Four-Door Monty – A Recapitulative Continuation
In our previous blog we maintained that the easiest way to understand how probabilities work at bridge is to imagine a simpler game of chance where choices are presented to a great number of participants. In this segment we return to our scenario where schoolchildren are used as contestants in a 4-four Monty Hall game. We assume they adhere strictly to our probabilistic model concerning how choices are made, although in practice we can never expect such perfection. The aim is to demonstrate simply the concepts behind Bayes’ Theorem and indicate how they apply to card play.
The box on the left below indicates the four doors behind one of which sits the prize. The bold letter denotes the door which hides the prize. Each line represents 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. In an ideal experiment the children choose exactly in accordance with a rule which makes one choice no more likely than another.
| Initial Conditions | Trial Results | Probabilities P (X | Y) |
| A | B | C | D | — | 30 | 30 | 30 | — | 1/4 | 1/4 | 1/4 | ||
| A | B | C | D | — | 0 | 45 | 45 | — | 0 | 3/8 | 3/8 | ||
| A | B | C | D | — | 45 | 0 | 45 | — | 3/8 | 0 | 3/8 | ||
| A | B | C | D | — | 45 | 45 | 0 | — | 3/8 | 3/8 | 0 |
| Totals | 120 | 120 | 120 | Chosen | B | C | D |
The symbol P(X | Y) represents the probability that the prize lies behind Door X after Door Y has been opened to reveal the prize is not behind that door. The probability of the prize being behind Door A is ¼ regardless of which door is chosen by the participants. Presumably is what commentators have in mind when they say, ‘the probabilities don’t change’. The message is that P(A), the a priori odds of the prize being behind Door A, equals P(A | Y) regardless of which of the 3 is Door Y. The validity of this statement depends on the way the choice of doors is made. If we introduce a bias in the choice, the probabilities P(A | Y) vary as indicated in the example from our previous blog where the doors are painted in different colors to induce various degrees of preference from unsuspecting juveniles, as is shown below.
| Initial Conditions | Trial Results | Probabilities P (X | Y) |
| A | B | C | D | — | 60 | 30 | 0 | — | 2/7 | 2/7 | 0 | ||
| A | B | C | D | — | 0 | 45 | 45 | — | 0 | 3/7 | 1 | ||
| A | B | C | D | — | 90 | 0 | 0 | — | 3/7 | 0 | 0 | ||
| A | B | C | D | — | 60 | 30 | 0 | — | 3/8 | 3/8 | 0 |
| Totals | 210 | 105 | 45 | Chosen | B | C | D |
Under the stated rules of preference, P(A | B) and P(A | C) are equal even though they represent vastly different numbers of choices, and they are not equal to P(A).
The sum of the columns must equal 1. That is represents the fact that after Door Y is opened the prize must sit behind one of the other 3 doors. The sum of the rows is not equal to 1. We note that the sum of the choices along a row must equal 90 as a consequence of our choice of presenting 90 prizes for each door. (We could have biased that as well.) In order to recover the a priori probability of the prize being behind Door A, one must factor in the probabilities of each door being chosen. Thus,
P(A) = P(A | B) · Q(B) + P(A | C) · Q(C) + P(A | D) · Q(D) ,
where Q(X) represents the probability overall that Door X would be chosen. For an unbiased choice of doors, Q(X) is the same 1/3 for all doors. For a biased choice, the Q(X) vary in such a way that the a priori odds, P(A) through P(D), are recovered.
The above arguments are intended merely to present a reasonable scenario as to how the various probabilities apply in a Platonic experiment. Mathematicians would prefer a more rigorous (and obscure) argument. A statistical approach will be considered later, but for now, we shall pass quietly on to the consideration of how these ideas apply to bridge.
Application to Card Play
We can translate the door matrix into a configuration in which a suit played by declarer has 4 missing cards, ♣Q-8-6-2. It is assumed the LHO holds 3 of the 4 cards. For this trial the participants play the role of the LHO and are asked to choose any card they wish excepting the queen. We aim to estimate the probability that the RHO holds the ♣Q. First, we assume the spot cards are chosen at random. We isolate the ballots that have chosen the ♣8.
| LHO has these cards | RHO must have | Plausible Plays | Number of ballots |
| ♣8 | ♣6 | ♣2 | ♣Q | 3 | 30 |
| ♣Q | ♣6 | ♣2 | ♣8 | 0 | 0 |
| ♣8 | ♣Q | ♣2 | ♣6 | 2 | 45 |
| ♣8 | ♣6 | ♣Q | ♣2 | 2 | 45 |
Given that the LHO cannot choose the ♣Q, we arrive at the result that the ♣Q sits on the right in 30 cases, and on the left in 90 cases, so the odds of the ♣Q being on the left is unchanged from the probability associated with the initial distribution (the deal). Of course, the chance of the ♣8 being on the right is now zero. If we use the same ballots to calculate the odds that the ♣2 is on the left, we find the probability has changed from 3:1 to 5:2. As a bridge player one is more interested in the location of the ♣Q than that of a spot card.
