Specificity and Probability
Normal English conversation tends more towards vagueness than specificity. ‘What exactly do you mean by that?’ is considered an impolite challenge to the speaker. The requirements of mathematics demand exactitude; the primary tasks of a mathematician are to define his terms and to limit his frame of reference. Unlike in political discussions hand waving appeals to general principles that may or may not be applicable to the matters under consideration are to be avoided. Unfortunately, common usage applied to mathematical problems may sometimes mislead the untrained mind along the wrong path. This is especially so with regard to probability, as words like ‘probably’ and ‘likely’ are vaguely understood. When asked for a definition of ‘a miracle’, Enrico Fermi replied, ‘anything with a probability of less than 20%’. Whether of not you can agree with that, it does present a firm basis for discussion.
Here is an example that illustrates how specificity can mysteriously transform probabilities. The key conclusion is that probability in card play depends not only on combinations, but also on permutations in play, or as we put it, ‘plausible plays’. Suppose we have two packs of cards. From each we draw a card at random. The question is: what is the probability that 1 red card (r) and 1 black card (b) have been drawn? Most will know that the probability of that occurring is 50%. Both red and both black draws each have a 25% chance of occurring. Let’s look at the 4 possible permutations in the draw.
bb br rb rr
Checking the r’s and b’s we can see how the expectations are manifest in the possible draws.
Next we ask the question, ‘if one of the cards is red, what are the chances that the other card is black?’ Note we have added information by restricting the results of the draw. Without much thought one might conclude that the other card stands to be red, because the a priori odds of bb was only 25% whereas the odds of one black and one red was 50%. The ratio of 50 to 25 is 2:1. Let’s look at the remaining permutations:
br rb rr
Sure enough, there are 2 permutations for mixed colours and only one for the same colour.
One might conclude that the information provided has not changed the probabilities one iota. On this basis many bridge writers wrongly tend to rely on a priori odds in their analysis of card play.
Let’s now be as specific as we would at the card table and say that one of the cards is the ♥ Q. What is the probability that the other card is black? A false argument is that it doesn’t matter which red card was drawn from the one deck, the draw from the other deck is independent of that, so the odds must not be affected. This is where Galileo got it wrong. To get it right look at the possible permutations:
b♥ Q ♥ Qb r♥ Q ♥ Qr
The chances that the other card is black is now 50%. What has happened? The rr permutation has been split into 2 equally probable draws: ♥ Q on the right and ♥ Q on the left. If the ♥ Q has been drawn on the left, it is 5050 that a black card has been drawn on the right, and similarly if the ♥ Q has been drawn on the left. It is obvious that the probability of a black card being drawn is 50%, provided you think of it in the proper way. The surest way to the correct answer is to consider the possible permutations in the draw. Specify.
Specificity in Card Play
With reference to the previous blog, let’s consider the case of declarer playing a club to his ace when the opponents hold 5 clubs. We shouldn’t think of the cards as ♣QTxxx, we have to be specific: ♣QT842. Next we assume the defenders would not follow with a club honour unless they have no choice, but that the other cards can be played equally at random. Consider the case where the clubs are split 23. There are 10 combinations for 6 of which the ♣Q on the right and for 4 of which the ♣Q is on the left, so the odds are 3:2 the ♣Q was dealt to the RHO. Here are those combinations.
LHO 
RHO 
LHO 
RHO 
Q4 
T82 
QT 
842 
T4 
Q82 
Q8 
T42 
84 
QT2 
T8 
Q42 


Q2 
T84 


T2 
Q84 


82 
QT4 


42 
QT8 
Suppose that on the first round of clubs the LHO has followed with the ♣4 and the RHO has followed with the ♣2. The 7 combinations listed on the right are no longer possibilities, leaving us with just 3 possible combinations listed on the left. Before play in the suit they were equally likely. Is it still so? That depends on the probability that the sequence ♣2♣4 would have been chosen at random. With the first 2 combinations, the RHO could have equally played the ♣8 or the ♣2, so there are 2 equally likely permutations in the play ♣8♣4 or ♣2♣4. With the third combinations the RHO was obliged to play the ♣2, but the LHO could have played the ♣8 instead of the ♣4, so again there are 2 equally probable permutations in the play. All 3 combinations have 2 possible variations in play to choose from, so they can be treated as equals in the calculation of the probabilities. The probability of ♣Q on the right is now twice the probability of ♣Q on the left, because there are 2 equally probable combination of that sort to 1 of the other.
A False Argument
Let’s now put forth a false argument of the kind one might encounter at a casual post mortem in the pub. Let the clubs be designated as ♣Qxxxx, where x represents a low card. On the first round of clubs the RHO and LHO follow with low cards. That leaves us with 3 possibilities remaining: Q opposite xx (1 case) or x opposite Qx (2 cases). So it appears the chance of ♣Q on the right is twice that of ♣Q on the left. This was true when the cards missing included the ♣T and the number of plausible plays is the same for each combination remaining, but it is not true for ♣Qxxxx. Let’s lift the play restriction on the ♣T by converting it to the ♣6. Again there are shown on the left the 3 combinations still in the running after one round of clubs is played.
LHO 
RHO 
LHO 
RHO 
Q4 
862 
Q6 
842 
64 
Q82 
Q8 
642 
84 
Q62 
86 
Q42 


Q2 
864 


62 
Q84 


82 
Q64 


42 
Q86 
With the first possible combination the RHO could have equally chosen to play any of 3 low cards, so the chance of seeing ♣2♣4 is 1 in 3. With the other 2 combinations, the RHO had 2 equal choices and the LHO had 2 equal choices, so all together there are 4 available permutations (plausible plays) to choose from, so the chance of seeing the ♣2♣4 is only 1 in 4 for each. The fewer the number of choices available the greater the probability that a specific choice has been made. As a consequence, the existence of the first combination is greater than the second or the third in the ratio of 4:3. However, there are 2 combinations with the ♣Q on the right, so the probability that the ♣Q is on the right is 3:2, just as it was before a club was played.
This demonstrates that under the circumstances the play of 2 low cards has not changed the a priori odds that the ♣Q was dealt to the RHO. To achieve this result we had to consider the number of plausible plays for each combination, otherwise we might conclude wrongly that odds have changed to 2:1. On the other hand, if the number of plausible plays is the same for the remaining combinations then a direct comparison of the number of combinations remaining is a valid procedure for obtaining probabilities.
Consequences
The a priori odds are subject to change. If one is to calculate the a posteriori odds on the location a queen, it is not sufficient to compare solely on the number of combinations remaining for the queen on the right and the queen on the left. It is necessary to allow for the number of plausible plays available for each combination. Only if the number of plausible plays is the same for each combination can a direct comparison be made.
I rarely drop comments, however I looked through some
comments here Bob Mackinnon