The K M Links Game - May 2024 Week 4
Quizzes & Puzzles2 mins ago
Children
A couple have two children (and they are a normal couple with a 50:50 chance of having a girl or a boy). You are intorduced to one of their children who is a girl. What is the probability that the other child is also a girl? There is a QI claxon waiting for the unwary.....
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For all you non-believers:
http://forums.xkcd.com/viewtopic.php?f=3&t=311
The following is an extract copied from the above thread:
(it assumes that the first child is a boy, rather than a girl - so just switch everything round)
The question doesn't specify how you know one of the children is a boy. If you grabbed the first child and looked at its genitals, and it happened to be a boy, then the chances that the second child is a girl are 50%, since it's an independent question. If, on the other hand, all the question is saying is that you checked to see if there was a boy in the pair, that's a different question. It implies you checked the first child, and if it was a girl, then you went on to check the second. In that case, the probability that one of the children is a girl is 2/3.
You can think of it this way: You know you found a boy. Either the first child you checked was a boy, or the first child was a girl and the second was a boy. Given the three possibilities (b-b, b-g, and g-b), all with equal probability, there is a 2/3 probability that the first child you checked was a boy. So 2/3 of the time there's a 50% chance the second child is a girl, plus the 1/3 of the time the first child is certainly a girl, for a total of 2/3 of the time one of the children is a girl.
............ continued in next post .........
http://forums.xkcd.com/viewtopic.php?f=3&t=311
The following is an extract copied from the above thread:
(it assumes that the first child is a boy, rather than a girl - so just switch everything round)
The question doesn't specify how you know one of the children is a boy. If you grabbed the first child and looked at its genitals, and it happened to be a boy, then the chances that the second child is a girl are 50%, since it's an independent question. If, on the other hand, all the question is saying is that you checked to see if there was a boy in the pair, that's a different question. It implies you checked the first child, and if it was a girl, then you went on to check the second. In that case, the probability that one of the children is a girl is 2/3.
You can think of it this way: You know you found a boy. Either the first child you checked was a boy, or the first child was a girl and the second was a boy. Given the three possibilities (b-b, b-g, and g-b), all with equal probability, there is a 2/3 probability that the first child you checked was a boy. So 2/3 of the time there's a 50% chance the second child is a girl, plus the 1/3 of the time the first child is certainly a girl, for a total of 2/3 of the time one of the children is a girl.
............ continued in next post .........
........ continued .........
The hard part here is accepting that the relative probabilities of the three combinations don't change when you remove g-g. It's easier to understand when you phrase it in terms of coin flips: if you perform 100 double coin flips, you'll come up with roughly equal numbers of each of the four possible combinations (h-h, h-t, t-h, and t-t). If you then discount all t-t combinations (presumably about 25) you'll still have equal numbers of the other 3. So if all we know about the result of a particular coin flip is that one of the coins came up heads (in other words, the result was not t-t), we know it's one of the 75 other results, 50 of which have one coin showing tails. Hence, the probability that one of the coins is tails is 50/75, or 2/3.
The hard part here is accepting that the relative probabilities of the three combinations don't change when you remove g-g. It's easier to understand when you phrase it in terms of coin flips: if you perform 100 double coin flips, you'll come up with roughly equal numbers of each of the four possible combinations (h-h, h-t, t-h, and t-t). If you then discount all t-t combinations (presumably about 25) you'll still have equal numbers of the other 3. So if all we know about the result of a particular coin flip is that one of the coins came up heads (in other words, the result was not t-t), we know it's one of the 75 other results, 50 of which have one coin showing tails. Hence, the probability that one of the coins is tails is 50/75, or 2/3.
When a couple ave twi children there are four equally likely possibilities::
- boy born first (B1) , girl second (G2)
- boy first (B1), boy second (B2)
- girl first (G1), girl second (G2)
- girl first (G1), boy second (B2)
As we have seen a girl we can rule out the B1, B2 combination. So the equally likely options now are:
- B1 &G2
- G1 & G2
- G1 & B2
The girl you have seen could gave been the older child or the younger child of the pair. It's reasonable to assume from the question that there's a 50/50 chance you saw the older child and a 50/50 chance you saw the younger child.
