You reach the excellent contract of 6S by East on this board.
South leads a club which you win with the ace, discarding a heart.
The next task is to draw trumps.
You don't know who has the queen, and, if they split 3-2, it is an even money guess.
But what if they split 4-1? The best hope seems to be to cash the ace then run the 10. This will be successful any time north has 4, or if North has the singleton queen. Overall this line avoids a trump loser 51% of the time.
In this case you will be unlucky as South wins trick 3 with the queen, returns a club which you ruff and draw the last trump.
The contract now depends on avoiding a diamond loser.
Again, if the diamonds split 3-2 you have an even money guess. So you consider the options if they split 4-1. You could lead the King and then run the Jack. This will be successful if South has either the singleton queen or nine, or if North has the singleton queen - 3 of 10 possible arrangements of a 4-1 split. So you lead the 2 from the East hand finessing the 10. If the finesse is successful you will return to hand to repeat the finesse. This will succeed if South has Qxxx - 4 of the possible 10 combinations. Overall this line has a 45% chance of avoiding a diamond loser.
- but in this case you will be unlucky again!!
Overall, the probability of success in this contract is about 73% - so you should certainly be there.
Before anyone else points it out, you already know that South has 3 spades and North has 2. This means that North has more chance of having a long diamond suit. This information slightly alters the probabilities, and may make the alternative line better. I will leave the calculations to others.
The basic probabilities of breaks in suits of various lengths may be found in Wikipedia under "bridge probabilities".
V
North
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D
NV
West
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Board12
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NV
East
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V
South
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7
11
16
6
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Some further calculations show that, after drawing trumps, the best play is to lead the king of diamonds.
ReplyDeleteTake into account that we know that south had 3 spades and north had 2, and that both had at least two clubs. We do not take account of north's discard, because he had a free choice.
Therefore of the 17 remaining NS cards, the total number of possible ways they could be distributed between the two players is 17C8 = 24310
Of these, the number which include 4 diamonds in the north hand is 5C4 x 12C4 = 2475.
the number which include a singleton diamond in the north hand is 5C1 x 12C7 = 3960.
So the probability that North has 4 diamonds is 2475/24310 = 0.1018, but the probability that south has 4 diamonds is 2475/24310 = 0.1629.
Assuming diamonds break 4-1:
The probability of achieving success by leading the king is:
(0.1629 x 2/5) + 0.1018 = 0.0814
The probability of achieving success by leading a low card towards the 10 is:
0.1018 x 4/5 = 0.0855
Therefore it is better to play the king - which on this occasion would have worked.
I wonder if anybody can do these calculations at the bridge table!!
Of course I forgot the most important factor to take into account when deciding how to play the diamonds. We now know the spade distribution - so even if the diamonds are 3-2, it is more likely that it is North who has 3, and thus more likely that he has the queen.
ReplyDeleteTo summarise, when the spades are played we have little information about the distribution in the defender's hands, hence we choose the line with the highest percentage probability of success simply based on combinations. When the diamonds are played later on, we have more information about the distribution - and that has a great effect on the probabilities.
I am not so sure about this. You need five diamond tricks, otherwise there is almost certainly a heart loser. Leading the king needs North to have Qxx, but low to the ten will succeed when South has Qxx or Qxxx. It is complicated a little by the chance of a stiff 9, but that does not change the principle much. I think low to the ten of diamonds is right after all.
ReplyDelete