Brothers And Sisters - An incorrect solution?
posted by SuperDuper
- May 04 2011 11:20:52 PM
So I was looking at this riddle on the website:
You and a friend are standing in front of two houses. In each house lives a family with two children.
"The family on the left has a boy who loves history, but their other child prefers math," your friend tells you.
"The family on the right has a 7-year old boy, and they just had a new baby," he explains.
"Does either family have a girl?" you ask.
"I'm not sure," your friend says. "But pick the family that you think is more likely to have a girl. If they do have a girl, I'll give you $100."
Which family should you pick, or does it not matter?
The official solution, according to the website, says (Warning: If you don't want to know the solution, don't look ahead.) that we should pick the house on the left, as there is a 2/3 chance that the house on the left has a girl, and a 1/2 chance that the house on the right has a girl, as the age of the children on the left house are never given.
They offer these charts for the houses:
Left house:
Younger Older
Girl Boy
Boy Girl
Boy Boy
Right house:
Younger Older
Girl Boy
Boy Boy
They also tell us:
"This is a very counterintuitive riddle. It seems like there should always be a 1/2 chance that a given child is a girl. And in fact there is. The key word there is "given". Because we are not asking about a "given" child for the house on the left. We are asking about what could be either child."
And I'll just stop it right there. In the riddle, you ask your friend "Does either family have a girl?", and the friend responds "pick the family that you think is more likely to have a girl."
This, to me, seems to be specifically asking the question "Is or is not the other child a girl?"
When the solution says "We are asking about what could be either child.", they seem to be suggesting that, at some point, they were asking the specific question "Is the other child a younger girl, a younger boy, an older girl, or an older boy?"
The riddle itself, however, was not asking this question; the riddle was asking if, yes or no, the other child was a boy or girl, and that the family had a child that was a girl, regardless of age.
If they were asking "what could be either child", they might as well have added in more factors, such as whether or not the other child could have been a long or short haired, green or brown eyed, taller or shorter, stronger or weaker, younger or older boy or girl.
There are a lot of things the other child could have been, but I don't think any of those other factors have anything to do with what the riddle asks, which is to "pick the family that you think is more likely to have a girl." And age doesn't have anything to do with whether the child is more likely to be a girl, correct?
The solution seems to me to be asking a question that the riddle itself was not.
And this doesn't seem to be an open-ended riddle either, that has a case of "we're both right"; the way this riddle is set up, and with the solution they give, they are saying that there is a single solution to this riddle of which house has a higher probability.
Please correct me if I'm wrong, and point out any flaws on my end. I honestly believe their solution is incorrect, or they changed around the original question to something that it actually wasn't. Something just doesn't seem right here. :/
You and a friend are standing in front of two houses. In each house lives a family with two children.
"The family on the left has a boy who loves history, but their other child prefers math," your friend tells you.
"The family on the right has a 7-year old boy, and they just had a new baby," he explains.
"Does either family have a girl?" you ask.
"I'm not sure," your friend says. "But pick the family that you think is more likely to have a girl. If they do have a girl, I'll give you $100."
Which family should you pick, or does it not matter?
The official solution, according to the website, says (Warning: If you don't want to know the solution, don't look ahead.) that we should pick the house on the left, as there is a 2/3 chance that the house on the left has a girl, and a 1/2 chance that the house on the right has a girl, as the age of the children on the left house are never given.
They offer these charts for the houses:
Left house:
Younger Older
Girl Boy
Boy Girl
Boy Boy
Right house:
Younger Older
Girl Boy
Boy Boy
They also tell us:
"This is a very counterintuitive riddle. It seems like there should always be a 1/2 chance that a given child is a girl. And in fact there is. The key word there is "given". Because we are not asking about a "given" child for the house on the left. We are asking about what could be either child."
And I'll just stop it right there. In the riddle, you ask your friend "Does either family have a girl?", and the friend responds "pick the family that you think is more likely to have a girl."
This, to me, seems to be specifically asking the question "Is or is not the other child a girl?"
When the solution says "We are asking about what could be either child.", they seem to be suggesting that, at some point, they were asking the specific question "Is the other child a younger girl, a younger boy, an older girl, or an older boy?"
The riddle itself, however, was not asking this question; the riddle was asking if, yes or no, the other child was a boy or girl, and that the family had a child that was a girl, regardless of age.
If they were asking "what could be either child", they might as well have added in more factors, such as whether or not the other child could have been a long or short haired, green or brown eyed, taller or shorter, stronger or weaker, younger or older boy or girl.
There are a lot of things the other child could have been, but I don't think any of those other factors have anything to do with what the riddle asks, which is to "pick the family that you think is more likely to have a girl." And age doesn't have anything to do with whether the child is more likely to be a girl, correct?