So the girl you have seen could have been G1 (the older child) or G2 (the younger child).
If the girl you saw was G1 then the other child is equally likely to be G2 or B2.
If the girl you saw was G2 then the other child is equally likely to be B1 or G1.
In either case the chance of the sibling being a girl is 50/50.
Another way of looking at it is that when a couple have 2 kids there is a 50/% chance both will be the same sex and 50% chance that they will be of opposite sex. The sex of the child you meet (provided it was at random) doesn't tell you anything about the sex of the other- the chance of the other being of the same sex is always 50/50.
- boy born first (B1) , girl second (G2)
- boy first (B1), boy second (B2)
- girl first (G1), girl second (G2)
- girl first (G1), boy second (B2)
As we have seen a girl we can rule out the B1, B2 combination. So the equally likely options now are:
- B1 &G2
- G1 & G2
- G1 & B2
The girl you have seen could gave been the older child or the younger child of the pair. It's reasonable to assume from the question that there's a 50/50 chance you saw the older child and a 50/50 chance you saw the younger child.
So the girl you have seen could have been G1 (the older child) or G2 (the younger child).
If the girl you saw was G1 then the other child is equally likely to be G2 or B2.
If the girl you saw was G2 then the other child is equally likely to be B1 or G1.
In either case the chance of the sibling being a girl is 50/50.
Another way of looking at it is that when a couple have 2 kids there is a 50/% chance both will be the same sex and 50% chance that they will be of opposite sex. The sex of the child you meet (provided it was at random) doesn't tell you anything about the sex of the other- the chance of the other being of the same sex is always 50/50.
Factor30 - you've almost explained it correctly in your post - yet you still fail to believe that the answer is 1/3.
You've already said:
"As we have seen a girl we can rule out the B1, B2 combination. So the equally likely options now are:
- B1 &G2
- G1 & G2
- G1 & B2 "
Now the important thing is, that we are not told whether the girl is the oldest or the youngest out of the 2 siblings. Therefore any one of the above 3 combinations is possible - and in 2 out of 3 occasions (B1 & G2 + G1 & B2), when one of the children is a girl - the other is a boy. Only in 1 out of 3 occasions (G1 & G2) is the other child a girl.
Now if we had been told that the girl was the eldest (or the youngest) of the 2 - then things would be different. B1 & G2 would no longer be an option (as G2 refers to the girl being younger than the boy), so the only remaining possibilities would be:
G1 & G2
G1 & B2
...... in which case the chance of the other child being a girl would be 50/50 ......... but, very imporatantly, we are told nothing about the identity of the girl, so the chance of the other child being a girl are 1 in 3.
You've already said:
"As we have seen a girl we can rule out the B1, B2 combination. So the equally likely options now are:
- B1 &G2
- G1 & G2
- G1 & B2 "
Now the important thing is, that we are not told whether the girl is the oldest or the youngest out of the 2 siblings. Therefore any one of the above 3 combinations is possible - and in 2 out of 3 occasions (B1 & G2 + G1 & B2), when one of the children is a girl - the other is a boy. Only in 1 out of 3 occasions (G1 & G2) is the other child a girl.
Now if we had been told that the girl was the eldest (or the youngest) of the 2 - then things would be different. B1 & G2 would no longer be an option (as G2 refers to the girl being younger than the boy), so the only remaining possibilities would be:
G1 & G2
G1 & B2
...... in which case the chance of the other child being a girl would be 50/50 ......... but, very imporatantly, we are told nothing about the identity of the girl, so the chance of the other child being a girl are 1 in 3.
No gizmonster- in my explanation, and in reality, the GG option is available twice in that when there are two girls you could be seing either of the girls.
The sex of two children are independent events. Just like tossing 2 coins in a dark room- if I shone a torch and saw one was a head the information would not change the fact that the chance of the other being a head is still 50/50.
