mathematical basis the odds change from 5:4 in favor of a singleton spade to 5:6 in favor playing for a singleton club if we take the ♣K into account.

But we should take into account the opening lead too.

If you believe a singleton club opening lead would be forthcoming more often than one time in six, you are again better of playing for a singleton spade.

Normally that will be the case.

 

Rainer Herrmann

[rbforster]

I would just add that the inference about leading a stiff (or not) is much stronger if the actual lead chosen is a clear one (A from AK, solid sequence) vs a less clear one (low in one of his suits). If he made an unconvincing lead, I would play him for the doubleton club. If he made a clear safe lead, I would go with the odds above.

[jogs]

I'm confused. Relative to the declarer, the michaels bidder

is LHO and opening leader? LHO lead a red suit?

Would this not suggest LHO may be 5530?

[BillHiggin]

I'm confused. Relative to the declarer, the michaels bidder

is LHO and opening leader? LHO lead a red suit?

Would this not suggest LHO may be 5530?

As I read the OP, there is one line that works if LHO has 2 small clubs and another that works if he has 2 small spades, and by inference 3-0 or 0-3 holdings mean failure - so clearly we proceed on the assumption that there is some way to succeed.