A simple logic problem

[timotinbowen]

The question doesnt ASK you whether the boy is the oldest or youngest...it is irrelevant to the solution....it just says that he has 2 kids one is a boy....then asks...of the other child...what are the odds that it is a girl.... 50/50

Again, we are not lying to you. This is incorrect. It is very relevant. You are quite likely to have a male out of 2 kids. You are also far more likely to have one of each than 2 males. This alone should be enough to show that it is not 50/50. We understand that the gender of an individual is 50/50. That is not what is being asked. Please read one of the articles, perhaps they can explain it better.

My final question: Do you still think we are all wrong? The proofs, the logic, the articles, the fact that it is taught in intro level set theory courses?

 

[Geoff101]

We've explained how to do these problems about 12 different ways. This thread is riddled with links. If we know someone has 5 children, and either 4 or 5 boys, the permutations are:

BBBBB
BBBBG
BBBGB
BBGBB
BGBBB
GBBBB

5/6 chance that the "other" child is female. We know that because the "other" child could fill any of the 5 spots in the birth order as a female, or the set could be all males.

Again, there are over 100,000 google hits for this problem. We understand that you do not believe the math. I posted this problem to be counterintuitive and get people thinking, but if you don't accept the reasoning by now you probably will not accept it. This problem has fooled many bright people because people often think terms of individuals, and not sets. In fact, this problem is often asked at the beginning of introductory set theory classes, to show the thinking required for such seemingly abstract problems.

Ok, ok...now given your math....that there is a 5 out of 6 chance, that's 83.3333% chance that the 5th person is a girl....so, let me show you how your math is wrong....your wife has a baby...boy, has another boy etc...she's pregnant with her 5th child...you are going to tell me that there is an 83.3% chance that the next child she gives birth to is going to be a girl????????????????

 

[quantumleap]

The question doesnt ASK you whether the boy is the oldest or youngest...it is irrelevant to the solution....it just says that he has 2 kids one is a boy....then asks...of the other child...what are the odds that it is a girl.... 50/50

I believe whether the boy is the oldest or the youngest should be taken into consideration as a possibility because it is explicitly given in the problem.

 

[timotinbowen]

Ok, ok...now given your math....that there is a 5 out of 6 chance, that's 83.3333% chance that the 5th person is a girl....so, let me show you how your math is wrong....your wife has a baby...boy, has another boy etc...she's pregnant with her 5th child...you are going to tell me that there is an 83.3% chance that the next child she gives birth to is going to be a girl????????????????

I honestly can't tell whether you're joking. This is A SET THEORY PROBLEM. There is no order to the children. Obviously, if we know the gender of the individual children then that changes things. But we do not know the order. Just because we know the sex of 4 children doesn't mean we know the sex of the first 4. You have said the same thing about 30 times, and each time it shows that you are looking at the problem incorrectly. Do not look at it by the individuals. Look at the set of children.

Suppose you play a 5-game parlay. If you are told that you hit at least 4 games in your parlay, is there a 50% chance you hit all 5?? NO! By your logic, there should be the same amount of 5-team parlays that hit all 5 as those that hit 4 out of 5, but in reality there are 5 times as many that hit 4 out of 5.

Again, please look at the links. You seem upset that we don't get it, but I promise you are not correct. I would not have posted the problem if it had such an easy answer.

 

[Geoff101]

I honestly can't tell whether you're joking. This is A SET THEORY PROBLEM. There is no order to the children. Obviously, if we know the gender of the individual children then that changes things. But we do not know the order. Just because we know the sex of 4 children doesn't mean we know the sex of the first 4. You have said the same thing about 30 times, and each time it shows that you are looking at the problem incorrectly. Do not look at it by the individuals. Look at the set of children.

Suppose you play a 5-game parlay. If you are told that you hit at least 4 games in your parlay, is there a 50% chance you hit all 5?? NO! By your logic, there should be the same amount of 5-team parlays that hit all 5 as those that hit 4 out of 5, but in reality there are 5 times as many that hit 4 out of 5.

if you play a 5 game parlay, and you have won 4 games, the odds on you winning the final play are 50/50....therefore at that point in time, the odds of you winning all 5 and winning the last game are the same, because you already have won the first 4....50/50. So after winning 4 games, the odds of you winning your parlay, is in fact 50/50....the odds certainly change as you receive your results...when you have 5 unknown games pending, your odds of winning your parlay are much less than 50/50...if you hit 4 of the 5, and there is only one game that you dont know the results of, you have a 50/50 chance of completing or winning that parlay at that point in time, the odds vary as outcomes are discovered...

 

[quantumleap]

This couldn't be more wrong. Please read the question again and just think about it. You have to remember, we know it's a SET of children. We KNOW one is a boy. We DO NOT KNOW whether he is the oldest or youngest.

