2 ms·
>Before you are told anything about the children we have four equally likely possibilities: BB, BG, GB, GG. After one is eliminated, why should the others still
by gametheoretic 13y ago
>Before you are told anything about the children we have four equally likely possibilities: BB, BG, GB, GG. After one is eliminated, why should the others still not be equi-probable?
Because in Monty Hall, the effect of changing probability is produced not just via elimination but via process of elimination - which is produced only because sufficient dependencies exist for it: a) the guaranteed 1:2 ratio, b) Monty must eliminate a goat, and c) the goat he eliminates can't be behind your door. These dependencies don't exist in the children problem. Unless Monty's choice occurs under the constraint of avoiding the door we chose, nothing changes. Monty: "You door has a goat behind it" -- so what? Now both of the other doors are 50/50. No similar conflict exists in the other problem. No conflict at all exists in the other problem. Elimination occurs, but it doesn't allow you to do anything further. "No GG" is just one rule, and if one elimination rule isn't sufficient for Monty Hall probabilities to change, why here?
And elimination of what? The elimination not of a possibility, but of a probability? Does that statement even have rational meaning? I'm being dead serious - we can give it mathematical meaning quite easily, but does it even have meaning to begin with?
Yes, I apparently misread that one too. But again, I disagree on the issue of whether the form of the proposed analogue matches the original problem.
- ColinWright 13y ago>> OK. Now B: >> > B: Now I hunt among couples with two children >> > until I find one that doesn't have two girls, >> > and I ask: what are the odds they have two >> > boys? >> Do you agree that it's 1/3? > Yes, I apparently misread that one too. OK, so you agree that this is 1/3. Now on to C: >> > C: I hunt among couples with two children until >> > I find one that has at least one boy, and I >> > ask: what are the odds they have two boys? Do you agree that this is 1/3? Let's assume for fun that you do agree with that. The original problem was: A: I met a man who said that he has two children, at least one of which is a boy. What is the probability that both children are boys? Let's compare that with this one: B: Among all the couples who have exactly two children, and such that at least one of their children is a boy, what is the probability that both are boys? Reading this: > But again, I disagree on the issue of whether > the form of the proposed analogue matches the > original problem. It seems like you're going to claim that these aren't the same problem. I've made it very clear exactly what my model is for problem B. If you claim that problem A is different, then you need to provide me with a very clear model for your interpretation.