A FourDoor Monty Hall Game
Bridge is a game of choices. A player ideally makes the choices that give him the best chance of success within the rules of the game. Decisions are based on evidence and prior knowledge, and usually there is a large degree of uncertainty involved. How to make decisions in the face of uncertainty is a topic shared by many games, some serious, some not, but the theoretically optimum approach can be expressed in terms of probabilities. It is probability theory that links many games. In order to understand the optimum decisionmaking process in a very complex game, one may investigate how probabilities apply in a simpler game, then seek to generalize. For this purpose the game often chosen by bridge writers is the socalled Monty Hall scenario, which is supposedly well understood.
The link between Monty Hall and bridge is the Bayes’ Equation, named after the British clergyman, Thomas Bayes (17021761), who first stated the ideas in the language of the time. As is common with entirely new ideas, the current language was inadequate for their expression. Intuitively Bayes knew he was right, but he just couldn’t find the phrases to convince the world at large. It was the language of 20th Century mathematics that was required to give a full and clear expression of the ideas he first conceived. Ironically, British academics were among the last to concede that their countryman Bayes had been right. Is it only a problem with the vagaries of the English language?
Bridge writers want to make matters as simple as possible for their readers, most of whom are not familiar as they should be with mathematical expressions. So they resort to common language, and avoid equations, even though they themselves may be familiar with Bayes’ Equation, as was Hugh Kelsey the great bridge author. Very often their explanations fail through being oversimplified, inexact and unconvincing on close inspection. Generalizations become hazardous. In this segment we shall introduce the exactitude of mathematical terminology for interested readers, as applied to the more complex scenario which we call ‘FourDoor Monty Hall.’
The most famous demonstration of the validity of Bayes’ Equation can be credited to Marilyn von Savant, who in her column in the magazine Parade, applied Bayes’ reasoning to the popular TV game show Let’s Make a Deal hosted by MC Monty Hall. The circumstances are described by Jeffrey S. Rosenthal in his entertaining and instructive book entitled Struck by Lightning: The Curious World of Probabilities. The setting for the game was a set of 3 doors behind one of which sat a large prize. A contestant was asked to choose one door in the hope of winning the large prize. Initially, the chance of receiving the large prize was 1 out of 3. The MC then opened one of the 2 doors remaining to show the prize did not sit behind it. The contestant was sometimes given the option of sticking with his first choice or switching to the other unopened door. Von Savant took the fun out of the game by pointing out that the odds were 2:1 in favor of switching doors. This claim was met with much opposition from her readers; even some professors, who argued it could not be true, on the grounds that Monty Hall always could open a door with no prize, so with 2 doors remaining the chances should be 5050 regarding which one hide the prize. Wrong! A ‘free’ choice has unexpected consequences.
The matter was resolved when Marilyn had teachers play the game with schoolchildren and report the results. The statistical evidence overwhelmingly supported Bayes’ 2:1 odds. To their credit, many dissenting experts publicly apologized and it appeared the matter was settled once and for all. Civilization moved a step forward.
In the game of bridge the wellknown principle of restricted choice arises from the very same Bayes’ Equation that governs the probabilities in the Monty Hall game. In order to explain its application to their readers, bridge writers discuss the solution to the Monty Hall problem in the hope that the reader is convinced and can make use of evidence that comes their way during the play of the hand when a defender follows with 1 of the 2 missing honors. Here we aim for a wider understanding and wider applications. So, we consider the Monty Hall Game where 4 doors are available. The contestant chooses one door in the hopes of winning a large prize, and the MC opens one of the 3 other doors to show that no prize sits behind it. We show how to calculate the probabilities that the prize sits behind each of the 3 remaining doors, and we express the process in mathematical terms that have a general applicability. The process can be extended to as many choices as we wish to incorporate.
The Statistical Approach to Probability
The easiest way to understand how probabilities work is to imagine an experiment where choices are presented to a great number of participants. The choices are counted and put in a table of results forming patterns from which conclusions can be drawn. This can be a convincing approach if the results are clearly in favour of one circumstance over another, as von Savant demonstrated to her readers when the odds were 2:1. We shall return to the idea of testing hypotheses from a collection of data, but first we shall assume that the contestants adhere strictly to our assumption concerning how their choices are made. This yields the expected or average numbers which one may or may not see reflected in the actual results that are subject to variation due sampling conditions. There are 2 assumptions to be studied for the purpose of exposition which are illustrated below.
