Restricted Choice: What Lies Behind It
This blog is in response to Linda Lee’s post Do you “believe” in Restricted Choice?
Bayes’ Theorem always applies where play probabilities are involved. Restricted Choice as generally understood is an application of Bayes’ Theorem where there is an equal chance of selecting 1 of 2 equal honor cards when following suit. The probability of having chosen that one particular card over the other is ½. There are cases where a player chooses 1 of 3 equal cards, in which the probability of choosing that one card is 1/3. These can be small cards, but we don’t think of that process as a ‘restricted’ choice, although Bayes’ Theorem applies equally to that situation.
By saying there was an equal chance of choosing 1 of 2 cards is equivalent to saying the choice was made randomly without bias. This is a theoretical assumption. Maybe some players will prefer playing the queen instead of the jack because they feel it is more likely to influence declarer’s play adversely. The choice is not completely random. Declarer must judge on the basis of experience whether he believes the choice is random. If not he must assign his own best estimate of the chances of the queen versus the jack, say, 2:1. He still applies Bayes’ Theorem on the basis of that bias.
There is sometimes a case for assuming a player must split his honors in front of a tenace in order to give declarer a guess on the next round. Does one always assume the defender will make the correct play and split? That is an interesting situation. If one believes a defender will always play one card rather than the other, or at the very least have a preference, then there is information to be got from the play other than a simple reduction based on random selection.
Dropping the Jack from QueenJack
Suppose declarer holding AT65 opposite K987 plays the ace and the jack drops from the RHO. What are the probabilities the jack was a singleton rather than from QJ? Let’s look at a particular situation and put aside for now consideration of the a priori odds. Declarer reaches 4♥ and a spade is led. They prove to be split 44. Declarer plays the ♥ A, LHO plays the ♥ 2 and the RHO drops the ♥ J. What are the odds the ♥ J was a singleton?
There are 2 cases to consider: Quwx opposite J (a 41 split) and uwx opposite QJ. Here u,w,x represent the missing low cards that can be freely played without loss.
Case 1
LHO

RHO


Spades 
4

4

Hearts 
4

1

Minors 
5

8

Minor Weights 
3

Case 2
LHO

RHO


Spades 
4

4

Hearts 
3

2

Minors 
6

7

Minor Weights 
4

The distribution of the minors is entirely unknown and assumed to be the result of a random deal. The number of combinations on a 67 split outnumber those on a 58 split in the ratio of 4 against 3, giving Case II an edge in that respect.
The play of the ♥ 2 was a choice of 1 of 3 equals so the probability of the ♥ 2 being chosen at random was 1 in 3. For Case I the play of the ♥ J was forced, probability of 1. The one remaining 32 combination is Case II with uwx opposite QJ. Again, the play of the ♥ 2 was a 1 in 3 chance, the same as with the 41 split. The play of the ♥ J was a 1 in 2 chance, as the ♥ Q could have been chosen equally on a random basis. Overall the chance of the appearance of ♥ 2 – ♥ J is 1 out of 3 for the 41 split and 1 out of 6 for the 32 split. The ratio of the probabilities on the play is 2:1 in favour of the 41 split.
We now combine this with the number of combinations of the minor suits yet to be played. The result is 3:2 odds in favor of the 41 split. As Reese may have put it, the odds are better that the jack was played of necessity rather than it resulted from a particular choice among alternatives. So on the next round of hearts declarer should finesse for the ♥ Q with a great degree of confidence. The rule of ‘eight ever’ applies.
The a priori Odds
These give slightly different numbers but the resulting decision is the same because the vacant places are evenly distributed between LHO and RHO. Here we don’t assume any cards have been played outside the heart suit, and that there is no clue as to how the other 3 suits are split. Outside hearts there are 21 cards in the defenders’ hands.
Case 1
LHO