Let's look at it another way.
*By seeing a girl you are deducing that there is a 2/3 chance the other is a boy- i.e. that two kids will be of different sex.
*Suppose you then go to another house which has 2 kids and you see a boy. By your logic there is a 2/3 chance the second will be a girl.
*So pooling the two results suggests there is a 2/3 chance that any pair of children will not be of the same sex. That seems wrong.
Rollo says he is a maths graduate- well so am I. I'm always keen to learn though and would appreciate Rollo's thoughts on where my logic is wrong.
The sex of two children are independent events. Just like tossing 2 coins in a dark room- if I shone a torch and saw one was a head the information would not change the fact that the chance of the other being a head is still 50/50.
Let's look at it another way.
*By seeing a girl you are deducing that there is a 2/3 chance the other is a boy- i.e. that two kids will be of different sex.
*Suppose you then go to another house which has 2 kids and you see a boy. By your logic there is a 2/3 chance the second will be a girl.
*So pooling the two results suggests there is a 2/3 chance that any pair of children will not be of the same sex. That seems wrong.
Rollo says he is a maths graduate- well so am I. I'm always keen to learn though and would appreciate Rollo's thoughts on where my logic is wrong.
Factor30, don't get confused, what we're talking about here is not about probability of people giving birth to girls or boys, its about knowing that there are two children & 1 is a girl then the probability of the other also being a girl. The relative ages do not come into it.
I am not only an engineering grad but also a 6 sigma black belt which involves a lot of stats & the answer my friend is not blowing in the wind but is 1/3!
I am not only an engineering grad but also a 6 sigma black belt which involves a lot of stats & the answer my friend is not blowing in the wind but is 1/3!
Factor30 you are almost answering the question - then you lose it ......
Let's go with the 2 coins 'cos it's easier to follow.
There are 4 ways to throw 2 coins:
HH
HT
TH
TT
Each one is as likely as the other, so say we throw 100 pairs of coins, probability says that we'll end up with:
2 heads 25 times
2 tails 25 times
1 head and 1 tail 50 times (25 times you'll throw a head followed by a tail - and the other 25 times, you'll throw a tail followed by a head).
If someone now picks one pair and they tell us that one of the coins is a head, for example, we know that we can't have 2 tails, so we are left with 75 possibilities:
25 ocassions where we threw a head followed by a tail
25 ocassions where we threw a tail followed by a head
25 ocassions where we threw a head followed by a head.
In these 75 pairs of coins, we have 50 ocassions where we have a head and a tail and 25 ocassions where we have 2 heads .......... i.e. where 1 coin is a head, the other will be a tail 2 out of 3 times (50 in 75) and the other will be a head 1 in 3 times (25 in 75).
What's very important here (and maybe this is where you're getting confused) is that we are not told which head we are looking at. If we are told, for example, that we are looking at the head that was thrown with the first coin, then we know that we can't have the TH (tails followed by a head), which leaves us only with 2 options:
HT
HH
..... each one being as likely as the other, so in this case the chance of the 2nd coin being a head would be as equally as likely as it being a tail.
Let's go with the 2 coins 'cos it's easier to follow.
There are 4 ways to throw 2 coins:
HH
HT
TH
TT
Each one is as likely as the other, so say we throw 100 pairs of coins, probability says that we'll end up with:
2 heads 25 times
2 tails 25 times
1 head and 1 tail 50 times (25 times you'll throw a head followed by a tail - and the other 25 times, you'll throw a tail followed by a head).
If someone now picks one pair and they tell us that one of the coins is a head, for example, we know that we can't have 2 tails, so we are left with 75 possibilities:
25 ocassions where we threw a head followed by a tail
25 ocassions where we threw a tail followed by a head
25 ocassions where we threw a head followed by a head.
In these 75 pairs of coins, we have 50 ocassions where we have a head and a tail and 25 ocassions where we have 2 heads .......... i.e. where 1 coin is a head, the other will be a tail 2 out of 3 times (50 in 75) and the other will be a head 1 in 3 times (25 in 75).