The solution seems to me to be asking a question that the riddle itself was not.
And this doesn't seem to be an open-ended riddle either, that has a case of "we're both right"; the way this riddle is set up, and with the solution they give, they are saying that there is a single solution to this riddle of which house has a higher probability.
Please correct me if I'm wrong, and point out any flaws on my end. I honestly believe their solution is incorrect, or they changed around the original question to something that it actually wasn't. Something just doesn't seem right here. :/
Reply by 21eyes
- May 05 2011 09:23:24 AM
Age is actually a factor, females (statisticaly in places without good medical care) have a higher survival rate than males, so the fact that the child survived that long shows means it is more likely a female.
Reply by Caprico
- May 05 2011 10:20:08 PM
@ 21eyes: I don't think that matters as such.. This is more of a maths riddle than a statistics one...
@ SuperDuper: This riddle is an example of Conditional Probability. If you've studied this any time of your life in detail, you'll understand the small implication that they have tried to hide very subtly. You are right to question it, but this riddle is, as you say, "set-up" in a manner to confuse you. It is also, as they say, counter-intuitive. The riddle is correct, and so is the solution... Your way of thinking isn't wrong either, but some branches of maths are such, that you have to adjust yourself to those laws, and then you'll understand why this is the solution. But an intelligent doubt nevertheless.
@ SuperDuper: This riddle is an example of Conditional Probability. If you've studied this any time of your life in detail, you'll understand the small implication that they have tried to hide very subtly. You are right to question it, but this riddle is, as you say, "set-up" in a manner to confuse you. It is also, as they say, counter-intuitive. The riddle is correct, and so is the solution... Your way of thinking isn't wrong either, but some branches of maths are such, that you have to adjust yourself to those laws, and then you'll understand why this is the solution. But an intelligent doubt nevertheless.
Reply by Ragib
- May 06 2011 09:51:15 AM
i totally agree with super duper.riddles are supposed to have clues in their narrative style.without any specific reason,why should we think different?science is full of things that are not scientific always.like physics is full of approximate values.velocity of light in vacuum is constant.but this fact is more statistic than scientific.assuming obama to be osama does not make osama bin laden alive.just like that,why should i assume that this specific riddle would be counter intuitive?
Reply by 21eyes
- May 06 2011 11:16:37 AM
Oh bugger, after reading the riddle at least ten times I finally get what Caprico means, although I don't really understand why it works.
Reply by Caprico
- May 07 2011 01:27:36 AM
@Ragib You are the last person to comment on that! In my Lying Chickens Riddle, you said that laws of maths shape the way for correct logic... Quoting you:
ya.logic is necessary for riddles but sometimes when you become ultralogical we may be unable to solve some riddles.there are in betweens in logic.einstein's neighbour's truth telling day riddle is like that.sometimes it depends on demand of situation.
So in this riddle, the situation demands conditional probability. It isn't assumption that it is counter-intuitive, it indeed is. Like .999... is equal to one. Instinct tells you that it isn't possible that an infinite fraction can ever be equal to an integer. But it is a mathematical proof, check it out on Wiki. Similarly, this specific riddle is a classical example of conditional probability. Physics may have approx values, but Maths is an exact science. You just can't trust your instinct always.
@21eyes: I'm glad you see my point! It is confusing, yes, but it requires some detail inspection of the riddle. Good to know.. :)
ya.logic is necessary for riddles but sometimes when you become ultralogical we may be unable to solve some riddles.there are in betweens in logic.einstein's neighbour's truth telling day riddle is like that.sometimes it depends on demand of situation.
So in this riddle, the situation demands conditional probability. It isn't assumption that it is counter-intuitive, it indeed is. Like .999... is equal to one. Instinct tells you that it isn't possible that an infinite fraction can ever be equal to an integer. But it is a mathematical proof, check it out on Wiki. Similarly, this specific riddle is a classical example of conditional probability. Physics may have approx values, but Maths is an exact science. You just can't trust your instinct always.
@21eyes: I'm glad you see my point! It is confusing, yes, but it requires some detail inspection of the riddle. Good to know.. :)
Reply by Caprico
- May 07 2011 01:34:10 AM
See the Monty Hall Problem. Classic reference for a probability riddle, and an end to this debate. It has a counter-intuitive solution. Its very very similar to this riddle.
