Logic problem (1 Viewer)

[GregR_2]

Okay I'm rethinking this;

The way I wee it, there are four possibilities; we're talking about a specific kid (TOM) and he has either

A - older sister

B - older brother

C - younger sister

D - younger brother

In two cases he has a sister, and two a brother. So now I'm thinking 50/50. What am I missing?

These are the possiblities of the SEQUENCE:BB

BG

GB

GG

The middle two are the same when talking about PROBABILTY.

Correct. There is an equal chance of each of those 2-child families resulting.If the question was, "What are the odds that any child in a 2-child family is a girl", then any of the children could be the one referred to and there are 4 chances at being a boy and 4 at being a girl, and the answer would be 50/50. (This wasn't the question asked).

If the question was "What are the odds that the younger child is a girl", then there are 2 younger boys and 2 younger girls it could refer to, and the odds would be 50/50. (This wasn't the question asked either).

The question asked had the father telling us that one of his children (we don't know which) is a boy. This means it can't be a GG family and the potential set of children has now become:

BB

BG

GB

GG

Further, we are asked the sex of the child that is not the boy the father referred to. So we can eliminate one boy from each family as being the boy that the father referred to:

BB

BG

GB

GG

There are only 3 children remaining with the same probability of being the one the father is referring to. 1 is a boy, and 2 are girls. Thus the 2/3 chance she is a girl, when we know the information that the father gave us. If we didn't have that information that he has a different child that is a son, then we couldn't eliminate those possibilities and it would just be a 50/50 chance.

 

[Big Dumb Ape]

My thought process here:

Observe argument. View which side Cavalier is on. Pick the other side.

 

[Sweet J]

Ivan has convinced me.

I'm still fighting it.No one has addressed my question of a guy having 100 kids with 99 boys. Once I hear that explanation I might come around.

Given 99 boys, there is 1 possibility of 100 boys and 100 possibilities of 99 boys and a girl (girl born first, second, third, etc.). The probability of the other child being a girl is 99% (100/101).

OK. I'm further from being convinced now.

I don't know whether this has been answered for you (I'm only on page 2) but your querry should be even MORE evidence that of the OP's problem.If we are talking about the 100th child born, you are right, it is 50/50.But if you are talking EITHER the 1st child, or the second, or ther third, or the forth . . . all the way to the 100th. Then the possibility that ONE of those children being a girl (i.e., one out of the 100) rises substantially. Same with the original problem.

 

[BoltThrower]

Cav, I just flipped a coin twice. Can you list all the possible outcomes for me?

any of the 50% crowd is welcome to answer this.

This is completely different because there is no known information. Without any information, the results could beHHHTTHTT

Do you agree that these are the 4 possibilities, and that they are equally likely? If i ran it a bazillion times we would get each of the above about 1/4 of the time, correct?

Yes (I think I know where this is heading, but let's go there)

So if I tell you one flip was heads, what are the possible outcomes (all still equally likely)?

If I know, IN ADVANCE, that one of the coins will be H, BUT NOT WHICH COIN IS H, then the options areH - ? or ? - HHH, HT or HH, THThe HH scenarios are NOT the same, because of timing.

In the OP, you don't know in advance. The kids are already born.Plus, even if you know there is going to be at least one H, there's still only one way to get HH.

 

[CBusAlex]

Okay I'm rethinking this;

The way I wee it, there are four possibilities; we're talking about a specific kid (TOM) and he has either

A - older sister

B - older brother

C - younger sister

D - younger brother

In two cases he has a sister, and two a brother. So now I'm thinking 50/50. What am I missing?

You're double-counting B and D. For every family with two boys, there will be a younger brother and an older brother.Whenever a father has a son with an older brother, that same father will also have a son with a younger brother. However, it's just one father!

 

[IvanKaramazov]

Now let's give those kids names

Tom

Joe

Mary

Sally

We can throw out Sally since we know it's not two girls. And we know Tom is included, so the remaining possible combinations are;

Tom, Joe

Joe, Tom

Tom, Sally

Sally, Tom

Do you agree with that?

Nope. You aren't allowed to arbitarily throw out Sally. Like you said, we know Tom is included, but we don't know whether Sally or Mary is included. Here are the possibilities:Tom-Joe

Tom-Mary

Tom-Sally

Joe-Tom

Joe-Sally

Joe-Mary

Sally-Tom

Sally-Joe

Sally-Mary

Mary-Tom

Mary-Joe

Mary-Sally

Of the remaining six possibilies, only two involve BB (Tom-Joe and Joe-Tom). The other four are either BG or GB. Again, 2/3 is the winner.

