One Girl - One Boy

[Guest]

Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Assume safely that the porbability of each gender is 1/2.

Ofcourse its not 1/2 else would make it a lousy puzzle...

Ans: 1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

[unreality]

The puzzle does not state that the oldest child is a girl and then ask what the probability is for the youngest child to be a girl. If that were the case, 50/50 would be correct. Instead the puzzle states that one of the children is a girl and then asks what the probability is for the other child to also be a girl. Since there are two children with two possible genders, there are four possible combinations. One of those combinations requires that both children be boys, which we know is not possible. So, there are three possible combinations. Only one of these three possibilities includes two girls. Therefore, the probability that the unspecified child is a girl is one in three.

let me quote again the important part of what you said:

the puzzle states that one of the children is a girl and then asks what the probability is for the other child to also be a girl.

then you said there are four possible combinations. YOU ARE WRONG! Yes there are four when looking at both children together. But look at my last quote of you...

for the other child to also be a girl

it doesnt matter what came before. The way the question is worded, we arent looking at as a whole. We're looking at one individual case. It's 1/2

reword the problem and then you will be right with 1/3

[Guest]

Probability does not work that way!

Correct definition:

P(A) = (The Number Of Ways Event 'A' Can Occur) /

(The Total Number Of Possible Outcomes)

Incorrect interpretetion:

P(A) = (The Total Number Of Possible Outcomes)- (The Number Of Ways Event 'A' Has Occured) /

(The Total Number Of Possible Outcomes)

If you toss a coin, and the first two tosses were heads, that does not influence(reduce/increase) the probability of future tosses: it is still 50% or 1/2

[some may argue that the amount of metal on one side is more, and there is greater chance of lighter side to come up]

If you throw a dice, the probability of any number after the 100th throw is still 1/6.

[again some may argue that the center of gravity of a dice is not precisely its center because of the different amount removed from each face]

Similarly in genetic mating, from statistical viewpoint the chance of a boy or a girl each time is 50% or 1/2. Having a boy or girl once does not reduce or increase the chance the next time [in fact some may argue some families have only girls and some have only boys].

The sum and substance is - if we see the first sibling, that in no way affects the probability of seeing the sex of the next sibling.

[unreality]

exactly what i said on the first page! its 1/2! I'm just saying, they could reword the problem to be looking in retrospect, saying "one of two children is a girl, what is the prob they are both girls?" would be a good rewording of it, where the answer is 1/3

the way they worded it, the answer is clearly 1/2

if they said this instead: "one of two children is a girl, what is the prob they are both girls?" the answer for that is 1/3

[Guest]

exactly what i said on the first page! its 1/2! I'm just saying, they could reword the problem to be looking in retrospect, saying "one of two children is a girl, what is the prob they are both girls?" would be a good rewording of it, where the answer is 1/3

the way they worded it, the answer is clearly 1/2

if they said this instead: "one of two children is a girl, what is the prob they are both girls?" the answer for that is 1/3

There is nothing wrong with the wording in the OP.

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Asking "what is the prob they are both girls?" is effectively the same as asking "what is the probability that the other kid is also a girl?"

Either way, the probability is 1/3.

[Guest]

If you toss a coin, and the first two tosses were heads, that does not influence(reduce/increase) the probability of future tosses: it is still 50% or 1/2

That is correct, but that's not a comparable scenario to the one in the OP.

Try the following real world experiment:

Grab a container of pennies and randomly lay out 100 pairs (200 pennies). Pennies that show heads we'll call boys, pennies that show tails we'll call girls.

Remove all pairs that show both heads, since couples that have two boys don't count.

Count the total number of pairs that you now have out and compare that to the number of tail-tail (girl-girl) pairs you have. About 1/3 of the pairs will be girl-girl.

So, for all pairs that have at least one girl, only 1/3 of the time will both kids be girls.

Or as unreality likes to phrase it: For all pairs that have at least one girl, 1/3 of the time the other child will be a girl.