Initial Conditions  Equal Choices  Biased Choices 
A  B  C  D  —  30  30  30  —  60  30  0  
A  B  C  D  —  0  45  45  —  0  45  45  
A  B  C  D  —  45  0  45  —  90  0  0  
A  B  C  D  —  45  45  0  —  60  30  0 
Chose B C D  Chose B C D 
The box on the left indicates the four doors, behind one of which sits the prize. The large letter denotes the door which hides the prize. Each line is given 90 samples, 360 in all, so the initial probability that the prize is hidden behind a given door equals ¼ for each door. The samples are presented to 360 school children who are asked to choose a door from B, C and D that does not hide the prize. Thus, from line 1 the 90 children have a choice of 3 doors, whereas for the other lines the other 270 children can choose from 2 doors only. First we consider the ideal experiment where the children choose exactly in accordance with a probability model. There are 2 models to consider.
Equal Choices Under this model the children have an equal chance of choosing any door that is presented to them. So with 3 choices they will line up with 30 choices each when there are 3 choices available and 45 choices each when there are 2 choices. The results are shown in the middle box. The lines top to bottom represent the samples where the prize sits behind Doors A, B, C and D, respectively. The columns left to right represent, respectively, the number of times Doors B, C, and D were chosen. The numbers in each column add to 120, but this is not the case in general. Now we may ask the question: what is the probability that the prize sits behind Door A if Door B is chosen?
Door B has been chosen 120 times out of 360, that is, for 1/3 of the samples. For 30 samples, the prize was behind Door A. Thus, the probability that the prize was behind Door A after Door B is chosen is given by the ratio of 30/120 or ¼. The same is true for opening Door C or Door D, but this result is not general, but arises from our assumption of equally likely choices. This special condition is important theoretically and is referred to as the condition of maximum entropy or maximum uncertainty.
Given that Door B has been opened, the probabilities that the prize lies behind C and D must be equal and their sum must be ¾ . This is so because the prize cannot lie behind Door B. We get the same result by adding alone the column 45 plus 45 and dividing by 120. Thus the probability that the prize lies behind Door C is 3/8, and similarly for Door D. Given that Door B has been opened, the chance that the prize lies behind Door C rather than Door A bears the favorable odds of 3:2 rather than 2:1 as in the 2door game.
Mathematical Notation We restate these evident truths in mathematical notation which provides the means to generalize. First we state that
P (A  B) + P(C  B) + P (D } B) = 1,
where P (A  B) denotes the probability that the prize sits behind Door A given that Door B has been opened. The equation states the fact that because the prize is not behind Door B, it must be behind one of the other doors. Next,
P (A  B) ∝ P (B  A),
This line simply claims that that probability of a prize being behind Door A is proportional to the probability that Door B would be opened if the prize were behind Door A. It is common sense to say that the more likely one is to choose Door B when the prize is behind Door A, the more likely that this is indeed the situation when Door B is chosen. More generally,
P (A  B) ∝ P (B  A) P (A),
P (C  B) ∝ P (B  C) P (C),
P (D  B) ∝ P (B  D) P (D),
where P (X) is the probability that the prize was initially behind Door X. It is not necessary to assume P (X) is the same for all doors, but that is an inherent condition of our experiment where each line represents 90 samples. The relevance to bridge is that any card initially has an equal chance of being dealt to any of the four players.
Biased Choices Randomness does not require that all possibilities are equally likely. It is easy enough to introduce a bias in our experiment by assigning colours, Door B being red, the most popular colour, Door C being chartreuse, and Door D being yellow, the least favoured colour. How exactly the addition of colours would bias the results is unknown, so we shall make assumptions that are reflected in the numbers given in the righthand box. Red is twice as popular as chartreuse, yellow is never chosen when red is available, but is an equal choice with chartreuse otherwise. This assumption could be the subject of a test on results obtained in practice.
If we now go through the process indicated above for equal choices, we find the following: P (A  B) = 2/7, P (C  B) = 3/7, and P (D  B) = 2/7. Thus, Thus, P (A  B) divided by P (C  B) is 2/3, as in the previous example, but P (A  B) is equal to P (D  B). Also, the probability that the prize sits behind Door A has changed from ¼ to 2/7, even though every child had a ‘free’ choice of a door behind which there is no prize. What we really should say with regard to the first case is that the child made an unbiased choice. Some children may prefer chartreuse to red, and they were free to make that choice based entirely on personal preference.
If Door D is chosen: P (A  D) = 0, P (B  D) = 1, and P (C  D) = 0.