WRONG, both sally and mary cannot exist. Choose one girl, delete the other and you finally get it.

They both can't exist simultaneously, but either was equally likely to have existed. You're getting messed up by a revelation that doesn't change anything about what you knew a priori. I think the mistake you're making is very similar to what leads people astray on the Monty Hall Paradox, but I'm not 100% sure.

 

[Follows Closely]

25 are boy-boy25 are boy-girl25 are girl-boy25 are girl-girl.

Have you considered33 boy boy33 boy/girl - order does not matter33 girl/girl

This guy gets it.Posters have said that you have 4 combinations:1) G G2) B G3) G B4) B BThis is flawed because sequence is not part of the question. If sequence was mention, such as: A man brought his first born child to town, a boy. What are the odds his child at home is a girl. You could eliminate two of the cases:[delete]1) G G[/delete]2) B G[delete]3) G B[/delete]4) B BLeaving a 50% 50% split.Since sequence is not mentioned you have three cases to startB BB GG GAfter elimination G GB BB G[delete]G G[/delete]You have a %50 %50 split.

 

[parrot]

There's 50 % chance he's talking about his youngest kidthat gives usBGBBThere's 50 % chance he's talking about his oldest kidthat gives usGBBBmultiplying the probablities:YBG = .25OGB = .25YBB = .25OBB =.25seems like 50/50...feel free to attempt to debunk

I'd like to see this debunked

The BB is the same in both sets, so it is reduced from four outcomes to three outcomes, with girls being in two out of three.

The BB is not the same, the kid in question can have an older brother, or a younger brother, just like he can have an older or younger sister.

Possible families with two kids:BBBGGBGGDo you agree or disagree with this?

Of course I agree with it. Now let's give those kids namesTomJoeMarySallyWe can throw out Sally since we know it's not two girls. And we know Tom is included, so the remaining possible combinations are;Tom, JoeJoe, TomTom, SallySally, TomDo you agree with that?

Tom/Joe and Joe/Tom are the same thing. It isn't as if the parents had the kid, looked at it and found it had a missing leg, then said, "Honey! Tom's here!" Or is it was a unibrow, said, "Honey! Joe's here!"

How are they the same? In one case Tom is the older sibling and in the other the younger. The fact that they're both boys doesn't mean you know longer have to consider the order. If the order doesn't matter with two boys, then it shouldn't matter whether his sister is older or younger either, which would make Tom/Sally the same as Sally/Tom. Tom has a sibling, it's either Sally or Joe. Joe is either older or younger, just like Sally.

 

[Verbal Kint]

Is this about testicular feminization?

 

[videoguy505]

25 are boy-boy25 are boy-girl25 are girl-boy25 are girl-girl.

Have you considered33 boy boy33 boy/girl - order does not matter33 girl/girl

This guy gets it.

 

[Cavalier]

Now let's give those kids names

Tom

Joe

Mary

Sally

We can throw out Sally since we know it's not two girls. And we know Tom is included, so the remaining possible combinations are;

Tom, Joe

Joe, Tom

Tom, Sally

Sally, Tom

Do you agree with that?

Nope. You aren't allowed to arbitarily throw out Sally. Like you said, we know Tom is included, but we don't know whether Sally or Mary is included. Here are the possibilities:Tom-Joe

Tom-Mary

Tom-Sally

Joe-Tom

Joe-Sally

Joe-Mary

Sally-Tom

Sally-Joe

Sally-Mary

Mary-Tom

Mary-Joe

Mary-Sally

Of the remaining six possibilies, only two involve BB (Tom-Joe and Joe-Tom). The other four are either BG or GB. Again, 2/3 is the winner.

WRONG, both sally and mary cannot exist. Choose one girl, delete the other and you finally get it.

Neither can Joe-Tom. Eliminate that and you finally get it.

You make a good point, I just don't agree. Since there can be two boys I leave that in the mix as probable. Since there can't be two girls you must lose one. I see what you are saying. I have the whole time, just not sure I agree.

