Oh, hurry on, lad, you can make it!

Sure, except for the heart disease and not being able to walk 3 feet without gasping for air.

I see a delightful comedy of ascending awareness in the three above comments, yours by far the best.

Coupled up or not, letting yourself go is fun and worthwhile sometimes. Depends on what you want at the time and in the long run.

I'm just saying, no one likes a fattie. I hope your 226 lbs. includes the weight of your big blimp head. 'cause then it makes sense.

Men are harvesting extra nutrients in case they have to go to war. Women are just lazy.

Well, Dark Angel, I guess you must be right and I am wrong. You are, no doubt, a very fine man, and I am perhaps "a dunce of the first class." Love, Doug

You should have just used the pair (D, H), and avoided all the unnecessary remarks about which doors are empty, making the proofe easier to follow. You also barely explained the use of Bayes' theorem. I have made a more readable version for my own benefit. An element of the probability space is a pair (D, H) where D and H are elements of {A, B, C}. D tells you which door has the prize. H tells you which door the host opens. For example, the pair (C, B) describes the situation in which door C has the prize and the host opened door B. You initially choose door C. It clearly has a 1 in 3 chance of containing the prize. Now the host may open either A or B with equal likelihood, making P(C, A) = P(C, B) = 1/3 * 1/2 = 1/6. Choosing C rules out (A, C) and (B, C). Since before any doors are opened all doors as just as likely to contain the prize, P(C, A) + P(C, B) = P(A, B) = P(B, A) = 1/3 = 2/6. The probability space is therefore: State --------- Probability (A, B) -------- (2/6) (B, A) -------- (2/6) (C, A) -------- (1/6) (C, B) -------- (1/6) The host then opens door B. (This rules out (C, A) and (B, A)). The probability now that the prize is behind C can be found using Bayes' Theorem: P(prize is behind C | host opens B) = P(prize is behind C & host opens B) / P(host opens B) = P((C, B) / (P(C, B) + P(A, B)) = (1/6) / (1/6 + 2/6) = 1/3 Therefore by sticking you have only a 1/3 chance of winning the prize. Therefore by switching to A you have a 2/3 chance of winning. This can be further shown by finding the chance that the prize is behind A using Bayes' theorem: P(prize is behind A | host opens B) = P(prize is behind A & host opens B) / P(host opens B) = P((A, B) / (P(C, B) + P(A, B)) = (2/6) / (1/6 + 2/6) = 2/3

Yes, the correct answer is that you should switch. Opening the empty door cannot increase the probability that your first door was the right door, because you can always open an empty door from the other two, no matter what door you chose. Your last point about having no information as to which door is correct isn't right. You are offered a choice between door A and door C, but the fact that the host opened door B actually teaches you something about door A because it rules out the case in which A is empty and B has the prize, leaving the cases in which both A and B are empty, or A has the prize and B is empty. This idea is formalised in the following analysis: An element of the probability space is a 4-tuple (W, X, Y, Z) where W,X,Y are elements of {E, P} and Z is an element of {A, B}. The first three (W, X, Y) tell you the state of the three doors (either empty E, or with prize P). Z tells you which door the host chose - either A, or B. For example, the 4-tuple (E, E, P, B) Describes the situation in which door A is empty, door B is empty, door C has the prize, and door B was chosen by the host. The probability space is given in the following table: State--------------- Probability (E, E, P, A)-------- (1/3)*(1/2) = (1/6) (E, E, P, B)-------- (1/3)*(1/2) = (1/6) (E, P, E, A)-------- (1/3) (P, E, E, B)-------- (1/3) The probability that the prize is in C given that the host opened B is obtained using Bayes' Theorem, and is: (1/6)/(1/6 + 1/3) = 1/3 The probability that the prize is in A, given that the host opened B is: (1/3)/(1/6 + 1/3) = 2/3 Had the host chosen his door at random, and then revealed that it happened to be empty, the choice would indeed be 50:50 so it would no difference.

I would have thought it didn't matter, but that is the obvious answer and so probably not it. Maybe it is switch, because whatever door you choose there is always an empty one that can be shown to you, so the probability you have chosen the right door is still 1/3 after door B is revealed, so the probability the prize is in the remaining door is 2/3. I mean, you either have chosen the right door initially (proability 1/3) or you haven't (probability 2/3). opening door B doesn't change anything, because whatever door you open, there is always an empty door that the host can reveal. It doesn't really tell you anything. But it would seem a bit odd that you could have two doors, one of which contains a prize, and you have no information as to which it is, and the probability (as far as you are concerned) is not 50:50.

Stick Stick Stick Stick. Is the gameshow host Bob Carolgies?