This is an interesting problem, and I think that Alf, ccvi, and Blauk have it right. The critical difference between the Monte Hall scenario and the one Mandoble proposed is that in the Monte Hall scenario, it is a given when one is partitioning the sample space that Monte will always select a door hiding a goat. That, of course, is not true in Mandoble's scenario--the bomb that was dropped might have been (with 1/3 probability) the live bomb.
The reason that this is important is that in the Monte Hall case, you know that you've got a 1/3 chance of having selected the correct door and a 2/3 chance that you're wrong. Once Monte shows you one of the two wrong choices, which he will always do, there's still just a 1/3 chance that your initial choice is correct. By switching, then, you'll increase your chances of being correct to 2/3.
We can represent the difference mathematically using a priori sample spaces, in which we assign probabilities to all of the permutations in the two cases. Here are the two a priori sample spaces, which show the initially selected choice followed by the revealed choice. The three options are labeled C (correct), I1 (incorrect choice #1), and I2 (incorrect choice #2).
Monte Hall:
C I1 1/6 chance
C I2 1/6 chance
I1 I2 1/3 chance
I2 I1 1/3 chance
Mandoble:
C I1 1/6 chance
C I2 1/6 chance
I1 I2 1/6 chance
I1 C 1/6 chance
I2 I1 1/6 chance
I2 C 1/6 chance
As you can see, the critical difference between the two cases is that in the Monte Hall case, when the initial selection is incorrect, the revealed choice will always be the other incorrect choice. In the Mandoble case, on the other hand, when the initial selection is incorrect, the revealed choice (i.e., the dropped bomb) may be either correct (i.e., live bomb) or incorrect.
If we define event A to be that the initial choice is correct and event B to be that the revealed choice is incorrect, our a priori sample spaces yield the following probabilies:
Monte Hall:
P(AB) = 1/3 (i.e., probability of both A and B happening, or the sum of the probabilities of the first two lines of Monte's chart)
P(B) = 1 (i.e., Monte ALWAYS reveals a door with the incorrect choice)
Mandoble:
P(AB) = 1/3 (same as Monte)
P(B) = 4 * 1/6 = 2/3 (i.e., sum of the rows 1, 2, 3, and 5 of Mandoble's chart)
From basic statistics, we know that the probability of A given B, which is often written P(A|B), can be calculated as follows:
P(A|B) = P(AB) / P(B)
Plugging in our numbers, we see that in Monte's case,
P(A|B) = (1/3) / 1 = 1/3 ==> 1/3 chance that our initial guess was correct, given that Monte revealed an incorrect choice. Thus, there is a 2/3 chance that if we switch, we'll win the car.
In Mandoble's case,
P(A|B) = (1/3) / (2/3) = 1/2 ==> there's a 1/2 chance that the initial selection was the live bomb, given that the revealed bomb was a dud.
Best regards,
JNOV