 

[videoguy505]

There's 50 % chance he's talking about his youngest kidthat gives usBGBBThere's 50 % chance he's talking about his oldest kidthat gives usGBBBmultiplying the probablities:YBG = .25OGB = .25YBB = .25OBB =.25seems like 50/50...feel free to attempt to debunk

I'd like to see this debunked

The BB is the same in both sets, so it is reduced from four outcomes to three outcomes, with girls being in two out of three.

The BB is not the same, the kid in question can have an older brother, or a younger brother, just like he can have an older or younger sister.

Possible families with two kids:BBBGGBGGDo you agree or disagree with this?

Of course I agree with it. Now let's give those kids namesTomJoeMarySallyWe can throw out Sally since we know it's not two girls. And we know Tom is included, so the remaining possible combinations are;Tom, JoeJoe, TomTom, SallySally, TomDo you agree with that?

Tom/Joe and Joe/Tom are the same thing. It isn't as if the parents had the kid, looked at it and found it had a missing leg, then said, "Honey! Tom's here!" Or is it was a unibrow, said, "Honey! Joe's here!"

How are they the same? In one case Tom is the older sibling and in the other the younger. The fact that they're both boys doesn't mean you know longer have to consider the order. If the order doesn't matter with two boys, then it shouldn't matter whether his sister is older or younger either, which would make Tom/Sally the same as Sally/Tom. Tom has a sibling, it's either Sally or Joe. Joe is either older or younger, just like Sally.

If Joe is older, and Sally is younger, Tom was never born.The combined odds of "Joe, Tom" and "Tom, Joe" are the same as the individual odds of "Tom, Sally" and "Sally, Tom"

 

[parrot]

25 are boy-boy25 are boy-girl25 are girl-boy25 are girl-girl.

Have you considered33 boy boy33 boy/girl - order does not matter33 girl/girl

This guy gets it.Posters have said that you have 4 combinations:1) G G2) B G3) G B4) B BThis is flawed because sequence is not part of the question. If sequence was mention, such as: A man brought his first born child to town, a boy. What are the odds his child at home is a girl. You could eliminate two of the cases:[delete]1) G G[/delete]2) B G[delete]3) G B[/delete]4) B BLeaving a 50% 50% split.

You can't consider sequence when the sister is involved, but not the brother. That's the hangup here.

 

[MikeMan]

Another way of saying exact same as has been said before:

In the world's population of parents with 2 kids:

A.50% have a boy and a girl

B.25% have 2 boys

C.25% have 2 girls.

If someone tells you one of their 2 kids is a boy, then you know that family is in group A or B. There are twice as many families in group A so the odds are 66.6/33.3 that the other is a girl/boy.

ETA: If you believe the answer is 50%, then you think the odds above are 33/33/33 which is totally incorrect no matter how the problem is worded.

 

[videoguy505]

Read the second Dr. Math again: http://mathforum.org/dr.math/faq/faq.boygirl.choose.html

 

[Cavalier]

Now let's give those kids names

Tom

Joe

Mary

Sally

We can throw out Sally since we know it's not two girls. And we know Tom is included, so the remaining possible combinations are;

Tom, Joe

Joe, Tom

Tom, Sally

Sally, Tom

Do you agree with that?

Nope. You aren't allowed to arbitarily throw out Sally. Like you said, we know Tom is included, but we don't know whether Sally or Mary is included. Here are the possibilities:Tom-Joe

Tom-Mary

Tom-Sally

Joe-Tom

Joe-Sally

Joe-Mary

Sally-Tom

Sally-Joe

Sally-Mary

Mary-Tom

Mary-Joe

Mary-Sally

Of the remaining six possibilies, only two involve BB (Tom-Joe and Joe-Tom). The other four are either BG or GB. Again, 2/3 is the winner.

WRONG, both sally and mary cannot exist. Choose one girl, delete the other and you finally get it.

They both can't exist simultaneously, but either was equally likely to have existed. You're getting messed up by a revelation that doesn't change anything about what you knew a priori. I think the mistake you're making is very similar to what leads people astray on the Monty Hall Paradox, but I'm not 100% sure.

This is where I think the constant of one son takes certain things out of play and you don't. If you know there is one son. GG should not be considered IMO but you still put that as part of you probability.

 

[DaVinci]

25 are boy-boy25 are boy-girl25 are girl-boy25 are girl-girl.