What's very important here (and maybe this is where you're getting confused) is that we are not told which head we are looking at. If we are told, for example, that we are looking at the head that was thrown with the first coin, then we know that we can't have the TH (tails followed by a head), which leaves us only with 2 options:
HT
HH
..... each one being as likely as the other, so in this case the chance of the 2nd coin being a head would be as equally as likely as it being a tail.
....except you are wrong Gizmonster.
Just tell me this... how can the result of one coin have anything to do with how the second coin fell?
So you are saying that if you see a head then there is a 2/3 chance the other will be tails.
By the same logic if on another day you repeat it and see a tail you'd argue that there is a 2/3 chance the other is heads.
So by your logic, 2 times out of 3 the two coins will always be different. But that contradicts your statement that initially there are 4 equally likely outcomes (HH,HT TH,TT).
You are still ignoring the fact that even though you have seen a Head there are the following equally likely options.
* Coin 1 landed on H, coin 2 on tails. You have seen the H so the other is a T
*Coin 1 landed on T, coin 2 on H. You have seen the H so the other is a T
*Coin 1 landed on H, coin 2 landed on H. You have seen
coin 1 so coin 2 is a Head (it doesn't matter that you don't know which is coin 1 or coin 2)
**Coin 1 landed on H, coin 2 landed on H. You have seen
coin 2 so coin 1 is a Head (again it doesn't matter that you don't know which is coin 1 or coin 2)
So that's still 4 outcomes, all equally likely, which for the other coin gives tails 50% of the time and Heads 50% of the time.
Simulate it and you'll see the flaw in the claim that 2/3 of the time the two coins will be different.
Just tell me this... how can the result of one coin have anything to do with how the second coin fell?
So you are saying that if you see a head then there is a 2/3 chance the other will be tails.
By the same logic if on another day you repeat it and see a tail you'd argue that there is a 2/3 chance the other is heads.
So by your logic, 2 times out of 3 the two coins will always be different. But that contradicts your statement that initially there are 4 equally likely outcomes (HH,HT TH,TT).
You are still ignoring the fact that even though you have seen a Head there are the following equally likely options.
* Coin 1 landed on H, coin 2 on tails. You have seen the H so the other is a T
*Coin 1 landed on T, coin 2 on H. You have seen the H so the other is a T
*Coin 1 landed on H, coin 2 landed on H. You have seen
coin 1 so coin 2 is a Head (it doesn't matter that you don't know which is coin 1 or coin 2)
**Coin 1 landed on H, coin 2 landed on H. You have seen
coin 2 so coin 1 is a Head (again it doesn't matter that you don't know which is coin 1 or coin 2)
So that's still 4 outcomes, all equally likely, which for the other coin gives tails 50% of the time and Heads 50% of the time.
Simulate it and you'll see the flaw in the claim that 2/3 of the time the two coins will be different.
I think gizmonster is correct and the answer is 2/3. And his post of 25/1/09 at 16.14 sets out a proof of that answer.
factor30 you obviously disagree. Since you think the answer is 1/2 then by definition you must think there is a flaw in gizmonsters proof. So I suggest you point to exactly where the error is in that proof, ie exactly which words are incorrect.
factor30 you obviously disagree. Since you think the answer is 1/2 then by definition you must think there is a flaw in gizmonsters proof. So I suggest you point to exactly where the error is in that proof, ie exactly which words are incorrect.
Thanks pulse, but when you ask me to pint out the flaw in Gizmonster's proof I cannot see which 'proof' you are referring to.
I have already outlined the flaws in Gizmonster's argument.
One flaw is that it is wrong to say that there are three equally likely possibilities remaining -GG, GB, BG and to therefore conclude that the second child is more likely to be a girl. As i have pointed out, there are still 4 equally likely outcomes:
Child A is a girl, child B is a boy. You have seen the Girl (must be child A) so the other is a Boy
*Child A is a boy. Child B is a girl. You have seen a girl (must be child B) so the other is a boy
Child A is a girl, child B is a girl. You could have seen child A in which case the other will be a girl . Or , equally likely, you've seen the girl who is child B and therefore the other child (child A) is also a girl.