The effect of our assumption with regard to selection due to colour is that Door D is chosen only when Door B is unavailable. In bridge this is equivalent to the conclusion that if a defender fourth to play takes a ten with a king, declarer may assume he was not dealt kingjack. It cannot be said that he was not dealt kingqueen.
We now present the tables of conditional probabilities under each condition defined by choice of door in the 2 cases considered, equal choices and biased choices.
Conditional Probability Matrix  Equal Choices  Biased Choices 
–  P (A  B)  P (A  C)  P (A  D)  –  2/8  2/8  2/8  –  2/7  2/7  0  
–  P (B  B)  P (A  C)  P (B  D)  –  0  3/8  3/8  –  0  3/7  1  
–  P (C  B)  P (C  C)  P (C  D)  –  3/8  0  3/8  –  3/7  0  0  
–  P (D  B)  P (D  C)  P (D  D)  –  3/8  3/8  0  –  2/7  2/7  0 
The initial probability, P(A), that the prize sits behind Door A is given by Bayes’ Equation:
P (A) = P (A  B) · Q (B) + P (A  C) · Q (C) + P (A  D) · Q (D),
where Q (X) represents the probability that Door X be chosen. We use Q (X) because we must make a distinction between the possibility that Door X might be chosen blindly and the probability that it would be chosen. This distinction is not necessary if the choice of doors is unbiased, but such an assumption doesn’t always apply. In bridge there is a difference between the probability of a card being dealt to a given defender, and the probability that he would play it under the given circumstances if he had it.
Thus, based on the number given in the righthand box,
P (A) = (2/7) · (7/12) + (2/7) · (7/24) + 0 = 1/4. Also,
P (B) = P (B  C) · Q (C) + P (B  D) · Q (D),
= (3/7) · (7/24) + (1/8) = 1/4.
Similarly, the initial probabilities for P (C) and P (D) are correctly given if we use Q (X), the probability of Door X being chosen, rather than P (X), the probability that the door might have been chosen if the choices were unbiased. In relation to cards, P (X) refers to the a priori odds of the deal where all cards are treated equally regardless of rank, and Q (X) refers to the probability that a card would be played by a defender from a selection of cards. During the play one must refer the conditional probabilities as they apply to the cards in the suit to which a defender must follow. Probabilities depend on the priori knowledge, which includes such information as an imbalance in the vacant places at the time of decision that imposes a bias on a defender’s choices.
Testing an Hypothesis Say that we enumerated 360 samples with coloured doors that were submitted by schoolchildren. We compare the matrix of selections with the box on the right and ask whether the results can be said to conform to our model. Well, we can say immediately that there will be some (probably among those who choose to sit at the back of the class) who will prefer a hideous shade of sickening yellow to a bright and cheerful crimson. So we should never model human behavior under an assumption of ‘never.’ Furthermore, even if we assumed 5% for yellow versus red, it would be hard to justify or reject that assumption on a small number of samples. To put a number to a rare occurrence requires many samples in order to achieve a high degree of confidence. In fact, the matrix of the results may fit more closely the box for equal choices (The school colours of one set of happy students was green and yellow, whereas at another school the unpopular home room teacher was wearing a red dress.) So we must always be suspicious of results from small samples where local conditions will affect the choices.
The scientific method is to gather a set of data, say, 360 samples, adjust one’s hypotheses, then test again on an entirely new set of data. As one plays hands one is gathering results to add to the collection of data stored in our memory. It is easy to be swayed by our emotions. Emotionally successes outweigh the failures, and we may continue to pursue bad practices based on the good feelings we felt when successful. It is difficult to adjust solely on the basis of one’s personal experiences. Rarely does one see a player start fresh and change his losing approach. With regard to conventions that rarely occur, one would have to play many years in order to establish on the evidence alone the efficacy of their operation. Computer simulations can be useful if done carefully.
It is fair to say that scant evidence potentially being highly variable will often appear to arise from conditions of maximum uncertainty. This leads to conclusions such as ‘system doesn’t matter’, or ‘luck plays the biggest part in determining victory’. By now we should know this isn’t so. However, when attempting to analyze the actions of a mediocre player at the table, partners included, it is often the safest approach to assume a condition of maximum uncertainty, but this may not be the optimum approach.
A Sequence of Choices Next we consider the situation in which the school children are asked to choose one door then another when there are 3 doors behind which there is no prize. It is possible, even likely that the choice of 2 doors may be biased in some way. If the doors are shown in a diagram, I would expect the middle door would be the most frequently chosen. However, let’s assume the sequences are chosen equally. There are 6 permutations of choices from 3 doors as shown below.
