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Yeah, I actually deleted it before seeing your response because I thought it was unclear. My bad. >The most obvious way to do that is to talk about the older a

Yeah, I actually deleted it before seeing your response because I thought it was unclear. My bad. >The most obvious way to do that is to talk about the older and younger. Exactly. The father doesn't talk about older and younger, but the table does. What it should look like, because you're correct that age is irrelevant: B (given) * [b|g] -> [Bb|Bg], or if you account for age: [B is first|B is second] * [b|g] -> [Bb|bB|Bg|gB]. But you have to pick one or the other! Either way, it works, but you have to account on both ends, not just one. He's calculating for BG as if age were relevant, but it's not. The computer "simulation" is wrong because it is grounded in the logic that one child's sex wasn't a given all along, when in fact it was. NB: if you're throwing results out, that's a bad sign. Yea/nay? Or, again, you can just step back and say: X is a boy. What is the probability that Y is a boy? Why on earth would it matter what X was? Do you meet 2/3 women because you are a man?

ColinWright 13y ago
After enumerating all the options and their respective probabilities, yes, you discard all those that are inconsistent with the information to hand. You bring up the Monty Hall problem. The options there are: [ c ] [ g ] [ g ] [ g ] [ c ] [ g ] [ g ] [ g ] [ c ] These are all equally likely. Then you choose a door. Let's assume you choose door 1. To keep things equally likely, let's suppose Monty flips a coin. Heads he opens the left-most unchosen door that doesn't reveal a car, Tails he opens the right-most. Now we have six options: Heads: [ c ] [ g*] [ g ] [ g ] [ c ] [ g*] [ g ] [ g*] [ c ] Tails: [ c ] [ g ] [ g*] [ g ] [ c ] [ g*] [ g ] [ g*] [ c ] As you can see, there are now 6 equi-probable choices, and in 4 of them it's better to switch. 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? 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? C: Finally, 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? This last is how the question is usually interpreted. There are three options. 1: Your answers to A, B, and C are not all the same; 2: They are all the same, and not 1/3; 3: They are all the same, and all 1/3. If your answers to A, B and C are not all the same, I'd like to know why. If they are not all 1/3, I'd like to play game A with you. If your answer to C is 1/3, then you've agreed with the usually interpretation and contradicted yourself.
- gametheoretic 13y ago
  > To keep things equally likely, let's suppose Monty flips a coin. Aha! But he can't! Do you see? He can't open a door with a car, so he can't flip a coin in the event that the car is behind a door other than yours. He is bound by the rules not to do so. 1/3 your door is good; 2/3 your door is bad. If bad, he MUST leave you with the good door to switch to. So in 2/3 of scenarios, you have a 100% likelihood of winning if you switch. The answer to all 3 is 1/2. GG + "at least one is a boy" is not a possible world, but you are calculating the probability of the others, initially, as if it were, and then asking another question on top of it. You can't do that. BB is only 1/2 as likely as BG/GB if GG is a possible world for the question to begin with.
  - ColinWright 13y ago
    >> To keep things equally likely, let's suppose >> Monty flips a coin. > Aha! But he can't! Do you see? He can't open a > door with a car, so he can't flip a coin in the > event that the car is behind a door other than > yours. He is bound by the rules not to do so. Read again ... I said: > To keep things equally likely, let's suppose > Monty flips a coin. Heads he opens the left-most > unchosen door that doesn't reveal a car, Tails he > opens the right-most. So I allow for that. I have him always flip a coin to ensure that every branch of the game tree is equally likely, even though in 2 out of 3 cases it doesn't actually affect his choice of door. It's a standard trick that is proven to work in the real world. More, it's supported by the appropriate measure theory versions of probability. Please, stop assuming everyone except you is an idiot.
  - phaemon 13y ago
    > The answer to all 3 is 1/2. No, it's 1/3 as everyone has pointed out. Go and play the coin game described. Seriously, stop arguing and go and play it. See?

ColinWright 13y ago

After enumerating all the options and their respective probabilities, yes, you discard all those that are inconsistent with the information to hand. You bring up the Monty Hall problem. The options there are: [ c ] [ g ] [ g ] [ g ] [ c ] [ g ] [ g ] [ g ] [ c ] These are all equally likely. Then you choose a door. Let's assume you choose door 1. To keep things equally likely, let's suppose Monty flips a coin. Heads he opens the left-most unchosen door that doesn't reveal a car, Tails he opens the right-most. Now we have six options: Heads: [ c ] [ g*] [ g ] [ g ] [ c ] [ g*] [ g ] [ g*] [ c ] Tails: [ c ] [ g ] [ g*] [ g ] [ c ] [ g*] [ g ] [ g*] [ c ] As you can see, there are now 6 equi-probable choices, and in 4 of them it's better to switch. 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? 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? C: Finally, 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? This last is how the question is usually interpreted. There are three options. 1: Your answers to A, B, and C are not all the same; 2: They are all the same, and not 1/3; 3: They are all the same, and all 1/3. If your answers to A, B and C are not all the same, I'd like to know why. If they are not all 1/3, I'd like to play game A with you. If your answer to C is 1/3, then you've agreed with the usually interpretation and contradicted yourself.

gametheoretic 13y ago
> To keep things equally likely, let's suppose Monty flips a coin. Aha! But he can't! Do you see? He can't open a door with a car, so he can't flip a coin in the event that the car is behind a door other than yours. He is bound by the rules not to do so. 1/3 your door is good; 2/3 your door is bad. If bad, he MUST leave you with the good door to switch to. So in 2/3 of scenarios, you have a 100% likelihood of winning if you switch. The answer to all 3 is 1/2. GG + "at least one is a boy" is not a possible world, but you are calculating the probability of the others, initially, as if it were, and then asking another question on top of it. You can't do that. BB is only 1/2 as likely as BG/GB if GG is a possible world for the question to begin with.
- ColinWright 13y ago
  >> To keep things equally likely, let's suppose >> Monty flips a coin. > Aha! But he can't! Do you see? He can't open a > door with a car, so he can't flip a coin in the > event that the car is behind a door other than > yours. He is bound by the rules not to do so. Read again ... I said: > To keep things equally likely, let's suppose > Monty flips a coin. Heads he opens the left-most > unchosen door that doesn't reveal a car, Tails he > opens the right-most. So I allow for that. I have him always flip a coin to ensure that every branch of the game tree is equally likely, even though in 2 out of 3 cases it doesn't actually affect his choice of door. It's a standard trick that is proven to work in the real world. More, it's supported by the appropriate measure theory versions of probability. Please, stop assuming everyone except you is an idiot.
- phaemon 13y ago
  > The answer to all 3 is 1/2. No, it's 1/3 as everyone has pointed out. Go and play the coin game described. Seriously, stop arguing and go and play it. See?