The other flaw is that it cannot be teh case with completely independent events that knowing the result for one child (or coin) tells you anything about the property of teh otehr child (coin)
Finally , as I have pointed out before, his argument, if true, would mean that with any pair of children (or two coins) there is always a 2/3 chance of different outcomes even though both oucomes are equally likely. And that can't be correct.
I have already outlined the flaws in Gizmonster's argument.
One flaw is that it is wrong to say that there are three equally likely possibilities remaining -GG, GB, BG and to therefore conclude that the second child is more likely to be a girl. As i have pointed out, there are still 4 equally likely outcomes:
Child A is a girl, child B is a boy. You have seen the Girl (must be child A) so the other is a Boy
*Child A is a boy. Child B is a girl. You have seen a girl (must be child B) so the other is a boy
Child A is a girl, child B is a girl. You could have seen child A in which case the other will be a girl . Or , equally likely, you've seen the girl who is child B and therefore the other child (child A) is also a girl.
The other flaw is that it cannot be teh case with completely independent events that knowing the result for one child (or coin) tells you anything about the property of teh otehr child (coin)
Finally , as I have pointed out before, his argument, if true, would mean that with any pair of children (or two coins) there is always a 2/3 chance of different outcomes even though both oucomes are equally likely. And that can't be correct.
I've just spent 30 mins or so, typing a reply and my PC went t*ts up lol ...... oh well :(
Anyway I think I may know where we are getting our wires crossed Factor30:
Quote from Factor30: "Just tell me this... how can the result of one coin have anything to do with how the second coin fell?
So you are saying that if you see a head then there is a 2/3 chance the other will be tails"
....... this is NOT what I'm saying. You are implying that we look at one coin, find it's a head, then try and work out the probabilty of the other coin being a head or a tail. This is not what I'm saying, 'cos in this case it's clearly 50/50.
What I am saying is that 2 coins are flipped and we are told that at least one of them is a head - this is completely different to flipping 2 coins and then turning the first one over to reveal a head. In my case, the person needs to look at both coins 'cos they may look at the 1st coin and find it's a tail. They then peek at the 2nd coin in order to check whether that is a head.
I posted this on another forum:
http://forums.xkcd.com/viewtopic.php?f=3&t=338 09
On the same forum, there's also another,more in-depth thread, also including the boy/girl scenario:
http://forums.xkcd.com/viewtopic.php?f=17&t=32 659
Anyway I think I may know where we are getting our wires crossed Factor30:
Quote from Factor30: "Just tell me this... how can the result of one coin have anything to do with how the second coin fell?
So you are saying that if you see a head then there is a 2/3 chance the other will be tails"
....... this is NOT what I'm saying. You are implying that we look at one coin, find it's a head, then try and work out the probabilty of the other coin being a head or a tail. This is not what I'm saying, 'cos in this case it's clearly 50/50.
What I am saying is that 2 coins are flipped and we are told that at least one of them is a head - this is completely different to flipping 2 coins and then turning the first one over to reveal a head. In my case, the person needs to look at both coins 'cos they may look at the 1st coin and find it's a tail. They then peek at the 2nd coin in order to check whether that is a head.
I posted this on another forum:
http://forums.xkcd.com/viewtopic.php?f=3&t=338 09
On the same forum, there's also another,more in-depth thread, also including the boy/girl scenario:
http://forums.xkcd.com/viewtopic.php?f=17&t=32 659
I've just been reading one of your earlier posts and you quote that there are 4 possibilities:
Quote from Factor30"
Coin 1 landed on H, coin 2 on tails. You have seen the H so the other is a T
*Coin 1 landed on T, coin 2 on H. You have seen the H so the other is a T
*Coin 1 landed on H, coin 2 landed on H. You have seen
coin 1 so coin 2 is a Head (it doesn't matter that you don't know which is coin 1 or coin 2)
**Coin 1 landed on H, coin 2 landed on H. You have seen
coin 2 so coin 1 is a Head (again it doesn't matter that you don't know which is coin 1 or coin 2)
All the above assume you see the heads first.
