3 Door Logic Problem

[tsg]

So if in the original problem had said that the game would be repeated, you would have answered differently?

Not necessarily. You changed the problem by specifying that person A answers "the second coin was a head" to a later flip.

Also do you agree that there is no way to test the original problem?

No. I've shown you how to test the original problem with the original information given. You've changed that problem by adding new information.

 

[tsg]

The equivalent using coins is to have a friend toss two coins, then pick one at random (and yes tsg I mean that there is an equal chance of picking either coin) and hand it to you. What is the chance the other coin has the same side showing?

This is very easy. 50% of the time the coins will match. But that's counting two tails as well as two heads. It's not the same problem.

 

[Wobble]

This is very easy. 50% of the time the coins will match. But that's counting two tails as well as two heads. It's not the same problem.

Compare the two:

A person tosses two coins and picks one randomly to give to you. You look at the coin and see it is a head. What is the chance the other coin is also a head?

A woman has two children and takes one of them, chosen at random, to the shop. You meet her there, and see that the child is a girl. What is the chance the other child is also a girl?

Now tell me what the difference is.

 

[Wobble]

Not necessarily. You changed the problem by specifying that person A answers "the second coin was a head" to a later flip.

So you would be happy to give a probability of 1/3 for the first round, not knowing then what the person would say in the following rounds?

No. I've shown you how to test the original problem with the original information given. You've changed that problem by adding new information.

I want a way to test the original problem by tossing coins, or simulating tossing coins. That requires being able to run it more than once, without changing the problem as you say my example does.

 

[CurtC]

tsg, I have two kids (really I do). The one named "Brady" is a boy. What's the probability that the other is a boy?

 

[CurtC]

I'll try this a different way. Again, with my old friend I run into at the grocery store with her kid, here are all eight equally likely possibilities. Here the suffix 1 denotes the older, suffix 2 is the younger.

1. B1/B2 : Has B1 with her
2. B1/B2 : Has B2 with her
3. B1/G2 : Has B1 with her
4. B1/G2 : Has G2 with her
5. G1/B2 : Has G1 with her
6. G1/B2 : Has B2 with her
7. G1/G2 : Has G1 with her
8. G1/G2 : Has G2 with her

If each kid has a 50/50 chance of being either sex, and she randomly picked one to take to the store with her, do you agree that each of these eight possibilities is equally likely, each representing 12.5% of the population?

If so, since we then see that she has a daughter with her, we can rule out possibilities 1, 2, 3, and 6. Therefore we are left with the equal probabilities of 4, 5, 7, and 8. Out of those remaining four, two of them (7 & 8) have the kid at home being a girl.

50%.

 

[tsg]

So you would be happy to give a probability of 1/3 for the first round, not knowing then what the person would say in the following rounds?

The original problem never said anything about following rounds. It didn't say anything about a game. It didn't say anything about why that person chose to give you the information he did. It simply asked you what the probability of him having two heads was given the information you had. That question is answerable. However you choose to complicate the problem doesn't change that.

I want a way to test the original problem by tossing coins, or simulating tossing coins. That requires being able to run it more than once, without changing the problem as you say my example does.

I've given you one twice. All you're doing in my example is counting the number of times it is possible for him to answer the way he did and counting the number of heads he has. Period. Why is this so hard to understand?

 

[tsg]

tsg, I have two kids (really I do). The one named "Brady" is a boy. What's the probability that the other is a boy?

100% or you wouldn't be asking me.

 

[tsg]

Compare the two:

A person tosses two coins and picks one randomly to give to you. You look at the coin and see it is a head. What is the chance the other coin is also a head?

A woman has two children and takes one of them, chosen at random, to the shop. You meet her there, and see that the child is a girl. What is the chance the other child is also a girl?

Now tell me what the difference is.

What is the probability the coins will match is not the same as what is the probability they are both heads.

 

[tsg]

and she randomly picked one to take to the store with her

This is never specified. We have no idea why she chose to bring the girl with her.

 

[CurtC]

100% or you wouldn't be asking me.

But you're evading the question - if you know the name of one kid (enough to tell it's a boy) but not its birth order, what is the chance that the other kid is also a boy?

 

[Rasmus]

But you're evading the question - if you know the name of one kid (enough to tell it's a boy) but not its birth order, what is the chance that the other kid is also a boy?

A - You could be meeting a parent of two boys
B - You could be meeting a parent of a younger boy and an older daughter
C -You could be meeting a parent of a younger girl and an older boy

So, in 2/3 of all cases, assuming full randomness on every step, the other kid would be a girl; and in 1/3 of all cases it would be another boy.

As soon as you know that the boy you met is older, case B is eliminated.
As soon as you know that the boy you met is younger, case C is eliminated.
Then, the other kid would be a boy in half of all cases.

 

[tsg]

But you're evading the question - if you know the name of one kid (enough to tell it's a boy) but not its birth order, what is the chance that the other kid is also a boy?

1/3

I have two children. One is a boy 5 years old, takes karate (has is high white belt), likes to fish, plays computer games, likes to watch SpongeBob and Dorah the Explorer. He has brown eyes, light brown hair and is way smarter than I am. None of this tells you anything about my other child. The chances that I have another boy, as far as you know, are 1/3.