[Guest]

you meet a family with one child. what are the odds that child's a girl?

you meet a family with 3 children and are informed that 2 of them are girls. what are the odds the last one's a girl?

you meet a family with 20 children and are informed that 19 of them are girls. what are the odds the last one's a girl?

common sense tells us the answer is 50-50 to all of these. statistical analysis tells us they aren't all equal. i think common sense is the right answer here. all of these say exactly the same thing as my first example. there is one child... odds are 50-50.

...

so you see that even in math circles, this question is still debatable. as i said... i'm going with common sense answer. but technically the answer is 1/3... too bad when math doesn't agree with common sense, isn't it?

No, not at all debatable. Read recursively till you understand the question. No claims are made to state that if 1st child BORN in a FAMILY is a GIRL then 2nd GIRL CHILD BORN in the same FAMILY is with probability 1/3. Cause that will be a FOOLISH claim. Pl read my previous post afresh and you should see the point..

[Guest]

If you toss a coin, and the first two tosses were heads, that does not influence(reduce/increase) the probability of future tosses: it is still 50% or 1/2

That is correct, but that's not a comparable scenario to the one in the OP.

Try the following real world experiment:

Grab a container of pennies and randomly lay out 100 pairs (200 pennies). Pennies that show heads we'll call boys, pennies that show tails we'll call girls.

Remove all pairs that show both heads, since couples that have two boys don't count.

Count the total number of pairs that you now have out and compare that to the number of tail-tail (girl-girl) pairs you have. About 1/3 of the pairs will be girl-girl.

So, for all pairs that have at least one girl, only 1/3 of the time will both kids be girls.

Or as unreality likes to phrase it: For all pairs that have at least one girl, 1/3 of the time the other child will be a girl.

yes, that is indeed the logical way to look at this problem!

[Guest]

we're all wrong. The initial problem clearly states ONE is a girl, so the other is a boy, with 0 chance that he is a girl.

[Guest]

we're all wrong. The initial problem clearly states ONE is a girl

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

And ONE child is a girl. That does not preclude the other child from also being a girl. If the problem stated that 'only' one one of them is a girl, then you'd be correct.

[Guest]

I was thinking along the lines of the 30? riddle, two coins total 30?, one isn't a nickel, the other one is. In this case, one is a girl, the other isn't.

And posting answers that stretches thing a little is a lot better than some of these topics where people feel compelled to create an account just to post the correct answer when it has bee clearly explained many other times.

[Guest]

I was thinking along the lines of the 30? riddle, two coins total 30?, one isn't a nickel, the other one is. In this case, one is a girl, the other isn't.

Mentioning the 30? riddle actually hurts your rationale that the answer to the OP's riddle is that there are zero chances that the other child is a girl.

"2 coins total 30?. one of them is NOT a nickel. What two coins do you have?"

The riddle states that one of the coins is not a nickel, which does not preclude the other coin from being a nickel.

Most people assume that when the riddle states "one of them is not a nickel", that both coins are being described. You are making the same mistake in regards to the OP's riddle. When the OP stated, "one of them is a girl" he was not quantifying the total amount of girls. He was describing ONE girl- which does not preclude the other child from being a girl also.

And posting answers that stretches thing a little is a lot better than some of these topics where people feel compelled to create an account just to post the correct answer when it has bee clearly explained many other times.

I never claimed that your type of post was worse than other types.

[Guest]

ok, now I'm being serious, I've been realy bored at work and now I think I finally got the math to back up the real solution.

too bad when math doesn't agree with common sense, isn't it?

I was a little rusty with my stat so thanks wikipedia.

A = child one is a girl

B = child two is a girl

P(A) = P(B)=1/2 [given]

P(AUB.) = 3/4 [A or B which in stat is one, the other, or both true]

P(A?B) = 1/4 [A and B both are ture]

P(A?B|AUB.) = P((A?B)?(AUB))/P(AUB.) [the probability of the intersection given the union]

P((A?B.)?(AUB.)) simplifies to P(A?B.)

P(A?B.)/P(AUB.) = (1/4)/(3/4)= 1/3!

I honestly didn't want to admit the answer was 1/3 until I could proove it with math including conditional probability so there it is.

