3 ms·
Before he asks the question, there are 4 possibilities: 1. A lives, A gets told B dies 2. A lives, A gets told C dies 3. B lives, A gets told C dies 4.
by gort 17y ago
Before he asks the question, there are 4 possibilities:
1. A lives, A gets told B dies
2. A lives, A gets told C dies
3. B lives, A gets told C dies
4. C lives, A gets told B dies
Since in fact A got told that B dies, we are either in situation 1 or 4. However, these did not start off at equal probability. Lets define P(x) as the probability that x is the actual situation. It is obvious that, before the question is asked:
P(1 or 2) == P(3) == P(4) == 1/3
Thus P(1) < P(4). This remains true once situations 2 and 3 have been ruled out.
- jonsen 17y agoInteresting setup. So P(1) == P(2) == 1/6 When A is told that B dies, it's down to 1. or 4. P(1 or 4) == 1/6 + 1/3 == 1/2 So chance A lives is now P(1) of P(1 or 4) == 1/6 of 1/2 == 1/3 And the chance C lives is now P(4) of P(1 or 4) == 1/3 of 1/2 == 2/3
- moo 17y agoA's survival chances don't change because the guard is filtering out the possible answers: B lives, A gets told A dies C lives, A gets told A dies