 

[tsg]

3. B1/G2 : Has B1 with her
4. B1/G2 : Has G2 with her
5. G1/B2 : Has G1 with her
6. G1/B2 : Has B2 with her

We don't have enough information to assume these are equally likely. Your statement was that I meet an old friend who has her daughter with her. I have no idea why her daughter is with her or why she chose to bring her. I can't assume that she would bring her son, if she had one, half the time so I can't eliminate those possibilities. I don't have enough information to do so. If she's buying shoes for her daughter, she would never bring her son, so in every case where she has a son she would bring her daughter, so the chances are 1/3 the other child is a girl. She could be doing something where she would only bring a daughter if she didn't have a son, in which case the probability of her having another daughter is 100%. It could be a case where she would have brought both daughters, if she had another one, so the chances are 0% she has another daughter. You're eliminating possibilities for which you don't have enough information to do so.

If each kid has a 50/50 chance of being either sex, and she randomly picked one to take to the store with her,

You're adding constraints that didn't exist in the original problem. This is only true if she is going to bring a particular child 1/2 the time and you never specified that, and it is not a reasonable assumption to make.

 

[CurtC]

I have two children. One is a boy 5 years old, takes karate (has is high white belt), likes to fish, plays computer games, likes to watch SpongeBob and Dorah the Explorer. He has brown eyes, light brown hair and is way smarter than I am. None of this tells you anything about my other child. The chances that I have another boy, as far as you know, are 1/3.

So according to your figuring, if I knew the age of this other child (I know the first, a boy, is 5), would the chance still be 1/3 that it's a boy?

What about if this five year old were a girl? Would the chance that the other kid being a boy now be 2/3? If so, how would the sex of your five year old change the chance of the other kid being a boy?

Does these questions not tell you that something is wrong?

Slicing the problem along the older/younger axis is one way to do it if that was is helpful, but you can't let it be the only way to look at it. In the case of your kids, the proper way would be to call the five year old who likes karate and SpongeBob kid "A" and the one whose sex I don't know kid "B".

So here are the (equally likely) possibilities, if I didn't already know kid A's sex:

A/B
---
B/B
B/G
G/B
G/G

Since I know that A is a boy, I can rule out the last two possibilites, therefore the other kid has a 50% chance of being a boy as well.

The whole older/younger thing is a red herring.

 

[tsg]

So according to your figuring, if I knew the age of this other child (I know the first, a boy, is 5), would the chance still be 1/3 that it's a boy?

No, because you could rule out an older/younger sibling.

Slicing the problem along the older/younger axis is one way to do it if that was is helpful, but you can't let it be the only way to look at it. In the case of your kids, the proper way would be to call the five year old who likes karate and SpongeBob kid "A" and the one whose sex I don't know kid "B".

However you slice it, you have to be sure the way you're slicing it leads to equally likely possibilities. The fact that a particular child happens to be with the parent is no more information than that the parent has at least one child of that sex. Having a name doesn't change anything. Having one age doesn't change anything. Knowing that the child is the oldest/youngest does because it eliminates possibilities that we know to be equally likely.

So here are the (equally likely) possibilities, if I didn't already know kid A's sex:

How do you know they are equally likely? You have no idea why I chose kid A and not kid B. If I always choose kid A because he's my favorite (not the case, it's just an example before I start getting parenting advice on a hypothetical probability problem), you are going to falsely eliminate half the time I would have chosen kid B when it would actually be never. You can't just assume that I chose them at random with equally likely probability unless I told you that I have.

 

[CurtC]

How do you know they are equally likely? You have no idea why I chose kid A and not kid B. If I always choose kid A because he's my favorite (not the case, it's just an example before I start getting parenting advice on a hypothetical probability problem), you are going to falsely eliminate half the time I would have chosen kid B when it would actually be never. You can't just assume that I chose them at random with equally likely probability unless I told you that I have.

True that I don't know what might have motivated you to pick one over the other, so I was going with the assumption that this was a random decision. However, even if I then find out whether this kid is older or younger, I still don't know your motivation for picking that one first, and the chance of the other being a boy cannot be different based on just that little fact.

Do you agree, that if we know that the "known" kid was selected randomly, that the chance of the other's sex is then 1/2 either way? I think I've seen you say that. Let's get agreement on that and then go from there.

 

[tsg]

True that I don't know what might have motivated you to pick one over the other, so I was going with the assumption that this was a random decision. However, even if I then find out whether this kid is older or younger, I still don't know your motivation for picking that one first, and the chance of the other being a boy cannot be different based on just that little fact.

Do you agree, that if we know that the "known" kid was selected randomly, that the chance of the other's sex is then 1/2 either way? I think I've seen you say that. Let's get agreement on that and then go from there.

I haven't committed to it yet, but I believe it may be the case. There's still something I'm missing somewhere...

 

[Lamuella]

OK, yet another scenario:
You run into an old friend at the store. She tells you she has two teenaged children. What are the chances of her having two girls if the odds of her having a girl or a boy are 50/50?

As you talk and walk around the store, she mentions something about getting a training bra. As boys do not as a rule wear training bras, you now know she has at least one daughter. To put it another way, you now know that she doesn't have two sons. What are the odds of her having two girls?

 

[Lamuella]

and another one:

you are at a vast social gathering that is attended by an entirely random cross section of the population. The only requirement is that you must be married and attend with your spouse.

The MC asks everyone to stand for a fun game.

The MC tells everyone with more than two children to sit down

The MC tells everyone with less than two children to sit down. The only people remaining are those with two children.

Of the people still standing, what are the chances of them having two daughters, assuming a 50/50 boy/girl split?

The MC then tells everyone who doesn't have a daughter to sit down. The only people still standing are people with at least one daughter.

Of the people still standing, what are the chances of them having two daughters, assuming a 50/50 boy/girl split?

The MC then tells everyone whose first child was a boy to sit down. The only people still standing are those whose first child was a girl.

Of the people still standing, what are the chances of them having two daughters, assuming a 50/50 boy/girl split?