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.
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.
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.