Have you considered33 boy boy33 boy/girl - order does not matter33 girl/girl

This guy gets it.Posters have said that you have 4 combinations:1) G G2) B G3) G B4) B BThis is flawed because sequence is not part of the question. If sequence was mention, such as: A man brought his first born child to town, a boy. What are the odds his child at home is a girl. You could eliminate two of the cases:[delete]1) G G[/delete]2) B G[delete]3) G B[/delete]4) B BLeaving a 50% 50% split.

:X You can't consider sequence when the sister is involved, but not the brother. That's the hangup here.

But a BG combination (regardless of order) is twice as likely to happen than either BB or GG. So it's not a 1/3 split among the 3; it's 1/2, 1/4, and 1/4.

 

[rock753]

The answer is 2/3. If a guy has two children, there are four equally-likely possibilities for their sequence of birth and genders:boy-boyboy-girlgirl-boygirl-girlWe know the last one didn't happen, but the remaining three are all still equally likely, so 2/3 of the time his other kid is a girl. And there's a probability of 1 that she can't drive as well as most men.

What if the question was phrased as:A man and his son walk up to you and say hello. The man says he has another child at home. What are the odds the other child is a girl?Wouldn't it be 50/50 in that case?

No, it's still 2/3. If the guy told you that his younger child was at home, then it would be 50/50.

Once you know the gender of the first child, that child is taken out of the question. The only question you have left is whether the second child is a boy or a girl, which is pretty close to a 50/50 split.

 

[fatguyinalittlecoat]

Once you know the gender of the first child, that child is taken out of the equation.

What equation?

 

[adonis]

There is only one way to have two brothers. BB can't be counted twice.

 

[videoguy505]

Now let's give those kids names

Tom

Joe

Mary

Sally

We can throw out Sally since we know it's not two girls. And we know Tom is included, so the remaining possible combinations are;

Tom, Joe

Joe, Tom

Tom, Sally

Sally, Tom

Do you agree with that?

Nope. You aren't allowed to arbitarily throw out Sally. Like you said, we know Tom is included, but we don't know whether Sally or Mary is included. Here are the possibilities:Tom-Joe

Tom-Mary

Tom-Sally

Joe-Tom

Joe-Sally

Joe-Mary

Sally-Tom

Sally-Joe

Sally-Mary

Mary-Tom

Mary-Joe

Mary-Sally

Of the remaining six possibilies, only two involve BB (Tom-Joe and Joe-Tom). The other four are either BG or GB. Again, 2/3 is the winner.

WRONG, both sally and mary cannot exist. Choose one girl, delete the other and you finally get it.

Neither can Joe-Tom. Eliminate that and you finally get it.

You make a good point, I just don't agree. Since there can be two boys I leave that in the mix as probable. Since there can't be two girls you must lose one. I see what you are saying. I have the whole time, just not sure I agree.

You're messing yourself up by giving everyone names, and then considering only some of the cases, and considering those cases to be equally probable.The first boy out of the womb is Tom. The second boy, if there is one, is Joe. But, if the parents decide to name the firstborn son Joe, then you met Tom, then you know it's a BB family. If htey had named the first son Joe, then Tom-Sally/Tom-Mary are impossible. If they had named the first son Joe, the odds of meeting a "Tom" have drastically reduced.

 

[Cavalier]

Once you know the gender of the first child, that child is taken out of the equation.

What equation?

Tom=SallyMarySquared

 

[Sweet J]

For people trying to look at it in gambling terms think of it in this way;I make you a bet;We both flip a coin, any time we both get heads (HH=BB), you keep the coins.Any time one of us flips a tail (HT=BG, TH=GB) I keep the coins.Two tails (TT=GG), nobody keeps the coin.Would you take that bet?

Sure when you always place one coin on heads. Like the senario.

But in his scenario you ARE always placing one coin on heads, because "nobody keeps the coin" when it comes out TT (i.e., you are "throwing it out" when it happens).

 

[bshipper]

Cav, I just flipped a coin twice. Can you list all the possible outcomes for me?

any of the 50% crowd is welcome to answer this.

This is completely different because there is no known information. Without any information, the results could beHHHTTHTT

Do you agree that these are the 4 possibilities, and that they are equally likely? If i ran it a bazillion times we would get each of the above about 1/4 of the time, correct?

Yes (I think I know where this is heading, but let's go there)

So if I tell you one flip was heads, what are the possible outcomes (all still equally likely)?

