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.
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 32, 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 14 club split is ♣2 opposite ♣QT84.
Furthermore, suppose South has followed with the ♣4 so the remaining 23 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 14 club split, but will succeed for all 18 combinations with 23 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 24 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 reallife 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.
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 .
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 socalled Girl Named Florida problem. (discussed in an earlier blog)
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 32, 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.)
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.
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.
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?
I love A Handmaid’s Tale. One thing of my favorite asecpts of the book was how unexpected little nuances were. ( not sure if these are spoilers, so viewer beware )For example, the Handmaid’ society seems to have generated from either the cultures of the US and/or Canada. I assumed that whatever caused this semiTaliban society to arise, occurred all over the world and not just in North America. However, I was shocked when people from an Asian culture visited the Handmaid society and not only were they more sexually liberated than the Handmaid society, but they were also depicted as more sexually liberated and free than their modern day counterpart. Such events as this caused me to reexamine certain assumptions about the longevity of a culture typically portrayed as surviving where others will not. It was refreshing in many ways and, for me, believeable but strange.
Babo, a ja znÃ³w zauwazyÅ‚em duÅ¼e uczucia HiszpanÃ³w wobec psÃ³w. ByÅ‚y wykarmione, zadbane, uÅ‚oÅ¼one. Mam nawet stosowne zdjÄ™cia. Natomiast fatalnie tam siÄ™ majÄ… koty. MyÅ›lÄ™, Å¼e niechÄ™Ä‡ do kotÃ³w to pokÅ‚osie arabskiego panowania.
It’s like you’re on a mission to save me time and money!
I found some really great projects recently for lighting. My apartment seems to be lacking the proper amount of light so I think there might be some new additions to the light family. I will take pictures as soon as I try this project. Â I found this project here onÂ heyÂ gorgeousÂ the blog.