posted
This seems to be making the rounds through the internets right now, much like the Monty Haul one did last year. I've seen it from two independent sources so far, so I figure I would beat the rush and post it here first.
Let's say you meet a father who tells you that he has two children and that one of these children is a boy and was born on a Tuesday. What would you say is the probability that his other child is also a boy?
The answer is actually 13/27. Weird, right? It's because knowing that the identified son was born on a Tuesday is actually important, because it let's you partially identify or fix your sample space.
There's a big difference in probability between ambiguous sets like "two kids and one of them is a boy" and a defined set like "two kids and the oldest is a boy".
In the first case, there's a 1/3 chance that the other child is a boy. Your sample space is: BG GB BB all of which are equally probable.
In the second case, there's a 1/2 chance that the other child is a boy. This sample space is: BG BB all of which are equally probable.
The real mind bender is that, though oldest and youngest are easy to use and make sense, any criteria that uniquely identifies a child has the same result.
So, if a parent tells you they have two kids and one of them is a boy, as stated the probability that other is also a boy is 1/3. However, if, right after he tells you that, his son comes up and says "Hi Dad!", the probability that the other is also a boy is now 1/2.
The "born on a Tuesday" case is kind of in between these two cases. You may notice that it is very close to the uniquely identifying cases as 13/27 is very close to 1/2. However, it doesn't fully uniquely identify the son and thus the boy you know about can't be completely fixed in the sample space. That's because the father could have two boys both born on a Tuesday. Because this is relatively rare (1 out of 7 days in the week), the probability is pretty close to 1/2.
However, if it was something where the probability of overlap was greater, the probability that the other child was a boy would be less. Say if he told you that his son was born in the morning, the probability of the other being a boy would be 3/7.
Posts: 10177 | Registered: Apr 2001
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quote:So, if a parent tells you they have two kids and one of them is a boy, as stated the probability that other is also a boy is 1/3. However, if, right after he tells you that, his son comes up and says "Hi Dad!", the probability that the other is also a boy is now 1/2.
I disagree. You don't know whether "hi dad" boy is the older or younger, and so you can't eliminate any of the three configurations of which he might be a part:
The likelihood that the child you haven't met is a girl is twice as high. Because the sample space doesn't change between "two children at least one of whom is a boy" and "two children one of whom I've met and therefore know is a boy."
I can't put this in strict mathematical terms, but I'm very nearly certain.
To illustrate:
You have a room full of fathers with two children at least one of which is a boy, who were otherwise selected randomly. Get enough fathers in there, and close to 1/3 of the fathers will be in there with two boys.
If you ask each father to step outside the room with one boy (leaving the other child behind), you can then satisfy the "child I've met" condition. It does not change how many of these fathers have two boys.
On the other hand, if you ask each father to step out with his older child if it is a boy, you will only get a subset of the fathers, and it changes the probability that each of those fathers has two boys to 50%.
Posts: 4287 | Registered: Mar 2005
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quote:Originally posted by MrSquicky: Let's say you meet a father who tells you that he has two children and that one of these children is a boy and was born on a Tuesday. What would you say is the probability that his other child is also a boy?
Me, I would play trick question word games with this and say that the chances of the other child being a boy is 0 since he stated that ONE was a boy.
Now if he said "at least one" is a boy, then we can play the game Posts: 891 | Registered: Feb 2010
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quote:If you ask each father to step outside the room with one boy
The problem with this is that you are filtering the sample space based on a dependent condition.
To preserve an independent sample space, you'd have to ask the fathers to leave the room with one child.
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posted
What do you mean by "filtering the sample space based on a dependent condition"? It's the same sample space (the exact same set of fathers that existed before they stepped out of the room).
I guess I better clarify: is your argument that having a boy show up and say "hi dad" reduces the sample space? If so, how? How does it eliminate 1 of the 3 possibilities that exist before he pops in?
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quote:Originally posted by MrSquicky: However, if it was something where the probability of overlap was greater, the probability that the other child was a boy would be less. Say if he told you that his son was born in the morning, the probability of the other being a boy would be 3/7.
What? I'm not a probability expert, but if you have one child who's a boy, the chances of your other child being a boy are 50%, if you assume that the chances of having either a boy or girl are equal. How does knowing anything else about the one child change anything about that? I'm lost.
Posts: 9912 | Registered: Nov 2005
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quote:Originally posted by scifibum: What do you mean by "filtering the sample space based on a dependent condition"? It's the same sample space (the exact same set of fathers that existed before they stepped out of the room).
I guess I better clarify: is your argument that having a boy show up and say "hi dad" reduces the sample space? If so, how? How does it eliminate 1 of the 3 possibilities that exist before he pops in?