Suppose we next impose upon the participants the severe restriction of always playing the highest spot card (giving false count.) The numbers of ballots on which the ♣8 has been chosen are as follows:
| LHO has these cards | RHO must have | Plausible Plays | Number of ballots |
| ♣8 | ♣6 | ♣2 | ♣Q | 1 | 90 |
| ♣Q | ♣6 | ♣2 | ♣8 | 0 | 0 |
| ♣8 | ♣Q | ♣2 | ♣6 | 1 | 90 |
| ♣8 | ♣6 | ♣Q | ♣2 | 1 | 90 |
The odds of the ♣Q (or the ♣2) being on the left are now 2:1, so the odds on the location of the ♣Q have been altered by the rules which have greatly restricted the number of plausible plays. At the other extreme in the scale of permissibility, each card can be chosen without restriction as with the dealing of the cards, which rule leads to the following configuration.
| LHO has these cards | RHO must have | Plausible Plays | Number of ballots |
| ♣8 | ♣6 | ♣2 | ♣Q | 3 | 30 |
| ♣Q | ♣6 | ♣2 | ♣8 | 0 | 0 |
| ♣8 | ♣Q | ♣2 | ♣6 | 3 | 30 |
| ♣8 | ♣6 | ♣Q | ♣2 | 3 | 30 |
The number of plausible plays is the same for all remaining conditions. This is reflected in the number of vacant places. The process began with 3 vacant places on the left and 1 on the right. After a card is played by the LHO, the vacant places are reduced to 2 on the left and 1 on the right. Thus, when the plausible plays have reached a status of equality between the remaining conditions, the vacant place ratio under the assumption of a given split equals the probability of the location of a particular missing card. Note that this same property is possessed as well by the results obtained under the previous rule, because, there too, the plausible plays attained equality.
Distinguishable Spot Cards
Spots cards cannot be said to be indistinguishable. Fans of Right Through the Pack have happily recognized this for a long time. Given a choice of cards from ♣8-6-2, most bridge players initially would play the ♣2 without giving the matter much thought. Some seeking to deceive would try the ♣8. Those with an innate love of obscurity are drawn inexorably to the ♣6. Taken together this unequal treatment of the spot cards alters the probabilities. Let’s suppose the ♣8 appears with a frequency of 1 in 6, the ♣6, 1 in 3 and the ♣2, 1 in 2 times. Here is the matrix for the play of the ♣8.
| LHO has these cards | RHO must have | Plausible Plays | Number of ballots |
| ♣8 | ♣6 | ♣2 | ♣Q | 6 | 15 |
| ♣Q | ♣6 | ♣2 | ♣8 | 0 | 0 |
| ♣8 | ♣Q | ♣2 | ♣6 | 2 | 45 |
| ♣8 | ♣6 | ♣Q | ♣2 | 2 | 45 |
Under the assumption that the spot cards are indistinguishable in the 3 combinations of Q-x-x, when the ♣8 appears the odds become 6:1 that the ♣Q is on the left. When the ♣6 appears the odds are unchanged from the a priori odds of 3:1. When the ♣2 appears, the odds are 2:1. So, when the ♣2 appears we are less confident than initially that the ♣Q is on the left, whereas when the ♣8 appears we are more confident in that regard. When the ♣6 appears we are left to puzzle its significance. Here are full results in matrix form with the first column representing the cards held by the RHO.
| Initial Conditions | Trial Results | Probabilities P (X | Y) |
| Q | 8 | 6 | 2 | — | 15 | 30 | 45 | — | 1/7 | 1/4 | 1/3 | ||
| 8 | Q | 6 | 2 | — | 0 | 45 | 45 | — | 0 | 3/8 | 1/3 | ||
| 6 | 8 | Q | 2 | — | 45 | 0 | 45 | — | 3/7 | 0 | 1/3 | ||
| 2 | 8 | 6 | Q | — | 45 | 45 | 0 | — | 3/7 | 3/8 | 0 |
| Totals | 105 | 120 | 135 | Chosen | C8 | C6 | C2 |
Plausible Plays In situations where the card choices are of equal probability, the number of plausible plays equal to the number of choices. Therein lies the origin of the terminology. In the combination of ♣Q-8-2, it is assumed the ♣8 and the ♣2 would be chosen equally. There are 2 real and immediate alternatives, and the probability of either being chosen is the reciprocal of the actual number of apparent choices. However, the probability of appearance of the ♣8 from ♣8-6-2 is 1 in 6. In this case there are not 6 immediate alternatives, there are only 3, and the various choices have differing probabilities of being chosen. However, in a virtual experiment choices are made repeatedly as many times as we specify. The expected number of ballots expressing the choice of the ♣8 is one-sixth of the total number cast, in the above case 15 out of 90. The expected number of ballots expressing a particular choice is the total number cast divided by the number of ‘plausible plays’, where fractional plays are allowed. If the ♣2 was expected over a long sequence of choices to be chosen at a frequency of 2 out of 3 times, say, the effective number of plausible plays would be 1.5, which is the reciprocal of 2/3. The ♣2 is chosen on average once out of every 1.5 opportunities.
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