Bob Mackinnon

A Best Laid Plan

The following problem was suggested by mystery man Jim Priebe who in a team game defended a slam in which both declarers went down.  The problem involves a fundamental probability calculation after a number of cards have been played. It illustrates the difference between a priori probabilities and a posteriori probabilities.

In order to calculate probabilities in 2 suits after something is known about the other 2 suits from the bidding and play, we assume a random distribution in the unplayed suits. This means that probabilities can be calculated exactly form the numbers of possible card combinations. This relates to the probability of the deal. Sometimes a refinement must be added that complicates matters as on the following example where the distributions of spades and hearts become known, and a decision must be made on best play in clubs and diamonds when there is a wide discrepancy in the number of vacant places.

W
 
KQ1086
A9
J108
A75
 
E
 
A972
A973
KJ963
West
East
1
4
4NT
5
5NT
6
6
Pass

 

North leads the J heart, ruffed in dummy. Declarer leads the 7 towards his hand and is much surprised to see South show out. He wins in hand finesse in trumps, cashes the A and return to hand with a club to the A in order to draw the last trump. South has discarded hearts throughout. When he draws the last trump he shall have to discard a card from a minor suit in the dummy, so before that he must decide how he will play the minors for 1 loser. There are 2 apparent choices:
play North for at least one of the missing diamond honors, roughly a 75% chance a priori;
play for the clubs to have been split 3-2, a lesser a priori probability .

Let’s see if the bidding and play have changed the preference for the play in diamonds. The discards by South indicate he began with 5 hearts, leaving North with 6 hearts. North had 4 trumps, so the vacant places available for the accommodation of the 11 minor suit cards is 3 in the North and 8 in the South.  These are the possible splits remaining.

Cards

North

South

North

South

North

South

6 diamonds

0

6

1

5

2

4

5 clubs

3

2

2

3

1

4

Combinations

 

 

 

 

 

 

Diamonds

1

 

6

 

15

 

Clubs

10

 

10

 

5

 

Product

10

 

60

 

75

 

Now we must take into account that one round of clubs has been played in which North followed with a low club, but not just any low club, but with the 2 specifically. This means that the only possible remaining 1-4 club split is  2 opposite QT84.
Furthermore, suppose South has followed with the 4 so the remaining 2-3 club combinations have been reduced to just 3 in number: 82  opposite QT4, T2 opposite Q84, Q2 opposite T84. At this point the combinations remaining are as follows:

Cards

North

South

North

South

North

South

6 diamonds

0

6

1

5

2

4

5 clubs

3

2

2

3

1

4

Combinations

 

 

 

 

 

 

Diamonds

1

 

6

 

15

 

Clubs

3

 

3

 

1

 

Product

3

 

18

 

15

 

To simplify the calculation we assume that South would have played differently if he had been dealt 6 diamonds to go along with the 5 hearts, so this possibility can be neglected, leaving us with 2 cases to consider. The club play will fail for all 15 combinations with a 1-4 club split, but will succeed for all 18 combinations with 2-3 splits.

If South had been dealt 4 diamonds, the best decision would be to play on diamonds hoping for split honours there. The numbers of successful conditions for leading the J from hand planning to run it if North plays low are given below.

 

North

South

North

South

Diamonds

xx

KQxx

x

KQxxx

 

Kx

Qxxx

K

Qxxxx

 

Qx

Kxxx

Q

Kxxxx

 

KQ

xxxx

 

 

Combos

15

 

6

 

Sucessful

9

 

2

 

Clubs

1

 

3

 

Product

9

 

6

 

 

 

 

 

 

 

The number of combinations for which the diamond play will succeed is 15, so the club play is favoured in the ratio of 6:5. Note that taken in isolation the chance of the diamond finesse succeeding when the diamonds split 2-4 is not 75%, it is only 60% (9 out 15 possible combinations), so it is dangerous to generalize from the a priori expectation.

There is one further refinement to be considered, and that is the number of plausible plays in the club suit. The plausible plays determine the probability that the plays of the 4 and the 2 would be chosen by the South and North players under the various conditions shown above. It so happens that there are 2 plausible plays for each combination shown, so a direct comparison of the number of club – diamond combinations is justified in the determination of the relative probabilities. This comes about because neither defencer would part with either the Q or the T if there were an alternative play available. Thus we are in a restricted choice situation, and what I have called the Extended Kelsey Rule can be applied. (That is, in the calculation of probabilities it is correct to compare combinations directly when there is equality in the number of plausible plays.)

The Unexpected Ending
It remains to give the solution to the real-life mystery: the winning play at the table was to go for split honours in the diamond suit. Against the odds clubs were dealt QT84 to the South, the only losing combination for the club play. Here are the hands in full.

 
Both
South
N
 
J543
J107654
K4
2
 
W
 
KQ1086
A9
J108
A75
 
E
 
A972
A973
KJ963
 
S
 
KQ832
Q652
Q1084
 

It might be said that the declarer who played on clubs, not diamonds, like Brutus at Philippi, could feel he’d earned the right to fall honourably upon his sword.


10 Comments

Antony LeeNovember 8th, 2012 at 8:22 pm

The fact that the clubs played first are the two and the four cannot be used in the analysis, see http://www.rpbridge.net/7z75.htm .

Bob MacKinnonNovember 9th, 2012 at 1:42 am

The outstanding clubs are QT842, and when clubs are played defenders must choose between them. Since the QT can’t be played without loss, the choice is narrowed to 842. As the choices become known, the probabilities change. It seems totally irrelevant which cards appear first, but that is not the way it goes. The choices are randomly selected, but the fact they are registered must be taken into account when calculating probabilities. We cannot say simply, ‘2 low cards were played.’ An equivalent problem is the so-called Girl Named Florida problem. (discussed in an earlier blog)

Don (down one) KerseyNovember 9th, 2012 at 2:42 pm

In the preliminary analysis you say: “There are 2 apparent choices: play North for at least one of the missing diamond honors, roughly a 75% chance a priori; play for the clubs to have been split 3-2, a lesser a priori probability.”

But that’s not quite right – playing on clubs succeeds when North has at least two clubs (or, of course, the singleton queen or ten), and this is a greater a priori probability than a diamond honour onside. (If North follows to the second club, you cover in dummy; if South wins, the rest of the clubs are good, and if South shows out, you cash the king and ruff out the clubs, with the diamond ace for entry to the fifth club.)

jeff akerNovember 9th, 2012 at 6:20 pm

Two other considerations: From KQ of diamonds the lead would often (to say the least) have been a diamond honor. Additionally, even holding jxxx of trumps the opening lead might have been a singleton club. Both of these argue to play on clubs rather than diamonds.

Bob MacKinnonNovember 10th, 2012 at 5:30 am

Thanks, Guys.
You make good points. There is another: what are the discards telling declarer? Inferences are a part of the decision process, and one of the most entertaining aspects of a game between good players. Against lesser players it is dangerous to assume they would play as one might expect under given circumstances.
I oversimplified the argument as many declarers might consider it as a straightforward choice based on the a priori odds.

Alvin P. BluthmanNovember 22nd, 2012 at 3:18 pm

I’m not sure why South must hold exactly five hearts. Granted with more than four diamonds, South might have discarded one, but can the same be said of clubs? Why is it impossible for South to have started with, say, 0=6=4=2 or 0=6=3=4? Also, is it certain, for example, that South would give accurrate count in discarding hearts on the trumps after declarer ruffs the opeong lead in dummy, preserving his known ace for later? Or, after drawing trumps, did declarer cash the heart ace, rather than preserve it – and if so, which minor suit card did he discard?

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