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>That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about

>That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about it, try it. I misread A. I agree that it's 1/3 if you flip pairs until at least one is heads. I disagree that living in a world where pairs of tails don't exist and living in a world where pairs of tails do exist but are ignored are the same thing. >Are you somehow suggesting that the probability distribution of the other options is changed by this? Yes. I'm saying it's an unwitting double Monty Hall. The fact that he can't give you 3 cars changes the odds. Going (invisibly) from 4 possible worlds (including GG) to 3 changes the prob dist of the remaining 3, by which I mean 2. If Monty must guarantee you at least one boy, then there aren't really 3 remaining possibilities. The unguaranteed can be a boy, or it can be a girl. What is the third?

ColinWright 13y ago
>> The father says that at least one child is a boy. >> That eliminates the option of GG. Are you somehow >> suggesting that the probability distribution of >> the other options is changed by this? > Yes. ... Going (invisibly) from 4 possible worlds > (including GG) to 3 changes the prob dist of the > remaining 3, by which I mean 2. If Monty must > guarantee you at least one boy, then there aren't > really 3 remaining possibilities. The unguaranteed > can be a boy, or it can be a girl. What is the > third? The standard reasoning is that there are two options, but they are not equally likely. Equally likely options are, labelling oder child first, BB, BG, GB. These were equally probable before we gained the extra information, and the standard interpretation is that they remain equally probable after. In the Monty Hall case you choose a door, and it has probability 1/3 of being right. It remains 1/3 probable after he chooses a door. The other two doors together have probability 2/3 of concealing the car. It remains 2/3 after he has chosen and opened a door. The probabilities of the branches don't change in any of these cases. 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? > A: So now let's take another experiment. > I tell you I'm going to flip two coins > until at least one of them is heads. I > do that. Now I ask you to bet on whether > or not there is a tail. What do you > think are fair odds? And you say: > I misread A. I agree that it's 1/3 if you > flip pairs until at least one is heads. 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?
- 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.

ColinWright 13y ago

>> The father says that at least one child is a boy. >> That eliminates the option of GG. Are you somehow >> suggesting that the probability distribution of >> the other options is changed by this? > Yes. ... Going (invisibly) from 4 possible worlds > (including GG) to 3 changes the prob dist of the > remaining 3, by which I mean 2. If Monty must > guarantee you at least one boy, then there aren't > really 3 remaining possibilities. The unguaranteed > can be a boy, or it can be a girl. What is the > third? The standard reasoning is that there are two options, but they are not equally likely. Equally likely options are, labelling oder child first, BB, BG, GB. These were equally probable before we gained the extra information, and the standard interpretation is that they remain equally probable after. In the Monty Hall case you choose a door, and it has probability 1/3 of being right. It remains 1/3 probable after he chooses a door. The other two doors together have probability 2/3 of concealing the car. It remains 2/3 after he has chosen and opened a door. The probabilities of the branches don't change in any of these cases. 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? > A: So now let's take another experiment. > I tell you I'm going to flip two coins > until at least one of them is heads. I > do that. Now I ask you to bet on whether > or not there is a tail. What do you > think are fair odds? And you say: > I misread A. I agree that it's 1/3 if you > flip pairs until at least one is heads. 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?

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.