[Guest]

you meet a family with one child. what are the odds that child's a girl?

you meet a family with 3 children and are informed that 2 of them are girls. what are the odds the last one's a girl?

you meet a family with 20 children and are informed that 19 of them are girls. what are the odds the last one's a girl?

common sense tells us the answer is 50-50 to all of these. statistical analysis tells us they aren't all equal. i think common sense is the right answer here. all of these say exactly the same thing as my first example. there is one child... odds are 50-50.

i'm familiar with skale's analysis and can confirm that most statisticians would agree. but there is even debate among mathematicians as to what the correct answer is. i personally reject the accepted answer and content this problem is too vague to be solved even by math standards.

the problem is a question of the sampling. does the fact that you know the sex of the first child reduce your sample to just families with one daughter or is the original sample of a totally random population still hold. statistics says to use the original sample giving the 1/3 answer but i say that's wrong. as soon as you know the first child is a girl, your sample changes to only families with at least one girl leaving only 2 possibilities. girl/girl and girl/boy and the 1/2 answer. so you see that even in math circles, this question is still debatable. as i said... i'm going with common sense answer. but technically the answer is 1/3... too bad when math doesn't agree with common sense, isn't it?

Silliest way of trying to rebut, by the way.

Think of it this way, if you have 20 children, and 19 of them are girls, then the last one (very specific here, the LAST ONE!!) has a 19/20 chance of being a girl. It's the same if it's the FIRST one, where the 1st has a 19/20 chance of being a girl. Being specific makes it impossible to be 1/2 chance, because it's known that 19 of them are girls, and 1 is a boy. 19/20 chance of being a girl. Endof.

The question can be easily explained this way. I have 2 apples. The apples can be either red/green. The situation is such that i can either have R/R, R/G, G/R, and G/G.

if one apple is Green (EITHER THE FIRST OR THE SECOND, it's not SPECIFIED!! REFER the top example for clarification), then R/R is not applicable. The remaining options are R/G, G/R, G/G. out of these options, if one is green, the chances for the other one to be green is 1/3 (G/G).

There. Lots of luck gambling, if you still can't agree to this!!

[Guest]

There are two aspects.

1] We are confusing sequential events with simultaneous events.

If we toss a coin, the probability of getting a head is 1/2. If we toss two coins together, the probability of getting 2 heads is 1/3 (3 possibilities, head-head, tail-tail and head-tail), but if we toss them one at a time, the probability of getting head each time is 1/2. If we get head the first time, it does not affect the probability of head in the second coin tossed separately!

If we toss a collection of coins simultaneously, the chance of getting half of them head is 1/2. But if we toss each coin from a collection sequentially, the chance of each being head is half, and is not influenced by the number of heads in the other coins.

As we are not dealing with twins (simultaneous events), the probability of each being a girl or a boy (born separately or sequentially) is 1/2. Seeing them separately or together has no influence on actual birth!

Now if the problem was like this:

We have 6 girls and 6 boys in a room. They are walking out of the room one by one. The first is a girl. What are the chances of the second walking out... Then the author's logic is ok.

In a population of girls and boys, unless it is specified that there are equal number of girls and boys, we cannot assume it! So if a family has 5 girls, what are the chances the last one born is a boy? 1/2!! (some may argue the chance of boy is even less!)

Coming back to our current problem, in a married couple, nowhere it is mentioned the children are twins.

2] Now I am going to point out the flaw in the answer provided:

The following are possible combinations of two children that form a sample space in any earthly family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Observe 2nd and 3rd options (Girl - Boy, Boy - Girl) indicate a sequence (1st child and 2nd child), otherwise both are the same! A permutation is taken, not a combination! In that case, if the first is girl, there exist only the first two options - probability is 1/2!

If you are really talking about combination, then the following three (not four) exists:

Girl - Girl

Girl - Boy

Boy - Boy

If you take out the last, only two possibilities remains!

I do not know why nobody observed this earlier.

[You cannot combine permutation and combination as it pleases you - you must select one or the other].

[Guest]

There are two aspects.