Prize  BC  CB  BD  DB  CD  DC 
A  15  15  15  15  15  15 
B  0  0  0  0  45  45 
C  0  0  45  45  0  0 
D  45  45  0  0  0  0 
The requirement to choose 2 doors results in a greater resolution of the location of the prize. Given the sequence Door B followed by Door C leaves only 2 possibilities remaining, with the probability of the prize being behind the door not chosen 3 times greater than behind Door A. The same holds for any possible permutation.
Next we assume there is a bias introduced into the choice of doors, as defined by our previous assumption. Yellow door is never chosen if it can be avoided.
Prize  BC  CB  BD  DB  CD  DC 
A  60  30  0  0  0  0 
B  0  0  0  0  45  45 
C  0  0  90  0  0  0 
D  60  30  0  0  0  0 
The prize sits behind Door A only when B and C are chosen, in which case there is an equal probability that the prize lies behind Door D, the yellow door.
An EightEver Example Let’s see what is the effect of a false card on the probabilities under this condition: AT86 opposite KJ94. Declarer plays the ace followed by the 6 towards the KJ9, the defenders following with small cards all the way. The missing cards we denote as Quwyz and the observed sequence has been card u followed by card z, then card w from the LHO on the second round, the sequence being denoted as uz;w. What are the current relative probabilities of these 2 conditions:
Condition 1

Quw opposite yz  Condition II  uwy opposite Qz 
If all low cards are chosen with equal probability, the probability of Condition I, the probability of Q on the left (QL), proportional to ¼ the probability of the 32 split, because there 4 equally probable permutations. The probability of Condition II, the probability of the queen on the right (QR) is proportional to 1/6 times the probability of the 32 split because there are 6 equally probable permutations. The relative probability in favour of the Q being on the left is 3:2. In mathematical terminology:
P (QL  uz; w) = (3/2) · P (QR  uz; w)
This is part of the justification for the eightever rule of always finessing for the queen. However, what if the probability of playing card z rather than card y on the first round was 1/3 rather than 1/2 due to a known tendency of the RHO to false card? Then the current probability of QL and QR would be equal after the given sequence. Furthermore, if the RHO would always false card from a combination of yz, or would always give the correct count playing card z, card y would be characterized by the same certainty of being withheld as the queen from the combination. In this case,
P (QL  uz; w) = 3 · P (QR  uz; w).
This merely expresses the result that Condition I is characterized by 2 possible permutations and Condition II by 6 under the assumption that the LHO needn’t consistently favour one low card over the other in a 3card combination. Let’s now consider an example of biased selection arising from a story by David Bird, who delights in situations where the probabilities are influenced by psychological factors that produce the effect that P (X) ≠ Q (X). The tale is funnier than the equation suggests.
A Muzzy Monk’s Muddled Math
Humor often educates as it entertains; such is the case with the amusing series of monastic stories by David Bird which continue to open our eyes to a dazzling display of squeezes and endplays of which Abbot Hugo YorkeSmith is the usual victim. It is clearly the author’s design to show in subtle ways that the sophistic Abbot gets it wrong for the wrong reasons, but when Bird’s preferred alter ego, the clever Brother Lucius, speaks, one assumes the emphasis has shifted from spoof to proof.
In the June 2009 ACBL Bulletin Lucius has again overreached to 6NT and needs to bring in a club suit consisting of ♣QJ95 in his hidden (South) hand opposite ♣A742 in the dummy. After successfully running the ♣Q on the first round, to which Brother Xavier (West) follows with the ♣3 and the Abbot (East) with a highly suspicious ♣8, declarer has to decide whether on the second round to start with the ♣5 hoping to draw the ♣K from West, or to lead the ♣J hoping to pin a now bare ♣10 in the East.
As part of the prior knowledge stored over several years of play, the Abbot is known for his penchant for ‘false’ cards, so he would normally play the ♣8 from 3 possible holdings, ♣108, ♣1086, or ♣1083. As 2 of those combinations are 3card holdings, Brother Lucius erroneously concludes the odds are 2:1 that the Abbot was dealt a tripleton and Xavier a doubleton club. He must have spent considerable time in the cellar that day reclassifying the monastery’s vast reserves of Home Counties claret because his thinking was rather muzzy. Why? Answer: because one of those tripletons was ruled out when Xavier followed with the ♣3. Does that mean the odds are now 5050? No, not even the Abbot would think so as he is an avid reader of the works of Terence Reese.