If I know, IN ADVANCE, that one of the coins will be H, BUT NOT WHICH COIN IS H, then the options areH - ? or ? - HHH, HT or HH, THThe HH scenarios are NOT the same, because of timing.

In the OP, you don't know in advance. The kids are already born.Plus, even if you know there is going to be at least one H, there's still only one way to get HH.

I don't see this ever ending, but it is a fun discussion nonetheless.After more thought, I am willing to concede that based on mathematical probability, you are correct, there is only one combination for HH. However, I still believe that the two flips are completely independent of one another, and knowing that one is H, does not change the fact that the other coin has a 50/50 chance of being H.Think of the problem that asks what is the probability of two coins being flipped and both are H? The answer is 50% * 50% = 25%. The answer is not 50% * 33% = 17% (since I know one is H then the other only has 33% chance of being H).

 

[Cavalier]

Now let's give those kids names

Tom

Joe

Mary

Sally

We can throw out Sally since we know it's not two girls. And we know Tom is included, so the remaining possible combinations are;

Tom, Joe

Joe, Tom

Tom, Sally

Sally, Tom

Do you agree with that?

Nope. You aren't allowed to arbitarily throw out Sally. Like you said, we know Tom is included, but we don't know whether Sally or Mary is included. Here are the possibilities:Tom-Joe

Tom-Mary

Tom-Sally

Joe-Tom

Joe-Sally

Joe-Mary

Sally-Tom

Sally-Joe

Sally-Mary

Mary-Tom

Mary-Joe

Mary-Sally

Of the remaining six possibilies, only two involve BB (Tom-Joe and Joe-Tom). The other four are either BG or GB. Again, 2/3 is the winner.

WRONG, both sally and mary cannot exist. Choose one girl, delete the other and you finally get it.

Neither can Joe-Tom. Eliminate that and you finally get it.

You make a good point, I just don't agree. Since there can be two boys I leave that in the mix as probable. Since there can't be two girls you must lose one. I see what you are saying. I have the whole time, just not sure I agree.

You're messing yourself up by giving everyone names, and then considering only some of the cases, and considering those cases to be equally probable.The first boy out of the womb is Tom. The second boy, if there is one, is Joe. But, if the parents decide to name the firstborn son Joe, then you met Tom, then you know it's a BB family. If htey had named the first son Joe, then Tom-Sally/Tom-Mary are impossible. If they had named the first son Joe, the odds of meeting a "Tom" have drastically reduced.

So In reality, when I meet someone who says they have a son and they have two kids. 67% of the time, the other child will be a girl.

 

[Sweet J]

By the way, does anybody else want a poll:

I'm a socially concervative Christian and I believe the answer is 50%.

I'm a socially Concervative Christian and I believe the answer is 66%.

I'm not a socially concervative Christian and I believe the answer is 50%

I'm not a socially concervative Christian and I believe the asnwer is 66%

For the life of me, I'm convinced the Christians are skewing the results to the 50% number.

 

[rock753]

Here are the coin flip analogies depending on how the question is asked/interpreted:

You flip a pair of coins 100 times. How many sets have a "head" in them if you only use the sets with at least one head in them - 66.66%

You flip a pair of coins and look at one of them and its a "head". Whats the odds the other one is a "head" - 50.00%

You flip a coint the first time, it is heads. That flip is over. Second flip is either heads or tails, 50%.You have a son, and you knock up the wife again. Son is already born, odds on the second child being boy or girl is 50%.

Think of it with your wallet. I tell you I have a son and I want you to guess the gender of my second child. Am I going to give you even money or 2:1. Even money, the only question left is the gender of my second child.

It would be different if I didn't already tell you I had atleast one son. At that point it would be 4:1 to name the gender of both of my children.

 

[Franknbeans]

I feel like there's a lot of talking in here, but nobody's getting any smarter

Welcome

 

[bshipper]

Read the second Dr. Math again: http://mathforum.org/dr.math/faq/faq.boygirl.choose.html

Great find!I think this just about sums it up, everyone is right! I guess this question falls under "choosing the family first", therefore the probability of the other child being a girl is 2/3. As opposed to "choosing the child first", which has a probability of 1/2.

 

[Cavalier]

Two schools of thought when using the coins. One has to be heads.

First you have two coins. You flip tails the second coin is always heads TH.

