Principle of Proportionality
The Principle of Proportionality (see references below) is an immediate consequence of Bayes' Theorem. It reads
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If various alternatives are equally likely, and then some event is observed, the updated probabilities for the alternatives are proportional to the probabilities that the observed event would have occurred under those alternatives.
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The formal derivation is simple. Assume
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P(A1) = P(A2) = ... = P(An) > 0 and P(B) > 0.
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Then P(Am|B) = K P(B|Am), for all m = 1, 2, ..., n, where K > 0 does not depend on m. Indeed, by Bayes' theorem,
| | P(Am|B) | = P(Am ∪ B) / P(B) |
| | | = P(Am) P(B|Am) / P(B) |
| | | = (P(Am) / P(B)) P(B|Am). |
The assertion holds, with K = P(Am) / P(B) - constant from (*).
Let's examine several examples.
Bear Cubs Problem
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There are two bears - white and dark. Assume it is known that one of the bears is male. What is the probability that both are males?
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Solution. With the common assumtion that sexes are evenly distributed among the bears as among the humans, at the outset, there are four equally probable variants: A1 = (female/female), A2 = (female/male), A3 = (male/female), A4 = (male/male). Event B is the acknowledgement that one of the bears is male. Conditional probabilities of B assuming one of the A's are as follows:
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P(B|A1) = 0, P(B|A2) = 1/2, P(B|A3) = 1/2, P(B|A4) = 1.
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The conditional probabilities of A's assuming B are proportional to the above but must add to 1. So they are 0, 1/4, 1/4, 1/2. Only in the last event the second bear happens to be male, thus the probability of the latter happening is 1/2.
Three Pancakes
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You have a hat in which there are three pancakes: One is golden on both sides, one is brown on both sides, and one is golden on one side and brown on the other. You withdraw one pancake, look at one side, and see that it is brown. What is the probability that the other side is brown?
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Solution. At the outset, there are three likely events to extract a pancake: brown/brown, brown/golden, golden/golden. A pancake is taken out and shows a brown side. Relative to the three possibilities, the probabilities of this happening are 1, 1/2, and 0. Since the probabilities ought to add up to 1, the conditional probabilities become 2/3, 1/3, 0. Only in the first case the chosen pancake will have the second side brown. This will happen with the probability of 2/3.
Monty Hall Problem
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Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
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Solution. Originally the car was equally likely to be behind door #1 or #2 or #3, and assume you selected door #1. Under this circumstances, the probabilities of Monty opening the door #3, say, when it is actually behind door #1, or door #2, or door #3 are 1/2, 1, 0. Since the probabilities must add up to 1, these must be prorated, after which we obtain 1/3, 2/3, 0. Thus your chance of winning the car is 1/3 if you stick with your original choice and 2/3 if you switch.
References
- J. S. Rosenthal, Struck by Lightning: The Curious World of Probabilities, Joseph Henry Press (March 28, 2006)
- J. S. Rosenthal, Monty Hall, Monty Fall, Monty Crawl, Math Horizons, MAA, September 2008, pp. 5-7
Copyright © 1996-2008 Alexander Bogomolny
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