Older boy-younger girl
Older girl-younger boy
Older boy-younger boy
Older girl-younger girl

We can eliminate older girl-younger girl because we know one is a boy. Leaves 2/3 chance other child is a girl. Yes I know this has been repeated. I just think this is the easiest way to explain it.

Correct GFB. Here is a way of asking the question as it was presented in the original problem and applying it to your set of possibilities:

"What is the probability that his other child is a girl" if the boy is the oldest and the girl is the youngest?
1 of 2

"What is the probability that his other child is a girl" if the girl is the oldest and the boy is the youngest?
1 of 2

"What is the probability that his other child is a girl" if there are two boys?
0 in 2

The question "What is the probability that his other child is a girl" if the girl is the oldest and the girl is the youngest should not be included since it doesn't fit the original scenario.

We see in 2 out of 3 scenarios when asked the original question of "What is the probability that his other child is a girl" that the question is valid. Therefore the probability is 2/3.

 

[quantumleap]

I honestly can't tell whether you're joking. This is A SET THEORY PROBLEM. There is no order to the children. Obviously, if we know the gender of the individual children then that changes things. But we do not know the order. Just because we know the sex of 4 children doesn't mean we know the sex of the first 4. You have said the same thing about 30 times, and each time it shows that you are looking at the problem incorrectly. Do not look at it by the individuals. Look at the set of children.

Suppose you play a 5-game parlay. If you are told that you hit at least 4 games in your parlay, is there a 50% chance you hit all 5?? NO! By your logic, there should be the same amount of 5-team parlays that hit all 5 as those that hit 4 out of 5, but in reality there are 5 times as many that hit 4 out of 5.

if you play a 5 game parlay, and you have won 4 games, the odds on you winning the final play are 50/50....therefore at that point in time, the odds of you winning all 5 and winning the last game are the same, because you already have won the first 4....50/50. So after winning 4 games, the odds of you winning your parlay, is in fact 50/50....the odds certainly change as you receive your results...when you have 5 unknown games pending, your odds of winning your parlay are much less than 50/50...if you hit 4 of the 5, and there is only one game that you dont know the results of, you have a 50/50 chance of completing or winning that parlay at that point in time, the odds vary as outcomes are discovered...

Correct, that scenario is 50%. However, as we have been saying all along that is a different scenario than what the problem is asking. The problem isn't asking "What is the probability that the remaining child is a girl".

I think you are confusing "other child" with "remaining child". When the question asks "other child" it opens up the case where the child can be oldest or youngest as explicitly stated in the problem and these outcomes should be considered.

 

[MrMore]

Originally Posted by MrMore
I flip 2 coins in the air and catch one in each hand, then slam them down on the table.

I turn over my left hand and reveal heads.

What are the chances the coin under the my right hand is tails?

this example, no doubt 50%

the part that is confusing is that if you peek at both and tell us ONE of them is heads (not revealing which hand), you are left with 3 possible scenarios:

right hand: heads ; left hand: heads
right hand: heads ; left hand: tails
right hand: tails ; left hand: heads

they are saying that in 2/3 of the scenarios, the "other" coin is tails.....

Winner. And that's my point. The difference in the two situations is what people are having trouble wrapping their minds around.

 

[christof]

if you play a 5 game parlay, and you have won 4 games, the odds on you winning the final play are 50/50....therefore at that point in time, the odds of you winning all 5 and winning the last game are the same, because you already have won the first 4....50/50. So after winning 4 games, the odds of you winning your parlay, is in fact 50/50....the odds certainly change as you receive your results...when you have 5 unknown games pending, your odds of winning your parlay are much less than 50/50...if you hit 4 of the 5, and there is only one game that you dont know the results of, you have a 50/50 chance of completing or winning that parlay at that point in time, the odds vary as outcomes are discovered...

In your example, you know the outcomes as they happen (the order). I skipped pages 5-12, so I don't know if it has been explained this way:

100 fathers in a room.
25 have GG
25 have BB
50 have BG

Everyone without a B is told to leave. You have 75 fathers left:
25 have BB
50 have BG

So of the fathers who have at least one B (75), 50 have a girl. 50/75 = 2/3.

 

[quantumleap]

the part that is confusing is that if you peek at both and tell us ONE of them is heads (not revealing which hand), you are left with 3 possible scenarios:

right hand: heads ; left hand: heads
right hand: heads ; left hand: tails
right hand: tails ; left hand: heads

they are saying that in 2/3 of the scenarios, the "other" coin is tails.....

That's the key, we don't get to know which coin is heads or which child is a son. If we knew, then we could know that the probability of the "remaining" child is 50%. That's not what the problem is asking. Since we dont know which child is the son we have to take into consideration 3 other possibilities. Of those 3 other possibilities, 2 have the chance to be true.