You flip heads then you flip the second coin for either HH or HT. 67%

You have two coins. You know one is heads. You can only flip the other for either HH or HT. 50%

 

[Men-in-Cleats]

This 2/3 thing is taking sequence into play and it shouldn't be. There are only 3 possibilities if sequence is not important, 2 boys, 2 girls, and one of each. 2 girls has been eliminated. Sequence is completely irrelevant to the question so you are doubling up if you consider BG and GB separately. Absolutely the only way you come up with 2/3 is if you consider two sequences of the girl and boy separately.

 

[Dave Baker]

Think of the problem that asks what is the probability of two coins being flipped and both are H? The answer is 50% * 50% = 25%. The answer is not 50% * 33% = 17% (since I know one is H then the other only has 33% chance of being H).

You're actually proving the opposite.If you flip two coins, there are only four possibilities that result after the flips.

You flip heads first, then heads again.

HH

You flip heads first, then tails

HT

You flip tails first, then heads

TH

You flip tails first, then tails.

TT

So:

HH

HT

TH

TT

That is your data set.

Then, AFTER the data set is in place, you are given more information. You are revealed one of the flips, and it was heads.

Now, given the the other flip has already occurred, you know that the end result from the options above was not TT. So therefore you know that in the other three options, tails is included two times, there 2 of 3 = 2/3. You can choose to ignore the additional information and pretend that this is somehow a new, future event (50%). Or you can use the additional information to your benefit (let's say it was a bet) and go with 66%.

Since we know (statistically) that 50% of the time you will have one heads and one tails, and you know that 25% of the time it willl be both tails and another 25% of the time it will be both heads, you know the answer is 66%.

 

[fatguyinalittlecoat]

This 2/3 thing is taking sequence into play and it shouldn't be. There are only 3 possibilities if sequence is not important, 2 boys, 2 girls, and one of each. 2 girls has been eliminated. Sequence is completely irrelevant to the question so you are doubling up if you consider BG and GB separately. Absolutely the only way you come up with 2/3 is if you consider two sequences of the girl and boy separately.

Not you too. What is going on here?

 

[Cavalier]

Read the second Dr. Math again: http://mathforum.org/dr.math/faq/faq.boygirl.choose.html

Great find!I think this just about sums it up, everyone is right! I guess this question falls under "choosing the family first", therefore the probability of the other child being a girl is 2/3. As opposed to "choosing the child first", which has a probability of 1/2.

 

[CBusAlex]

You flip a coint the first time, it is heads. That flip is over. Second flip is either heads or tails, 50%.

The first flip isn't guarenteed to land on heads, though.If the first flip is heads, the second flip is 50/50 for heads or tails.If the first flip is tails, the second flip is 50/50 for heads or tails.So 75% of the time, you will get at least one heads. And for 2/3 of that 75%, the other flip will be tails.

 

[FatMax]

Cav, I just flipped a coin twice. Can you list all the possible outcomes for me?

any of the 50% crowd is welcome to answer this.

This is completely different because there is no known information. Without any information, the results could beHH

HT

TH

TT

Do you agree that these are the 4 possibilities, and that they are equally likely? If i ran it a bazillion times we would get each of the above about 1/4 of the time, correct?

Yes (I think I know where this is heading, but let's go there)

So if I tell you one flip was heads, what are the possible outcomes (all still equally likely)?

If I know, IN ADVANCE, that one of the coins will be H, BUT NOT WHICH COIN IS H, then the options areH - ? or ? - H

HH, HT or HH, TH

The HH scenarios are NOT the same, because of timing.

In the OP, you don't know in advance. The kids are already born.Plus, even if you know there is going to be at least one H, there's still only one way to get HH.

I don't see this ever ending, but it is a fun discussion nonetheless.After more thought, I am willing to concede that based on mathematical probability, you are correct, there is only one combination for HH. However, I still believe that the two flips are completely independent of one another, and knowing that one is H, does not change the fact that the other coin has a 50/50 chance of being H.

Think of the problem that asks what is the probability of two coins being flipped and both are H? The answer is 50% * 50% = 25%. The answer is not 50% * 33% = 17% (since I know one is H then the other only has 33% chance of being H).

But that isn't what the question is asking. The coins have already been flipped. The stipulation is that you already know what one outcome was. You're being asked to put odds on the other.

 

[BoltThrower]

Think of the problem that asks what is the probability of two coins being flipped and both are H? The answer is 50% * 50% = 25%. The answer is not 50% * 33% = 17% (since I know one is H then the other only has 33% chance of being H).

