On June 8, Steven Allen posted "An interesting play problem" which presented, in part, how to play the combination of the club Ax (North) opposite the club KQTxx (South) for 5 winners given the number of open spaces which may be held by the defenders. (His problem was more complex but it caused me to consider this basic combination.)

Initially, I embarrassed myself by guessing that a two card differential was the key. Then Richard Pavlicek posted that declarer should finesse if East has at least twice as many open spaces as West. I chided myself for being lazy. I didn't doubt Richard but the solution wasn't intuitive to me, so I spent a large part of my weekend at our mountain cabin armed only with a pen, pad of paper and the calculator app on my cell phone exploring the mystery of this problem. It proved to be an interesting search.

Declarer has two choices, "the bang" (cash A-K-Q) or "finesse" (cash A and then play to the T.) If declarer bangs, he picks up 1/3 of the 4-2's and all of the 3-3's. If declarer finesses he picks up 2/3 of the 2=4's (2 West, 4 East), 1/2 of the 3-3's and 1/3 of the 4=2's. So, the problem reduced to determining the probability distribution of 2=4, 3-3 and 4=2 as a function of the number of open spaces for East and West.

To set the stage, assume you know nothing about the East and West hands (they both have 13 open spaces.) 4-2's occur 48% (24% for 2=4 and 24% for 4=2) and 3-3's occur 36%. If declarer bangs, he picks up 1/3×48+36=52%; if he finesses, he picks up 2/3×24+1/2x36+1/3×24=42%. In the abstract, banging is the winner by a significant margin.

As the number of open spaces West has becomes less than the number of open spaces East has, the success of the finesse relative to the bang increases. After some trial and error, if East has 13 open spaces and West has 7 open spaces, declarer picks up 5 tricks 41.09% of the time with either line. If West has 8 open spaces, declarer should bang; if West has 6 open spaces, declarer should finesse (as Richard predicted.)

A peculiar thing happens as you reduce the number of East's open spaces. If East has 11 open spaces, the break even occurs when West has 6. Declarer picks up 42.22% with either the bang or the finesse. If East has 9 open spaces, declarer is indifferent when West has 5 (success rate of 43.95%.) If East has 7 open spaces and West has 4, either choice wins 46.97%. And if East has 5 open spaces and West has 3, both plays win 5 tricks 53.57%. Note that declarer's success increases as the number of open spaces declines.

But when East has an even number of open spaces, there is no number of open spaces for West where the bang and the finesse yield the same result. For example, if East has 6 open spaces and West has 4, banging wins 54.77% and finessing wins 50.00%; but if East has 6 and West has 3, the pendulum swings in favor of the finesse 47.62% to 41.04%.

[To give you a flavor of the tedium in the calculations, consider East with 6 open spaces and West with 4. That means there are 6 clubs remaining and 4 non-clubs. West can have 2 clubs

6x(6/10×5/9×4/8×3/7)=2160/5040 (42.86%)

where 6=4!/2!x2! (the number of ways of choosing two clubs with four chances), 6/10 is the chance of the first club, 5/9 is the chance of the second club, 4/8 is the chance of the first non-club and 3/7 is the chance the second non-club.

Similarly, West can have 3 clubs

4x(6/10x5/9x4/8x4/7)=1920/5040 (38.10%)

and West can have 4 clubs

4x(6/10x5/9x4/8x3/7)=360/5040 (7.14%)

If declarer bangs, he picks up

1/3x(42.86+7.14)+38.10=54.77%

If declarer finesses, he picks up

2/3x42.86+1/2×38.10+1/3×7.14=50.00%

Banging wins 54.77% to 50.00%.]

Why there is a break even when West has an odd number of open spaces but not when he has an even number is still a mystery to me. But the takeaway is "Richard's rule", "Finesse only if East has at least twice as many open spaces as West" but the numbers for one off are very close and may be a break even, so if one opponent yawns or you need a swing, this is a position where you can taunt the percentage gods at your own risk.

And I don't feel lazy any more.

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