Problem no. 35, 2008

Färgglad linje

The problem

Sp 2 Table Sp A Q 10 9 6 5 4

In a newly published book, I read about a deal where East was declarer in 4S with this trump suit. There were two sure losers in the side suits, so East only could afford one loser. Declarer had plenty of entries to both hands. The author wrote that the "best way to play the trump suit was to finesse the queen", and added that you then have a "56% chance to succeed".

If we forget the risk of the defenders' getting a ruff, is it as the author writes? I myself don't think so, but why?


It is true that the the chance of playing the suit for one loser is 56% initially, and the right way to play the suit is to lead towards East, with the intention of finessing the queen, unless North plays the king. Then, you can handle three 4-1 splits (king singleton with North or jack singleton in each hand) plus 14 3-2 splits (the king with North, or jack doubleton with North, or king-jack doubleton with South). If we add everything, we get to 55.95%, so the author's 56% is approximately correct.

A minor error is that he doesn't mention what to do if North plays the king, either because he had to (king singleton) or because he chose to play it from, say, king doubleton or king-jack third. To play the queen then, isn't exactly the best play...

I myself dislike careless writing, but I wouldn't call this an "error".

The error is to say you have 56% chance of only one trump loser if you finesse the queen, because you haven't. In that calculation, we have included king singleton onside, but once North doesn't show up with the king, and we finesse the queen, we can rule out that possibility. A specific 4-1 split will occur 2.83% of the time, so when we finesse the queen, our chance of getting only one trump loser is not 56%, but 53.12%.

The reason finessing the queen is better than finessing the ten is that you pick up jack singleton with South. If you finesse the ten, and South has king singleton, you still lose two trump tricks. Otherwise, those two plays are equally good.

Initially, the chance of one spade loser is 55.95%. When North does not follow with the king, the chance drops to 53.12%. And if he doesn't follow with the jack either (when he does, our problems are solved, and the chance of success goes up to 100%), it falls below 50%. Then, we can also rule out North's having jack singleton (2.83%) and king-jack dobuleton (3.39%), so we will have one spade loser only 46,9% of the time.

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