Reply by Ragib
- May 07 2011 02:44:03 AM
of course there are in betweens.but the logics are sometimes based on just statistical experiments.like speed of light.ya,you have to accept it to solve problems but your conscience won't let you completely believe it.there are exceptions.but exceptions must have some clue that helps an reader understand that it is actually exceptional.we,who try to solve riddles,are actually like lizards who change colours.we use logic to solve one, then use antilogic to solve another.i clearly understood what you and solution say but i think it's not actually riddle.it's a problem of high class probability.like i used to post problems earlier which made you angry.like i said earlier,riddles should have answers in themselves,the problem nowhere asks for elder or younger girl,it just asked for girl.i will tell you about a well known mathematical problem.'an 8*8 square can be shaped into a 13*5 rectangle.but their areas aren't same.why does that happen?'in this problem,i clearly say that the 2nd one is rectangle.answer would be that the diagonal of it won't be straight line and hide a little area.then actually the 2nd quadrilateral isn't a rectangle.so it's not just to name it as rectangle in riddle.you can use confusing words but can't use wrong or false things.
Reply by Caprico
- May 08 2011 10:00:31 AM
Yes, I do agree this is actually not a riddle, and if it is, its not of our level. Here, the probability distribution changes in a complex manner, that is difficult for us to grasp...
Reply by Ragib
- May 08 2011 11:24:04 AM
that's my boy.you really understand now what i am trying to say.it's way unthinkable for some 17 year old teenagers.let me ask you a ques-a bulb can be lightened up by connecting it to either negative-positive or vice versa.but a clock can't be run without proper contact of positive or negative points.why?
Reply by Ragib
- May 08 2011 11:24:58 AM
i forgot to mention battery.
Reply by Caprico
- May 10 2011 12:28:16 AM
A clock runs on a motor. It also has a diode connected so that current can't run the other way. Technically, the clock will run, but backwards, if I remove the diode that is... A bulb just needs current to run through the tungsten wire inside it.
Reply by Ragib
- May 10 2011 12:59:16 AM
you are right.
Reply by mikeland86
- May 14 2011 11:59:24 PM
@Caprico The problem with the solution is that it assumes that the boy on the left is either the youngest xor (exclusive or) oldest but doesn't count both possibilities. Given at least one of the two children on the left is a boy, I would think the possible distributions are
[b-h, b-m]: 25%
[b-m, b-h]: 25%
[b, g]: 25%
[g, b]: 25%
Using h and m to denote history and math preferences. This would mean that either house has a 50% of having a girl.
A small detail that I thought of is that the house on the left has a probability of identical twins (both boys) that the right doesn't have, thus making the right slightly more likely to have a girl. A quick Google search puts the probability of identical twins at 0.4%.
[b-h, b-m]: 25%
[b-m, b-h]: 25%
[b, g]: 25%
[g, b]: 25%
Using h and m to denote history and math preferences. This would mean that either house has a 50% of having a girl.
A small detail that I thought of is that the house on the left has a probability of identical twins (both boys) that the right doesn't have, thus making the right slightly more likely to have a girl. A quick Google search puts the probability of identical twins at 0.4%.
Reply by GojanTorresque
- August 29 2011 04:33:24 AM
mikeland86 has it right, the essential problem is one of three things: claiming too many possibilities (the second child is a girl or is a boy, 50% chance of either), claiming too FEW possibilities (the second child is either a younger girl, older girl, younger boy, or older boy, 25% chance of each) or claiming that all three options are equally likely (either the other child is an older girl, younger girl, or any boy, 25% for each of the first two, 50% for the last).
To explain this in a simpler analogy, what is basically happening is this: "I flipped a penny and a dime. The penny is heads. What are the chances that one of them is tails?" Because you know the outcome of one of them, and they are not dependent on each other for their results, the answer is the same as if there was only one involved. The possibilities are as follows(considering order just like the riddle tries to do but fails):
Penny: heads, Dime: heads
Penny: heads, Dime: tails
Dime: heads, Penny: heads
Dime: Tails, Penny: heads
In 50% of these situations, you will have a tails for one coin, and this time, they ARE equally likely, assuming the coin flip ITSELF is equal.
This is not similar to the monty hall problem because the genders are NOT dependent nor will order matter to the end result.
To explain this in a simpler analogy, what is basically happening is this: "I flipped a penny and a dime. The penny is heads. What are the chances that one of them is tails?" Because you know the outcome of one of them, and they are not dependent on each other for their results, the answer is the same as if there was only one involved. The possibilities are as follows(considering order just like the riddle tries to do but fails):
Penny: heads, Dime: heads
Penny: heads, Dime: tails
Dime: heads, Penny: heads
Dime: Tails, Penny: heads
In 50% of these situations, you will have a tails for one coin, and this time, they ARE equally likely, assuming the coin flip ITSELF is equal.
This is not similar to the monty hall problem because the genders are NOT dependent nor will order matter to the end result.
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