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 Queen-Jack
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 4-4. 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 4-1 split) and uwx opposite QJ. Here u,w,x represent the missing low cards that can be freely played without loss.

Case 1

Case 2

The distribution of the minors is entirely unknown and assumed to be the result of a random deal. The number of combinations on a 6-7 split outnumber those on a 5-8 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 3-2 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 4-1 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 4-1 split and 1 out of 6 for the 3-2 split. The ratio of the probabilities on the play is 2:1 in favour of the 4-1 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 4-1 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

Case 2

Based on the (unrealistic) a priori conditions the odds in favor of the particular 4-1 split after the drop of the ♥ J is 5:3. The assumptions are unrealistic as we always know more than the ‘know-nothing’ 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 3-2 split is twice as likely as the 4-1 split.’ The table of odds states these probabilities: 4-1: 28.26%;   3-2:  67.83%. The figures include the number of possible combinations: 10 for 4-1 and 20 for 3-2. That is a 2:1 ratio in favor of 3-2. If we consider the probability of one particular 4-1 split and one particular 3-2 split we find the percentages to be 4-1: 2.826%, 3-2 3.3915%. The ratio is 6:5 in favor of 3-2, as reflected in the Others Weights given above. So comparing one 4-1 split against one 3-2 split is quite different from comparing all 4-1 splits against all 3-2 splits. The 2:1 odds that David G. quotes are largely, but not entirely due to double the number of combinations available for the 3-2 split. His argument is false when one is comparing just one combination against another.

One can say with greater accuracy, ‘3-2 splits are more likely than 4-1 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 4-3-3-3 shapes and 4-4-3-2 shapes. Table I in the Official Encyclopedia of Bridge (1984) shows both the total percentages and the specific percentages. We all know that 4-4-3-2 is more probable than 4-3-3-3 at the beginning (or rather before the beginning), however, one particular 4-3-3-3 is more likely than one particular 4-4-3-2. 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, ‘4-4-3-2 shapes are more likely a priori than 4-3-3-3 shapes solely because there are one-and-a-half 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 5-4-3-1 shapes than 5-4-2-2. 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.