Probability question

[CurtC]

I posted earlier that the answer is 1/3, but some here do have a point that the wording leaves something to be desired. I myself have two boys, but I would never tell someone "I have two children, and one is a boy." That to me would clearly imply that the other is a girl, which is false in my case.

The way I originally heard the problem makes more sense. A census worker questions a woman. How many children do you have? Two. Is one of them a boy? Yes. Now, what is the probability that both are boys? 1/3.

 

[Just thinking]

The way I originally heard the problem makes more sense. A census worker questions a woman. How many children do you have? Two. Is one of them a boy? Yes. Now, what is the probability that both are boys? 1/3.

That's more in line with how I suggested an alternate ...

A woman is in a store with her son. You begin a conversation with her and find out she has two children. What is the probability that her other child is also a boy? (1/3)

 

[69dodge]

Re: Re: Re: Probability question

Originally posted by Just thinking
A man states that he has one sibling -- what is the probability of him having a brother?

A woman is in a store with her son. You begin a conversation with her and find out she has two children. What is the probability that her other child is also a boy?

1/2, for both of those. The reason it's more than 1/3 is this: if the man's parents, or the woman, have one son and one daughter, there was a 50 % chance you wouldn't have been faced with the question to begin with, since you might have met the man's sister instead of him, or the woman might have taken her daughter shopping instead of her son. So, the fact that you did see a male after all increases the probability that both siblings are male.

Another way to look at it is simply that in these two cases, in contrast to those whose answer is 1/3, you do know which sibling is the male: it's the one standing right in front of you.

I like CurtC's census version. The answer there is unambiguously 1/3.

 

[treble_head]

I admit I didn't read the other posts before answering. (wanted to get it myself) It is 1/3, assuming that boys and girls come out in the same ratio, which they don't but for the sake of argument...

The possibilities are thus

Boy first Girl second
B G
Girl first Boy second
G B
Boy first Boy second
B B
Girl first and Girl second is eliminated... One is a BOY.

That leaves one out of 3 or 1/3

I'm sure someone came up with this, but,again, I wanted to figure this one out myself.

 

[Soapy Sam]

I am the younger of two children.
I am male.
My sister is three years older than me.

Before my birth, a stranger, knowing that my mother had one child, would have assessed that child's probability of being female at 1/2.

After my birth, meeting my mother with me and knowing my mother had two children, the stranger assesses the probability of my sibling being female as 1/3.

I'm baffled.

How does my existence alter the probability that the older child is equally likely to be male or female?

We're back to the guy on tv with the car and the goats.

Flip two coins.

One comes down heads. What are the odds of the other coming down heads? How are those odds affected by the first coin?

1/3 may be the right answer , but it's a stupid right answer.

 

[Drooper]

Actually, I have now revised my answer.

The answer is now zero. I have two children, one is a boy.

 

[Drooper]

Seriously though.

For those who don't believe the answer is 1/3 you can prove it to youself quite easily.

You need a friend to help, but you can set up a simple process using playing card (red v back) or coins (heads v tails). I think you could guess the process, have them randomise two selections and then inform you "one is black", say.

Record that information and then reveal the other outcome and make a note. Analyse the data and you will see that the frequency (of black being the other card in this exmaple) will gravitate to 1/3.


[edit]

I just realised wat I wrote there might not be clear enough. When I said "reveal the other outcome" I meant reveal both the coins (to the observer). It is important that the observer is only given the information that one is heads - nothing can be revealed about which of the coins is heads (if it is the probability collapses to a 1/2 chance).

 

[Just thinking]

Re: Re: Re: Re: Probability question

1/2, for both of those. The reason it's more than 1/3 is this: if the man's parents, or the woman, have one son and one daughter, there was a 50 % chance you wouldn't have been faced with the question to begin with, since you might have met the man's sister instead of him, or the woman might have taken her daughter shopping instead of her son. So, the fact that you did see a male after all increases the probability that both siblings are male.

Another way to look at it is simply that in these two cases, in contrast to those whose answer is 1/3, you do know which sibling is the male: it's the one standing right in front of you.

Why is it any different seeing one of the woman's offspring (a male) as opposed to her just telling you what one is? To me seeing one is a male is no different than being told one is a male; how is this different? There is no difference in the information given you. Mind, I'm not insisting I'm 100% correct, I just don't see the difference in your explanation. Also, do you really know which sibling is male -- the older or younger?

 

[CurtC]

After my birth, meeting my mother with me and knowing my mother had two children, the stranger assesses the probability of my sibling being female as 1/3.

I'm baffled.

Just Thinking was incorrect. If the question pertains to a particular kid, as in the grocery store example, the probability is 1/2. The trick with the 1/3 answer is that you're figuring the probability for either one of them being a boy.