 

[timotinbowen]

I honestly can't tell whether you're joking. This is A SET THEORY PROBLEM. There is no order to the children. Obviously, if we know the gender of the individual children then that changes things. But we do not know the order. Just because we know the sex of 4 children doesn't mean we know the sex of the first 4. You have said the same thing about 30 times, and each time it shows that you are looking at the problem incorrectly. Do not look at it by the individuals. Look at the set of children.

Suppose you play a 5-game parlay. If you are told that you hit at least 4 games in your parlay, is there a 50% chance you hit all 5?? NO! By your logic, there should be the same amount of 5-team parlays that hit all 5 as those that hit 4 out of 5, but in reality there are 5 times as many that hit 4 out of 5.

if you play a 5 game parlay, and you have won 4 games, the odds on you winning the final play are 50/50....therefore at that point in time, the odds of you winning all 5 and winning the last game are the same, because you already have won the first 4....50/50. So after winning 4 games, the odds of you winning your parlay, is in fact 50/50....the odds certainly change as you receive your results...when you have 5 unknown games pending, your odds of winning your parlay are much less than 50/50...if you hit 4 of the 5, and there is only one game that you dont know the results of, you have a 50/50 chance of completing or winning that parlay at that point in time, the odds vary as outcomes are discovered...

WOWOWOW

OK last try. You have not hit 4 and have 1 remaining. All the games are in. You hit at least 4. It is far more likely that you hit exactly 4 than 5. If you are told that you got at least 4 games, you probably got exactly four. Of all SETS containing at least 4 winners, 83% of them will have exactly 4 winners.

I understand that this is confusing (although not this confusing), but again I urge you to read the links. We are not wrong.

 

[timotinbowen]

In your example, you know the outcomes as they happen (the order). I skipped pages 5-12, so I don't know if it has been explained this way:

100 fathers in a room.
25 have GG
25 have BB
50 have BG

Everyone without a B is told to leave. You have 75 fathers left:
25 have BB
50 have BG

So of the fathers who have at least one B (75), 50 have a girl. 50/75 = 2/3.

This is a very nice way to visualize it. I don't know how it is possible to refute this example, but I'm sure someone will find a way.

 

[quantumleap]

WOWOWOW

OK last try. You have not hit 4 and have 1 remaining. All the games are in. You hit at least 4. It is far more likely that you hit exactly 4 than 5. If you are told that you got at least 4 games, you probably got exactly four. Of all SETS containing at least 4 winners, 83% of them will have exactly 4 winners.

I understand that this is confusing (although not this confusing), but again I urge you to read the links. We are not wrong.

timotin, I think shdw01's post explains why the scenario Geoff is bringing up regarding the parlays is not the same as the original scenario. A parlay that has already hit 4 winners has those winners known. In the original scenario we don't know which child is the son and therefore there are 3 possible scenarios left, not 2.

 

[djmatt]

14 pages for this??? WOW

Did anyone here take an introduction to statistics course?

 

[Geoff101]

This is a very nice way to visualize it. I don't know how it is possible to refute this example, but I'm sure someone will find a way.

because the scenario has 4 possibilities, not 3....

B1,B2
B1, G
B2,B1
B2, G

50/50

 

[RobFunk]

Probability is infinite. It has no 'hard' mathematical value nor does it have an absolute.

In the example given, you can only solve for the chances of what is 'probable'.

 

[timotinbowen]

timotin, I think shdw01's post explains why the scenario Geoff is bringing up regarding the parlays is not the same as the original scenario. A parlay that has already hit 4 winners has those winners known. In the original scenario we don't know which child is the son and therefore there are 3 possible scenarios left, not 2.

I'm not talking about a parlay that hit 4 specific winners already. I'm talking about making 5 picks, and later being told that you hit at least 4 (when the entire result is already known). It might be a hazy example though.

 

[timotinbowen]

because the scenario has 4 possibilities, not 3....

B1,B2
B1, G
B2,B1
B2, G

50/50

You are given B1's existence. Therefore, B2, G is not a possibility. Further, we are only looking at permutations of the set. So {B1, B2}, {B1, G1}, {B1, G2} are the possibilities.

Do you still really believe that you'd get rich off the earlier wager? Even after seeing the "100 parents in a room" example, all other evidence, and the knowledge that this is an elementary set theory problem?

 

[Geoff101]

You are given B1's existence. Therefore B2, G is not an option.

Are you just being difficult to frustrate me?

nope...B2 is absolutely an option...you can certainly have 2 boys per the puzzle B1 and B2, and you can have B2 an B1, just as you can have B1 and G and G and B1

 

[timotinbowen]

nope...b2 is absolutely an option...you can certainly have 2 boys per the puzzle b1 and b2, and you can have b2 an b1, just as you can have b1 and g and g and b1

2/3