What happens when the first coin you see is a tail??
There are 2 more options that you missed:
Coin 1 lands on a head and coin 2 lands on a tail and you see the tail first, then you see the head last.
Coin 2 lands on a head and coin 1 lands on a tail and you see the tail first, then you see the heads last.
You're missing these 2 options 'cos I think you're assuming that we pick the heads first (which we do not).
There are 6 options available - all with equal probability .... in 4 of these options there are head-tails pairings. In 2 options there are heads-heads pairings - giving probabilities of 2 in 3 that when 1 is a head the other will be a tail and 1 in 3 of both being heads.
Please tell me you agree now ????
Quote from Factor30"
Coin 1 landed on H, coin 2 on tails. You have seen the H so the other is a T
*Coin 1 landed on T, coin 2 on H. You have seen the H so the other is a T
*Coin 1 landed on H, coin 2 landed on H. You have seen
coin 1 so coin 2 is a Head (it doesn't matter that you don't know which is coin 1 or coin 2)
**Coin 1 landed on H, coin 2 landed on H. You have seen
coin 2 so coin 1 is a Head (again it doesn't matter that you don't know which is coin 1 or coin 2)
All the above assume you see the heads first.
What happens when the first coin you see is a tail??
There are 2 more options that you missed:
Coin 1 lands on a head and coin 2 lands on a tail and you see the tail first, then you see the head last.
Coin 2 lands on a head and coin 1 lands on a tail and you see the tail first, then you see the heads last.
You're missing these 2 options 'cos I think you're assuming that we pick the heads first (which we do not).
There are 6 options available - all with equal probability .... in 4 of these options there are head-tails pairings. In 2 options there are heads-heads pairings - giving probabilities of 2 in 3 that when 1 is a head the other will be a tail and 1 in 3 of both being heads.
Please tell me you agree now ????
Oh boy!! 33 replies to what I thought was a relatively simple riddle!
Think some are getting muddled up between the probability of giving birth to a girl or boy & the probability of the 2nd child being a girl, knowing
a) that the couple have 2 children &
b) that one of them is a girl
You have to take the 2 factors together & not in isolation.
Think some are getting muddled up between the probability of giving birth to a girl or boy & the probability of the 2nd child being a girl, knowing
a) that the couple have 2 children &
b) that one of them is a girl
You have to take the 2 factors together & not in isolation.
There is a difference between "They have two children, at least one is a girl, what is the chance they are both girls?" and "They have two children. THIS ONE is a girl, what is the chance THE OTHER ONE is a girl"
If you think the chance the other is a girl, then you also think (if you had met a boy) there was a 2/3rd chance the othjer was a boy - in other words, you think 2/3rd of pairs of children are the same sex. But they aren't, and neither is the answer to this puzzle 2/3. Or think of it another way: Imagine they only have one child - a girl. What is the chance their next child will be a girl? 2/3rds? Patently not!
If you think the chance the other is a girl, then you also think (if you had met a boy) there was a 2/3rd chance the othjer was a boy - in other words, you think 2/3rd of pairs of children are the same sex. But they aren't, and neither is the answer to this puzzle 2/3. Or think of it another way: Imagine they only have one child - a girl. What is the chance their next child will be a girl? 2/3rds? Patently not!
Quote form Ridcully:
"Imagine they only have one child - a girl. What is the chance their next child will be a girl? 2/3rds? Patently not! "
This is NOT what we're saying, 'cos clearly in this case the odds are 50/50.
The initial question states that there are 2 children and at least one of them is a girl (not specifically this one, or that one, i.e. we do not know which one).
Let's take things one step back form yours Ridcully and assume that a woman is pregnant with 2 children.