 

[Just thinking]

Just Thinking was incorrect. If the question pertains to a particular kid, as in the grocery store example, the probability is 1/2. The trick with the 1/3 answer is that you're figuring the probability for either one of them being a boy.

CurtC,

Do you really know (by seeing a boy) which one it is? The older or younger?

 

[Just thinking]

Re: Re: Re: Re: Probability question

1/2, for both of those. The reason it's more than 1/3 is this: if the man's parents, or the woman, have one son and one daughter, there was a 50 % chance you wouldn't have been faced with the question to begin with, since you might have met the man's sister instead of him, or the woman might have taken her daughter shopping instead of her son.

I don't believe you can argue my problem from that perspective since I already stated you see a boy. This now becomes a conditional probability -- probability B given A. We do not need to factor in the chances of her walking in with a girl. It's like asking what are the chances that I flipped a coin and got 2 heads -- and I show you one of the coins and it's heads. You see, you're still choosing from a possible outcome of 3 equally likely events, HH, HT or TH. My showing you one head does not make any one of the three less likely.

 

[Soapy Sam]

To remove the genetic problems (ie more than two possible sexual states exist) let's revert to coins.

Toss two coins. (No edge landings allowed).

4 possible groups exist:- HH HT TH TT

So let's say that we know coin 1 is H.
What is the probability that the group result is HH? 1/4
What is the probability that the group result is TH? 1/4
What is the probability that the group result is HT? 1/4
What is the probability that the group result is TT? Zero.

Now some will say that HT and TH are the same, but I don't see that specified in the original question- though it may be implied- and , frankly, I don't think they are the same. Two things that are the same, don't look different. *

Those are the answers to questions about groups.

But we are not asked about a group in the original question.
We are asked about the probability that one unspecified child is male or female.

The answer to that is 1/2. It doesn't matter how many possible groups there are. Weasel words like "also" don't work in this case. The kid either is, or is not, a boy. He is not "a boy also".

I have seen this question many times before and been argued at by many people. Given that I was behind the door when they gave out mathematical ability, I will concede defeat ahead of time, as I do with the stupid imaginary number nonsense.

I still think it's nonsense. And if it's not nonsense it darn well should be.


*(0.999... and 1 for example).
(Dons tin hat and climbs into tank)

 

[John Jackson]

It’s often the case with these little teasers that the question is harder to understand than the answer.

The probabilities change depending on the amount of information given in the question.

If a woman has 2 children, one of which is known to be a boy, then there is a 1/3 chance of the other child also being a boy.

If a woman has 2 children, the eldest of which is known to be a boy, then there is a 1/2 chance of the other child also being a boy.

There are 4 possible outcomes for 2 children. B-B, B-G, G-B, G-G.

In both cases G-G is eliminated as one child is known to be a boy.

If the birth order is not given this leaves B-B, B-G, G-B as the possibilities: 2 chances of the other child being a girl and one chance of it being a boy.

If the birth order is known, say the boy is the eldest, then this leaves B-B, B-G as the possibilities.

Knowing the birth order, i.e. having more information, changes the probability of predicting the sex of the sibling.

 

[casebro]

Drooper was almost correct with his deck of cards. Define first child as RED, what are chances next card is red? By the time you flip through the deck, you will get 26 reds, and 26 blacks....1/3???

 

[Drooper]

Drooper was almost correct with his deck of cards. Define first child as RED, what are chances next card is red? By the time you flip through the deck, you will get 26 reds, and 26 blacks....1/3???

That is not what I meant. Decks of cards don't come into it.

In my example, take two coins. Alternatively take 4 cards, two red and two black and have the tester choose two randomly. After each trial you do the same thing again with the same 4 cards.

Anyway it is easier with the coins.

 

[Drooper]

To remove the genetic problems (ie more than two possible sexual states exist) let's revert to coins.

Toss two coins. (No edge landings allowed).

4 possible groups exist:- HH HT TH TT

So let's say that we know coin 1 is H.
What is the probability that the group result is HH? 1/4
What is the probability that the group result is TH? 1/4
What is the probability that the group result is HT? 1/4
What is the probability that the group result is TT? Zero.

Now some will say that HT and TH are the same, but I don't see that specified in the original question- though it may be implied- and , frankly, I don't think they are the same. Two things that are the same, don't look different. *

Those are the answers to questions about groups.

But we are not asked about a group in the original question.
We are asked about the probability that one unspecified child is male or female.

The answer to that is 1/2. It doesn't matter how many possible groups there are. Weasel words like "also" don't work in this case. The kid either is, or is not, a boy. He is not "a boy also".

