cannabineer
Ursus marijanus
^^ Still not sure I get it, Heis. Would you post an analogy using strippers?
They won't show you the one who is the Tuckmaster. cn
Critical Thought Experiments
- Thread starter Heisenberg
- Start date
They won't show you the one who is the Tuckmaster. cn
Heisenberg
Well-Known Member
This is going to sound similar to a previous one, but pay attention.
You meet an old friend on the bus. He tells you he has two children. One is a girl born on a Friday. What are the odds that your friend has two daughters?
No trick of words or logic, I just want to know the mathematical chance based on the information given.
You meet an old friend on the bus. He tells you he has two children. One is a girl born on a Friday. What are the odds that your friend has two daughters?
No trick of words or logic, I just want to know the mathematical chance based on the information given.
Beefbisquit
Well-Known Member
The fact that he has a daughter already doesn't have any bearing on the odds of his second child being a female. 50/50 chance of it being a female.
Heisenberg
Well-Known Member
This would be correct if the question was, what are the chances of the second child being a girl. The question is, given what you know, what are the chances he has two daughters.
See this post for a previous similar question and answer.
Beefbisquit
Well-Known Member
There's a total of four outcomes for the two pregnancy's. Two outcomes for the first and two for the second. 1/2 x 1/2= 1/4 or .25%
Heisenberg
Well-Known Member
I am inclined to agree with your logic, though you need to take it a step further and eliminate the boy/boy outcome, so 1/3 chance.
But alas that is not the answer this time because it doesn't account for all the information given...
mindphuk
Well-Known Member
50%
You designated one as 'being born on a Friday.' That makes the other one either a boy or a girl. In the first scenario, you had the situation where child 1 could be a girl or a boy and child 2 a girl or a boy. Here, you have defined child 1 as a girl, leaving the only probability left with child 2. You eliminate one of the b-g combinations from the original 1 in 3 chance.
Heisenberg
Well-Known Member
I didn't mean to imply that the first born was a girl, though I see I have inadvertently suggested that in my response to BB.
We know he has two children, one of the children is female born on a Friday.
We know he has two children, one of the children is female born on a Friday.
NietzscheKeen
Well-Known Member
Why am I just now finding this thread!? It will take me a while to catch up on all the posts, but at least it is intriguing.
Beefbisquit
Well-Known Member
mindphuk
Well-Known Member
We don't need to know who was born first, only that a child was specifically identified. We can just call her Friday. Friday has 100% chance of being a girl. The other child has a 50% chance, which is the same as the odds that both are girls. I don't see how 2:3 would work, but if that's the right answer, please detail.
cannabineer
Ursus marijanus
I'll give this a whack. i am reading the setup to say that the other is not a girl born on a Friday. If the other is a girl, she'd have to be born on one of the remaining days of the week. A boy would be under no such constraint. I am assuming that the chances of a pregnancy being a boy or girl are at 50/50. So I'm calling a ~46% chance (six out of thirteen) chance that he has two daughters. cn
mindphuk
Well-Known Member
I don't see that as a valid assumption. It could have been, "I have one girl, born on a Friday, and the other girl also happened to be born on a Friday." We only have information about one girl, it doesn't eliminate anything about the second child.
Heisenberg
Well-Known Member
Well lets back up a bit. The chance that any man who has sired two children will have two daughters is 1/4. Each birth has a 1/2 chance of being female. 1/2 x 1/2 = 1/4. The possibilities are , MM, MF, FM, FF. Since we know at least one event resulted in female, we eliminate the MM, leaving us with 1/3 chance of the man having two daughters.
Now we are given additional information. The female was born on a Friday. Do we need to adjust our calculations to account for this new information, or is it irrelevant? Do the odds change?
I'll give a little more time before I reveal the answer.
Now we are given additional information. The female was born on a Friday. Do we need to adjust our calculations to account for this new information, or is it irrelevant? Do the odds change?
I'll give a little more time before I reveal the answer.
cannabineer
Ursus marijanus
I've been trying to see how this could be a Monty Hall problem but can't. If it were, then the answer (omitting the trickiness about weekdays) would be one out of three, since BB, GB, GG and BG each have equal probability of being a two-child father's family. (And of the four, only BB is disqualified by Dad's reply.) But I just cannot get past unity probability for one girl = 1/2 probability for two. Someone he'p! cn
<edit> lol. If I incorporate the weekday trickiness, then the chance of two girls is 12/20 or 60%.
Heisenberg
Well-Known Member
You do have to assume uniformity over gender and days of the week, if that helps.
cannabineer
Ursus marijanus
I would say No. The man volunteered that the girl was born on a Friday, discharging any selection bias for that day.
These father/child problems make me say Uncle. cn
Heisenberg
Well-Known Member
You are on the right track, but as MP said, you can not make any assumptions about the other child. It could be a boy or a girl, and could have been born on any day of the week.
cannabineer
Ursus marijanus
If that is so, I'll say two out of three.
If this is indeed a Monty Hall problem.
Is there a way to recognize those as a class? cn
mindphuk
Well-Known Member
Okay, let's break down the number of possibilities. For g-b we have 7 distinct possibilities, the girl on a Friday and the boy on one of the 7 days.
For the b-g, we have the same 7 possibilities.
For the g-g, we have only 13, not 14 as we shouldn't count Friday-Friday twice. So we have 27 possibilities and only 13 are two girls making 13/27 or 48.1%
If we take 'neer's belief that one and only one girl is born on a Friday, then we get 12/26 or 6/13 or 46.1%