From what I understand, there's a 2/3 chance that there is a boy and a girl child, and a 1/3 chance that there are too boys. However if an unidentified child approaches, the family with two boys would be twice as likely to have a boy show up than a family with only one boy, thus evening out the odds.
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posted
The problem is incomplete as stated. We don't know why the father is telling us the information. If he is just volunteering some information about one of his children then the probablity of the second one being a boy is 50/50.
If you restate the problem as follows: Mark: Hi I have two kids Me: Is one a boy born on Tuesday Mark: (Thinks a second) Why yes!
Then the probability of that the second is a boy is 13/27.
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posted
scifi, You can't pick the boys to leave with the father and keep an independent probability because the the fact that they are boys are part of the problem.
Like I said, to keep independent probabilities, you'd have to have the fathers leave with a random child.
If you do that, you'll end up with, on average, 66% of the fathers leaving with a son and 33% with a daughter. If you look at the fathers with sons, the chance that their other child is a son is 50%.
Posts: 10177 | Registered: Apr 2001
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quote:Originally posted by MrSquicky: However, if it was something where the probability of overlap was greater, the probability that the other child was a boy would be less. Say if he told you that his son was born in the morning, the probability of the other being a boy would be 3/7.
What? I'm not a probability expert, but if you have one child who's a boy, the chances of your other child being a boy are 50%, if you assume that the chances of having either a boy or girl are equal. How does knowing anything else about the one child change anything about that? I'm lost.
Think of it this way.
If you have two kids, the sample space is:
BB BG GB GG
and each has an equal chance.
Now, if you know that a parent has at least one boy, you can eliminate the last combination. Then you're left with:
BB BG GB
again, each is equally likely. So, you can easily see that the chance of of two boys (BB) is 1 out of 3 equally likely possibilities and having one boy and one girl (BG or GB) is 2 out of 3.
If we can fix the boy in a spot however, by finding out he's the oldest or the only one you met, then the only possibilities are
BB BG
So, in that case, the probability is 1/2.
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posted
To look at it another way, let's say we have 100 parents, each of whom have 2 kids.
25 have BB 25 have BG 25 have GB 25 have GG
Remove any that don't at least on boy.
25 have BB 25 have BG 25 have GB
and our total is 75 parents. You can easily see that only 1/3 of the parents have two boys.
The reason why the probabilities you expect get messed up is that we're removing parents based on a dependent condition - that is, they don't have at least one boy.
That moves us from 1/4 of parents having two boys to 1/3.
Posts: 10177 | Registered: Apr 2001
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quote:Originally posted by MrSquicky: scifi, You can't pick the boys to leave with the father and keep an independent probability because the the fact that they are boys are part of the problem.
Like I said, to keep independent probabilities, you'd have to have the fathers leave with a random child.
If you do that, you'll end up with, on average, 66% of the fathers leaving with a son and 33% with a daughter. If you look at the fathers with sons, the chance that their other child is a son is 50%.
This is sensible enough, I guess. I was not attaching any probability to the event of the child walking in and saying "hi". I think it works as long as you do so.
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posted
John's right. If the father is just saying if he has at least one boy and then supplying a day that one of them was born on you've gained no extra information from the day. If he's being asked specifically if one of them is born on Tuesday it goes to 13/27, and the intuitive reason is that the more boys he has the more likely one of them is born on Tuesday.
Same as how the Monty Hall problem doesn't work if it's not specified that the host knows the contents of the doors and always opens a bad one. If he opened a random door there's no reason to switch.
Posts: 148 | Registered: Feb 2000
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posted
I curious. I thought that my statement of the problem was pretty obviously indicating that a single child is identified as both a boy and born on a Tuesday. To wit:
quote:Let's say you meet a father who tells you that he has two children and that one of these children is a boy and was born on a Tuesday.
Are people saying that this is insufficient to get the 13/27 probability or was it just missed?
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posted
I think it depends on how those statements come about. Given two children, if the conversation goes:
- Do you have at least one boy? - Yes - Do you have a boy born on a Tuesday? - Yes
then 13/27, and it makes logical sense that the probability goes up because in the case with 2 boys there's more chance that one of them will be born on the day you chose to ask about. If the second part is instead:
- Randomly name a day of the week you have a son born on. - Tuesday.
then you've learned nothing useful and it's still 1/3. There are still 13 boy-boy combinations and 14 mixed combinations, but in the mixed combinations there's a 1/1 chance that he would say Tuesday, whereas in the boy-boy combinations there's a 6/13 chance he would give the birthday of the other boy instead. So the mixed combinations are 13/7 as likely and you're back to 1/3.
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