I have seen this question many times before and been argued at by many people. Given that I was behind the door when they gave out mathematical ability, I will concede defeat ahead of time, as I do with the stupid imaginary number nonsense.

I still think it's nonsense. And if it's not nonsense it darn well should be.


*(0.999... and 1 for example).
(Dons tin hat and climbs into tank)

You do the hard work and jump to the wrong answer.

You correctly identify the following

4 possible groups exist:- HH HT TH TT

So let's say that we know coin 1 is H.
What is the probability that the group result is HH? 1/4
What is the probability that the group result is TH? 1/4
What is the probability that the group result is HT? 1/4
What is the probability that the group result is TT? Zero.

But assign incorrect probabilities.

What you need to think is that with the new information the chance of TT is zero - easy.

But that means you are only left with three possible outcome (out of the original 4) of equal likelihood. 3 equally likely (and complete) events have a probability of 1/3 each (not 1/4) - you forgot to update your probabilities for the extra information you were given.

So the probability of another head is simply the probability of HH, which is 1/3.

If you don't trust the maths, the only way to prove it to yourself is do the experiment I suggest above. I guarantee it works.

 

[bmillsap]

At the risk of beating this to death, let me abstract this in case it helps to get it away from the sexes of children.

Imagine I have four bags into which you can't see. Into the four bags I put:

Bag number 1: two $1 bills
Bag number 2: two $100 bills
Bag number 3: a $1 bill then a $100 bill
Bag number 4: a $100 bill then a $1 bill

As a side (but important side) note, if I had a process for randomly picking $1 bills and $100 bills with equal likelihood, and I filled lots and lots of bags with random pairs picked this way, I'd expect 1/4 of the bags to fall into each of the four categories above.

Now, suppose I reach in and pull one bill out of one bag. I get a $1 bill. What do I know? I know I didn't reach into bag number 2, because it doesn't have any $1 bills. But it's possible that I reached into any of the other bags. If I reached into bag number 1, the other bill is a $1 bill. If I reached into bag number 3, the other bill is a $100 bill. If I reached into bag number 4, the other bill is a $100 bill.

So, even though overall the probability of getting a $1 bill is 50/50, with what I know after pulling one bill, if I pull the second bill out of the same bag, the probability is 1/3 that I'll get another $1 bill and 2/3 that I'll get a $100 bill.

The key is that pairing a $1 bill with a $100 bill is twice as likely as having two $1 bills or two $100 bills because there are two ways to do it.

In a sense, it's like rolling dice - each die is equally likely to come up any number 1 through 6, but when you start pairing them up the different totals are NOT equally likely because there are different numbers of ways to make them. This is like rolling two two-sided dice each with numbers 1 and 2 on their two faces. Getting a total of 3 (i.e. each die is different - call it 1 and 2, 1 head and 1 tail, 1 boy and 1 girl, 1 $1 bill and 1 $100 bill) is like getting a 7 with real dice - more ways to make it, more likely. Getting a total of 2 or 4 (i.e., each die comes up the same) is like rolling 2 or 12 with real dice - less ways to make them, less likely.

(edited for minor typo)

 

[casebro]

Let ME restate the question: There is a particular child who is not here.

1) What are the chances of this child being male ?

2) How could any sibling's sex have any possible effect on the sex of the child in question?

 

[John Jackson]

Let ME restate the question: There is a particular child who is not here.

1) What are the chances of this child being male ?

50%

2) How could any sibling's sex have any possible effect on the sex of the child in question?

It can’t.

Where there are 2 children, however, the probability of predicting the sibling’s sex depends on one’s knowledge of the first child’s sex and whether or not the birth order is known.

I think the playing card analogy works well. If you have 2 red cards and 2 black cards mixed randomly, then deal one card. You have a 50% chance of getting it right.

Now that you know what that card is, let’s say it was red, then there’s a 66% chance that the next one will be black as you have 2 black and one red card left.

Now if there are 2 children in a family and you know that one of them is a boy, then out of the 4 birth combinations, only 3 are possible: boy-boy, boy-girl, girl-boy (girl-girl is ruled out obviously). So if one child is known to be a boy then there is one chance that his sibling will be a brother but two chances that it will be a sister.

This is why the probability is 1/3.

 

[Drooper]

Let ME restate the question: There is a particular child who is not here.

1) What are the chances of this child being male ?

for that specific question the answers is 0.5 (1/2)

2) How could any sibling's sex have any possible effect on the sex of the child in question?

It can't and in this problem it doesn't.

However, given some limited knowledge about the nature of the event (i.e. the gender of two unknown children) modifies the number of possible outcomes - in this case from 4 equally likely outcomes to 3 equally likely outcomes.