Correct, because the only way to get HH is to flip H first so you eliminate TH and TT as possible outcomes after the first flip, leaving only HH and HT.There are only 4 possible outcomes for 2 coins. You listed them yourself. Three of those include at least one H, and you admitted they are all equally likely. Of those 3 equally likely outcomes, 2 of them have 1 H and 1 T, and one has 2H. So 2 out of every 3 times that an outcome contains at least one H, it will contain a T. Thus 2 out of every 3 two-child families that contain at least one son will contain a daughter.

 

[FatMax]

This 2/3 thing is taking sequence into play and it shouldn't be. There are only 3 possibilities if sequence is not important, 2 boys, 2 girls, and one of each. 2 girls has been eliminated. Sequence is completely irrelevant to the question so you are doubling up if you consider BG and GB separately. Absolutely the only way you come up with 2/3 is if you consider two sequences of the girl and boy separately.

Not you too. What is going on here?

 

[Dave Baker]

Read the second Dr. Math again: http://mathforum.org/dr.math/faq/faq.boygirl.choose.html

Great find!I think this just about sums it up, everyone is right! I guess this question falls under "choosing the family first", therefore the probability of the other child being a girl is 2/3. As opposed to "choosing the child first", which has a probability of 1/2.

Yeah, obviously that's where people get stuck. The first part of the problem begins "a man tells you he has two children." Thus creating your data set.

 

[CBusAlex]

This 2/3 thing is taking sequence into play and it shouldn't be. There are only 3 possibilities if sequence is not important, 2 boys, 2 girls, and one of each.

THESE THREE POSSIBILITIES DO NOT HAVE EQUAL PROBABILITIES

 

[bshipper]

We need an FFA poll to find some real results for those of us with only one sibling.

Is your sibling a different gender?

 

[Dave Baker]

This 2/3 thing is taking sequence into play and it shouldn't be. There are only 3 possibilities if sequence is not important, 2 boys, 2 girls, and one of each. 2 girls has been eliminated. Sequence is completely irrelevant to the question so you are doubling up if you consider BG and GB separately. Absolutely the only way you come up with 2/3 is if you consider two sequences of the girl and boy separately.

I think some are getting the word "sequence" confused here. What you're dealing with is a data set.If it's agreed that the odds of having a girl vs a boy is 50/50, then we know that the odds, in a two child family, of having a boy and a girl is 50%, the odds of having two boys is 25% and the odds of having two girls is 25%. Does anyone disagree with this?Once we're agreed to that, we can take the next step.

 

[videoguy505]

Now let's give those kids names

Tom

Joe

Mary

Sally

We can throw out Sally since we know it's not two girls. And we know Tom is included, so the remaining possible combinations are;

Tom, Joe

Joe, Tom

Tom, Sally

Sally, Tom

Do you agree with that?

Nope. You aren't allowed to arbitarily throw out Sally. Like you said, we know Tom is included, but we don't know whether Sally or Mary is included. Here are the possibilities:Tom-Joe

Tom-Mary

Tom-Sally

Joe-Tom

Joe-Sally

Joe-Mary

Sally-Tom

Sally-Joe

Sally-Mary

Mary-Tom

Mary-Joe

Mary-Sally

Of the remaining six possibilies, only two involve BB (Tom-Joe and Joe-Tom). The other four are either BG or GB. Again, 2/3 is the winner.

WRONG, both sally and mary cannot exist. Choose one girl, delete the other and you finally get it.

Neither can Joe-Tom. Eliminate that and you finally get it.

You make a good point, I just don't agree. Since there can be two boys I leave that in the mix as probable. Since there can't be two girls you must lose one. I see what you are saying. I have the whole time, just not sure I agree.

You're messing yourself up by giving everyone names, and then considering only some of the cases, and considering those cases to be equally probable.The first boy out of the womb is Tom. The second boy, if there is one, is Joe. But, if the parents decide to name the firstborn son Joe, then you met Tom, then you know it's a BB family. If htey had named the first son Joe, then Tom-Sally/Tom-Mary are impossible. If they had named the first son Joe, the odds of meeting a "Tom" have drastically reduced.

So In reality, when I meet someone who says they have a son and they have two kids. 67% of the time, the other child will be a girl.

Not exactly. In reality, when you meet someone who says they have two kids and one is a son, 67% of the time, the other child will be a girl.