The situation as Brother Lucius ponders his second lead in the club suit has been reduced to 2 potentially favorable combinations:
Xavier  Abbot  
Condition I  ♣ K3  ♣ 1086 
Condition II  ♣ K63  ♣ 108 
If with ♣K63 Xavier would always play the ♣6 on the first lead (as Bird seems to imply), then Condition II has been eliminated entirely as a possibility, so the ♣5 should be led on the second round, regardless of which club the Abbot has chosen. The ♣K will pop up to be taken by the ♣A in dummy. If Xavier only sometimes plays the ♣6 and the Abbot always plays the ♣8, Condition I is the more probable, and the same low club play is indicated. Lucius had no problem choosing to play a low club on the second round based on the known proclivities of his opponents.
On first take the theme of the episode appears to be a condemnation of defenders who always play ‘false’ cards thus becoming their own worst enemies, however, that may be said also of players who always choose ‘true’ cards. If Brother Xavier always gives true count with the ♣3 from ♣K63 and the Abbot similarly with the ♣6 from ♣1086, the card position becomes a certainty. The defect lies in the defenders’ predictability. Their best strategy is to follow with low cards at random, making plays of equal probability out of the appearances of the ♣3 on the left and the ♣8 on the right. This transmits the minimum possible amount of information to declarer. An insightful view of the defenders’ unbiased random selection of equivalent cards is that they are merely simulating the act of dealing, thus keeping the declarer in a state of maximum uncertainty to the extent possible under the circumstances.
The Inference from the Opening Lead The a priori odds are that the 32 and 23 club splits are equally probable. Conditions I and II each have 2 plausible permutations when the defenders follow randomly with low cards to the first round. To make a decision that has better odds than a coin flip, declarer must decide whether a 32 split is more probable than a 23 split on this particular deal, and to do so he needs to consider the other suits. The only clue is that Xavier has led the •Q placing him with touching honors in that suit and perhaps length. A 43 diamond split gives preference to a 23 club split, which, perhaps fortuitously, leads to the correct (Bird’s) decision.
Let’s look at this lead of the ♦Q with 7 diamonds to the QJ10 held by the defenders. Assume the lead against an uninformative auction to 6NT is from a sequence of QJ10. The fellow defender plays his most discouraging card. There are 2 cases to consider:
Diamonds 43  Diamonds 34  
QJ10x opposite xxx  QJ10 opposite xxxx  
Combinations  4  1 
Possible Plays  1  1 
On the basis that false carding is not an option it appear obvious that the diamonds are more likely to be split 43 than 34. Of course, there would be no funny story to tell if one thought that way….or maybe not. If the adventuresome Brother Cameron had been on lead and had chosen ♦Q from queen doubleton, the outcome might have been a happier one for the Abbot. Fans of the monks know that such a lead would be uncharacteristic of Brother Xavier, who, despite his perpetual criticism of the Abbot’s normal leads, wouldn’t himself dare to deviate from orthodoxy.
Against players who defend to best advantage, after the first round of clubs declarers should gather more information outside the club suit by cashing some top hearts and spades to see if the vacant places, like unreliable clarets, need to be reassessed. This may result in going down more than 1 trick in the slam, but it increases the probability that declarer will make the right decision in the club suit. By immediately playing a second round of clubs, Brother Lucius indicated he was pretty sure his opponents were acting predictably.
Happy anniversary Bob,
from what I can get, you started this blog exactly 1 year ago on December the 10th 2008.
And I feel like you are back on the same exact track, but correcting the conclusion you made when analyzing the hand 108 for KQ9xxx
Last year, you played 6 for K swallowed by RHO’s Ace, and wondered how to catch the J
You calculated that there would be 4 outcomes from Juw / Av and 6 from uvw / AJ
…but erred, I feel, in concluding that playing the Q was best.
Actually, assuming that RHO plays randomly A or v with Av, it’s 3 to 2 in favor of the finesse.
It looks like this is the message in “EightEver Example” today: restricted choice applies on small cards, but you have to compare
RHO: AJ = always A % Av = 1/2 A
LHO: Juw; always uw % uvw = 1/3 uw
so again 3 to 2
then you introduce the monks:
if the defenser’s Ace is not played exactly 1 time out of 2 with Av, then declarer’s strategy should adapt: if it’s played with probability p, you balance 1/2*p with 1/6
meaning:
p = 0 (always v from Av; AJ is certain): play the Q
p = 1/3 (2 times v, 1 time A): toss a coin
p = 1/2: finesse with ratio 3 to 2 as above
p = 1 (always plays A from Av) finesse with success 3 to 1
Cheers
Christophe
I have on it a little different view
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