Wonderfully logical illogical probability problem

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OSB

OSB

OSB

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I had to work this one out this morning, i first thought wtf!? Then i worked it out, see what you think:

You meet your new tutor in town accompanied by a young girl. He says to you "I have two children, this is my daughter Lisa". What is the probability that his other child is a boy. Would it make a difference if he had said "I have two children, this is my youngest child, Lisa"? (You may assume a 50% chance of any one birth being a boy or a girl.


Have fun!

OSB
 
Obviously it's more likely he has a son if he refers to the other child as his daughter. In the second instance he's just referring to children without bringing gender into it, it's probable that means that there's no gender difference to separate the children.
 
The possibility in both instances is 50%. No matter how badly worded the statements may be to try to persuade you to think one way or the other, there is still in both instances a 50% possibility that the other child is a boy. (Edit: as the person above me has just demonstrated, perfectly)
 
The obvious answer (to me anyway) would be that it's very probable he has a son. If he said "youngest child", it is 50/50.

I'm most likely wrong though.
 
Unless he imagines that people can't tell a girl from a boy, the fact he specified 'daughter' means nothing.

Either that or his daughter is pig ugly and looks like a boy.

50/50 anyway.
 
In terms of pure statistics it will only ever be 50% either way, yes. But as he was referring to different turns of phrase I assumed he meant to take them on-board too. How do you do a statistical analysis of inflection and phrasing influencing the result... I'd imagine it's possible somehow but I hated statistics at college.
 
There are four configurations of children:
Youngest first the possibilities are GG, GB, BG, BB (G - Girl, B - Boy) and all are equally likely.

First situation.

BB of them is ruled out by knowing that one is a girl.
Thus, we only have GG, GB and BG left.
Knowing that the child with him is a girl gives a 1/3 chance that he has another girl and 2/3 chance that the other child is a boy.

Second situation.
We now know that the youngest child is a girl, this leaves only GG and GB.
Thus the probability is 1/2 that the other child is a boy.

Hope that's right!
 
Haircut said:
There are four configurations of children:
Youngest first the possibilities are GG, GB, BG, BB (G - Girl, B - Boy) and all are equally likely.

First situation.

BB of them is ruled out by knowing that one is a girl.
Thus, we only have GG, GB and BG left.
Knowing that the child with him is a girl gives a 1/3 chance that he has another girl and 2/3 chance that the other child is a boy.

Second situation.
We now know that the youngest child is a girl, this leaves only GG and GB.
Thus the probability is 1/2 that the other child is a boy.

Hope that's right!
Click to expand...

Spot on!!!! Crazy how looking at it you would think 50-50 both times, stupid question! But actually when you do the maths its 2/3 first time and only 1/2 the second!!!
 
See, this is where maths confuses me. You meet a guy with his daughter and you know he has one other child. As long as he doesn't actually tell you the gender of the other child or give you any clues, how can anything he say make it anything other than 50/50?
 
Masterdog said:
See, this is where maths confuses me. You meet a guy with his daughter and you know he has one other child. As long as he doesn't actually tell you the gender of the other child or give you any clues, how can anything he say make it anything other than 50/50?
Click to expand...

Exactly, this is nothing to do with maths so you can't analyse it as such.
 
Masterdog said:
See, this is where maths confuses me. You meet a guy with his daughter and you know he has one other child. As long as he doesn't actually tell you the gender of the other child or give you any clues, how can anything he say make it anything other than 50/50?
Click to expand...

Probability can be counterintuitive, if you work it out you will come to the correct answer though.
This is a similar problem to the Monty Hall Problem, Wikipedia has a good page explaining it.
http://en.wikipedia.org/wiki/Monty_Hall_problem

I've also just found out that they have a page dedicated to this exact problem!
http://en.wikipedia.org/wiki/Boy_or_Girl

Is there anything left that's not on Wikipedia these days? :p
 
Haircut said:
There are four configurations of children:
Youngest first the possibilities are GG, GB, BG, BB (G - Girl, B - Boy) and all are equally likely.

First situation.

BB of them is ruled out by knowing that one is a girl.
Thus, we only have GG, GB and BG left.
Knowing that the child with him is a girl gives a 1/3 chance that he has another girl and 2/3 chance that the other child is a boy.

Second situation.
We now know that the youngest child is a girl, this leaves only GG and GB.
Thus the probability is 1/2 that the other child is a boy.

Hope that's right!
Click to expand...

Correctomundo!
 
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