 

[IvanKaramazov]

This 2/3 thing is taking sequence into play and it shouldn't be. There are only 3 possibilities if sequence is not important, 2 boys, 2 girls, and one of each.

As a million people have already pointed out, these three possibilities are not equally likely. "One of each" is twice as likely as "2 boys." A bunch of people have posted numerical illustrations that prove this.

 

[thesurfshop19]

Great link that really brings these two schools of thought together. (videoguy's)

Another thought that may show how a lot of the arguments in this thread on both side are correct ....

BB

BG

GB

GG

If you know that there are 2 kids and you know that there is a son, there's a 2/3 chance that the family has a girl in it as well.

However, if you ask a randomly chosen boy from this group of 8 to tell us what gender their sibling is, there's a 1/2 chance they will answer boy. This is because you could randomly ask either of the two boys in the first group.

Not sure if this helps.

 

[bshipper]

Think of the problem that asks what is the probability of two coins being flipped and both are H? The answer is 50% * 50% = 25%. The answer is not 50% * 33% = 17% (since I know one is H then the other only has 33% chance of being H).

You're actually proving the opposite.Since we know (statistically) that 50% of the time you will have one heads and one tails, and you know that 25% of the time it willl be both tails and another 25% of the time it will be both heads, you know the answer is 66%.

Here is what I am trying to prove/discuss.I flip two coins and don't know the results of either. I think we can all agree that at this point, there is a 25% chance that it is HH. The flips are treated as independent actions, hence 50% each. Why then, when one is uncovered, do we all of a sudden treat them as dependent actions? If that 2nd coin now has a 33% chance of being H, why isn't the original chance of HH 17%?

 

[JAA]

This 2/3 thing is taking sequence into play and it shouldn't be. There are only 3 possibilities if sequence is not important, 2 boys, 2 girls, and one of each. 2 girls has been eliminated. Sequence is completely irrelevant to the question so you are doubling up if you consider BG and GB separately. Absolutely the only way you come up with 2/3 is if you consider two sequences of the girl and boy separately.

I think some are getting the word "sequence" confused here. What you're dealing with is a data set.If it's agreed that the odds of having a girl vs a boy is 50/50, then we know that the odds, in a two child family, of having a boy and a girl is 50%, the odds of having two boys is 25% and the odds of having two girls is 25%. Does anyone disagree with this?Once we're agreed to that, we can take the next step.

jumping in here, but you are making assumptions. Re-read the question. He isnt inquiring about a set of children, simply 1 child.

 

[IvanKaramazov]

This 2/3 thing is taking sequence into play and it shouldn't be. There are only 3 possibilities if sequence is not important, 2 boys, 2 girls, and one of each. 2 girls has been eliminated. Sequence is completely irrelevant to the question so you are doubling up if you consider BG and GB separately. Absolutely the only way you come up with 2/3 is if you consider two sequences of the girl and boy separately.

I think some are getting the word "sequence" confused here. What you're dealing with is a data set.If it's agreed that the odds of having a girl vs a boy is 50/50, then we know that the odds, in a two child family, of having a boy and a girl is 50%, the odds of having two boys is 25% and the odds of having two girls is 25%. Does anyone disagree with this?Once we're agreed to that, we can take the next step.

jumping in here, but you are making assumptions. Re-read the question. He isnt inquiring about a set of children, simply 1 child.

You need to know the possible family structures and the probability of each to be able to answer the question. Also, Dave isn't making any assumptions. The 50-25-25 breakdown that he posted is indisputable fact.

 

[CBusAlex]

Here is what I am trying to prove/discuss.

I flip two coins and don't know the results of either. I think we can all agree that at this point, there is a 25% chance that it is HH. The flips are treated as independent actions, hence 50% each.

Why then, when one is uncovered, do we all of a sudden treat them as dependent actions? If that 2nd coin now has a 33% chance of being H, why isn't the original chance of HH 17%?

When one coin is discovered, it no longer has a 50/50 chance of being H. It is now 100%. Probability can only be used to discuss unknown quantities.

 

[FlapJacks]

Here is where I'm stuck.

if we just say he has two kids and one is a boy we get 2/3.

if we specify that the oldest is a boy we get 1/2

fine

But I'm stuck on the fact that age is such an arbitrary identifying trait.

what if that arbitrary identifying trait, instead of being age was

"the kid he talked about."

hence the paradox