RHO


Hearts 
4

1

Others 
9

12

Others Weights 
5


Adjusted 
5

Case 2
LHO

RHO


Hearts 
3

2

Others 
10

11

Others Weights 
6


Adjusted 
3

Based on the (unrealistic) a priori conditions the odds in favor of the particular 41 split after the drop of the ♥ J is 5:3. The assumptions are unrealistic as we always know more than the ‘knownothing’ odds assume. However, the method is the same and the conclusion is the same, the ♥ Q is more likely to be with LHO by a wide margin, 5:3 against the previous 4:3 where the spade suit was counted out.
The Specious Argument
Now let’s examine the argument based on a table of a priori probabilities that begins with, ‘the 32 split is twice as likely as the 41 split.’ The table of odds states these probabilities: 41: 28.26%; 32: 67.83%. The figures include the number of possible combinations: 10 for 41 and 20 for 32. That is a 2:1 ratio in favor of 32. If we consider the probability of one particular 41 split and one particular 32 split we find the percentages to be 41: 2.826%, 32 3.3915%. The ratio is 6:5 in favor of 32, as reflected in the Others Weights given above. So comparing one 41 split against one 32 split is quite different from comparing all 41 splits against all 32 splits. The 2:1 odds that David G. quotes are largely, but not entirely due to double the number of combinations available for the 32 split. His argument is false when one is comparing just one combination against another.
One can say with greater accuracy, ‘32 splits are more likely than 41 splits largely, but not entirely, because there are twice as many of them.’ The play of the cards eliminates some combinations that were included in the a priori tables, and that basically is why the a priori odds aren’t an infallible guide, and certainly aren’t accurate mathematically after cards have been played.
Probability of Distribution Patterns
The same is true of the comparison of 4333 shapes and 4432 shapes. Table I in the Official Encyclopedia of Bridge (1984) shows both the total percentages and the specific percentages. We all know that 4432 is more probable than 4333 at the beginning (or rather before the beginning), however, one particular 4333 is more likely than one particular 4432. That is, 4=3=3=3 is more likely that 4=4=3=2. If one is down to a choice between the 2, the a priori odds favor the 4=3=3=3 (2.634% versus 1.796%). So one can say with accuracy, ‘4432 shapes are more likely a priori than 4333 shapes solely because there are oneandahalf times more of them.’
Similarly one sees from the table that 5=4=2=2 is more likely than 5=4=3=1, even though there are overall more 5431 shapes than 5422. This is the basis for the argument that as play progresses, if one has to choose one particular shape over another, always choose the flattest shape regardless of the a priori odds. (Bob’s Blind Rule). There are more 2=2 combinations than 3=1 combinations, and 3=3 combinations than 4=2 combinations.
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Wouldn’t a better analysis involve a ninecard fit (instead of an eightcarder)? With only eight in the combined hands, the drop of the Q or J could be a falsecard.
Yes, it is possible that RHO would drop the Jack from QJx. Usually in these situtations one doesn’t include frivolous plays and one assumes the opponents are following a normal path and defending optimally.
However, to be complete mathematically one can include the case J from QJx, but one must assign a probability to that action. I would guess the probability is so low it can be ignored when we are looking for general guidelines.
With regard to a 9card fit, there is a difference between odd numbers and even numbers that is reflected in the rule ‘eightever, ninenever’. In the case of 5 cards missing, there are 2 even splits 32 and 23, whereas with 4 cards missing 22 stands alone as the most even split. The mathematical procedures are the same, so it is just a matter of working it through with pencil and paper (and eraser).
You say restricted cance is a consequence of Bayes’ Theorem. How do you deduce it from Bayes’ Theorem?
Bob…I’m afraid you missed the point of Bayes in restricted choice..It was “Bayes Postulate” which served as “evidence” that a player would play “randomly” from double honors. The RC theorists however leapfrogged ahead to presume random play at an aggregate level, then asserted that gave mathematical validity to “randomness” in plays from double honors at a single event level. That however is a fallacy akin to the Gambler’s Fallacy, which unfortunately characterizes all of “restricted choice”.
That “Bayes Postulate is known also as the “equidistribution of ignorance”, since it is based on guessing when there are two options and you don’t know how to distinguish…Sort of like a space alien choosing the winner of the Patriots’ playing local high schools by tossing a coin and getting it “right” half the time. The actual theory uses the 5050 presumption as a starting point and “Bayes updating” occurs after results come in to update their assumption (Hey, the Patriots always win!), which RC ignores or just did not understand. The claim that “restricted choice” is based on “Bayes” was a selling point for a colossal fallacy…A fallacy because there was and is a simple way to explain the odds the RC theorists were confused about. There is very much wrong there.