Problem nr 8, 2001


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The problem

The eigth problem is 25 years old. It was a very cold winter, and the captain for the Swedish junior team, Rolf Johansson, had arranged a trial for the Nordic junior championship for teams the following summer. In the middle of the tournament, I and my partner Ronny Blomdahl, met two other Scanian guys: Göran Ofsén and Leif Olofsson.
  On one of the deals, Göran opened with 3 diamonds first in hand (neither side vulnerable), but that didn't stop us from bidding a slam. It depended on clubs being 3-2, no more, no less. And when they duly did, our slam sailed home.
  The chance for a 3-2 split is 67.8% if you don't know anything about the opponents' distributions. But since we could expect Göran to have seven or eight diamonds, it looked like the chance for a normal club split had decreased – maybe to roughly 50%. Therefore, some of the players thought we had been very lucky and that the slam shouldn't be bid. No wonder they complained; we were the only pair to bid it.
  A couple of months ago I came to think about the deal again (don't ask me why) and I suddenly realized how stupid the "complainers" had been. And when I used my own Scania BridgeDealer program to see how often clubs really were 3-2, I got the result I expected, namely that it was even more likely that clubs were 3-2 than 67.8%. I've ran lots of statistical tests and they all show that the probabilities for a 3-2 club break were 75% and that Göran had two clubs per deal on average.
  Do you have an idea how this could be the case?


Solution

If West has seven diamonds, and East-West five clubs together, it is true that the chance for a 3-2 split no longer is 67.8%. If you don't know anything more about the deal, the chance for a 3-2 split is roughly 57%. So the wise guys were wrong. And they would have been wrong even if it had been close to 50%. Their error was that they forgot that they knew more about the distribution.
  Yes, North-South had eight clubs, but we didn't bid slam in clubs. We bid it in spades, the suit where we had ten trumps together. Since East-West only had three spades between them, West's shortest suit will be spades more often than not. And because of West's seven diamonds, his average length in spades is one card. Knowing that, it should be pretty clear that he will have two or three clubs more often than you first thought.
  The full deal was something like this:

S Q 7 6 2
H 10 4 3
D 5
C K Q 8 4 2
S 8 Table Sp 5 4
H J 7 2 H K Q 9 8
D A J 10 9 6 4 3 D K Q 8 2
C 9 6 C J 10 5
S A K J 10 9 3
H A 6 5
D 2
C A 7 4


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