For the sake of this exercise let's assume that same-sex twins are born with equal probability as different-sex twins, 'cos realistically same-sex twins are probably more common than different-sex twins.
So, there are the following options possible:
boy born first, followed by boy born 2nd
boy born first, followed by girl born 2nd
girl born first, followed by boy born 2nd
girl born first, followed by girl born 2nd
(Note that the boy-boy and girl-girl option are only used once)
Each of these is as likely as the other, with a 1 in 4 chance.
Agreed so far ???
Now, the doctor approaches you and informs you that one of them is a girl. Now, VERY IMPORTANTLY, he does not say whether it is the youngest or the oldest , only that he knows that there is a girl in the womb.
What are the odds that the other child will be a girl?? If you think it's 50/50, then read on 'cos you're wrong!!
The boy-boy option is now no longer an option, so the remaining possible births are:
boy born first, followed by girl born 2nd
girl born first, followed by boy born 2nd
girl born first, followed by girl born 2nd
Look at the options and you'll see that 2/3 of the time a boy and girl will be born and 1/3 of the time 2 girls will be born.
Please do not try and say that the
"Imagine they only have one child - a girl. What is the chance their next child will be a girl? 2/3rds? Patently not! "
This is NOT what we're saying, 'cos clearly in this case the odds are 50/50.
The initial question states that there are 2 children and at least one of them is a girl (not specifically this one, or that one, i.e. we do not know which one).
Let's take things one step back form yours Ridcully and assume that a woman is pregnant with 2 children.
For the sake of this exercise let's assume that same-sex twins are born with equal probability as different-sex twins, 'cos realistically same-sex twins are probably more common than different-sex twins.
So, there are the following options possible:
boy born first, followed by boy born 2nd
boy born first, followed by girl born 2nd
girl born first, followed by boy born 2nd
girl born first, followed by girl born 2nd
(Note that the boy-boy and girl-girl option are only used once)
Each of these is as likely as the other, with a 1 in 4 chance.
Agreed so far ???
Now, the doctor approaches you and informs you that one of them is a girl. Now, VERY IMPORTANTLY, he does not say whether it is the youngest or the oldest , only that he knows that there is a girl in the womb.
What are the odds that the other child will be a girl?? If you think it's 50/50, then read on 'cos you're wrong!!
The boy-boy option is now no longer an option, so the remaining possible births are:
boy born first, followed by girl born 2nd
girl born first, followed by boy born 2nd
girl born first, followed by girl born 2nd
Look at the options and you'll see that 2/3 of the time a boy and girl will be born and 1/3 of the time 2 girls will be born.
Please do not try and say that the
......... ran outta room lol
....... Please do not try and say that the girl-girl option should be used twice. It wasn't used twice when there were all 4 options available - so why now ??
All the evidence is there, plus them links that I gave in an earlier post ....... if you don't understand now - you're obviously misunderstanding/misreading the question
....... Please do not try and say that the girl-girl option should be used twice. It wasn't used twice when there were all 4 options available - so why now ??
All the evidence is there, plus them links that I gave in an earlier post ....... if you don't understand now - you're obviously misunderstanding/misreading the question
Ha- because the girl the doctor has seen could turn out to be the first born or the second born- both are equally likely.
So if the doctor has seen the girl who turns out to be born first then as you have said the other baby could be a boy or a girl (your options 2 and 3- which are both equally likely.)
If instead the doctor had seen the girl who then turns out to be born second then you'll see under your options 1 and 3 there are two equally likely outcomes- a boy or a girl for the first born. So it's 50/50
You are looking at this from the wrong angle!
So if the doctor has seen the girl who turns out to be born first then as you have said the other baby could be a boy or a girl (your options 2 and 3- which are both equally likely.)
If instead the doctor had seen the girl who then turns out to be born second then you'll see under your options 1 and 3 there are two equally likely outcomes- a boy or a girl for the first born. So it's 50/50
You are looking at this from the wrong angle!