1] We are confusing sequential events with simultaneous events.

If we toss a coin, the probability of getting a head is 1/2. If we toss two coins together, the probability of getting 2 heads is 1/3 (3 possibilities, head-head, tail-tail and head-tail)...

No, you're the one who's confused. When you toss two coins, one at a time or together, the probability of getting one H and one T is not 1/3. You are assuming that HH, TT, and HT (as combinations, as you put it) are equally likely. They are not. HT/TH is twice as likely as the others. If you don't believe it, just toss two coins a hundred times, sequentially or together. I guess you didn't bother with the coin experiment I posted for you earlier?

Your whole post is infused with silliness about the magical relevance of simultaneity. (Do you think it makes a difference if events are almost, but not quite, at the exact same time; say, within 5 ms of each other? At what point do things switch to the simultaneous distribution? Does this happen discontinuously?)

[Guest]

I will use your coin flip example just because I am sick of regering to children and want some variety. ;p

Anyway, let say you flip two coins, consecutively or sequentially as you like, and record the results. You might get something like this:

TH,HH,HT,TT,TT,TT,TH,HH,TH,HT (This is actual results that I recieved.)

Two of those ten combinations are HH, three of them are TT and five of them are mixed. This actually falls very close to what one would expect.

Now, let say we throw out all of the TT combinations because we only want instances that include a head. So, HH constitutes two of the seven remaining instances, or `28.6%. Inversely, lets say we throw out the HH combinations because we only want instances with a tail. The TT groups would constitute three of eight instances, or 37.5%. Remarkably enough, the average of those two results is ~33.0%

Mock up the numbers as you like and you will end up with similar results. The worst you can really do is force all instances to be the same and that does not support a 25% answer any more than a 33% answer. If you go unreasonable to one side, it will balance on the other.

HH,HT,HH,HT,HH,HT.HH,HT,HH,HT would give you 50% for HH or 0% for TT, but it is a ridiculous distortion of expectations.

HH,HT,HH,HT,HH,TT.HH,HT,HH,HT gets 55% for HH and 20% for TT, averaging to ~37.8%.

As with all the previous compelling arguments for the 33% answer, I am feeling pretty solid with on the logic behind this.

We are not talking about the possible gender of a single child, but rather the possible genders of a set of two children with the specific criteria that at least one of them is female.

[Guest]

We are not talking about the possible gender of a single child, but rather the possible genders of a set of two children with the specific criteria that at least one of them is female

Ok, I understand your point of view, but you do not note the fallacy that creeps in when you combine both permutation and combination! When you create your set, you are using permutation. When you are pointing out the girl, you are using combination. Let me ellaborate the fallacy:

What are the chance the other coin is H if the 1st coin is H? 50% or 1/2 (out of HT and HH).

What are the chance the other coin is H if the 2nd coin is H? 50% or 1/2 (out of HH and TH).

Combining the above two,

What are the chance the other coin is H if the 1st coin or the 2nd coin is H? How can the answer be less than either of the above two, as it includes both the cases??? That is the hidden fallacy of combining permutation and combination.

What are the chances of the other child being a girl if the 1st child is a girl? 50% or 1/2 (out of GG and GB)

What are the chances of the other child being a girl if the 2nd child is a girl? 50% or 1/2 (out of GG and BG)

What are the chances of the other child being a girl if the 1st or the 2nd child is a girl? How can the chance be less than the above two, especially as both the above cases are included??? The fallacy disappears if we say:

What are the chances of the other child being a girl if one of the child is a girl?

(in combination, we have these members in the set: GG, GB, BB. So the answer needs to be 1/2).

If you can solve the fallacies pointed out above, I shall accept your point of view.

Your whole post is infused with silliness about the magical relevance of simultaneity

I understand what you mean - maybe I have used the wrong terms or expressions. What I meant to point out is probability is a tricky thing which may easily give wrong answers if used in the wrong context.

[Guest]

How can the answer be less than either of the above two, as it includes both the cases???

Because there isn't any inclusion going on here; you've re-conditionalized your probability. First, you were conditioning on the 1st coin being H. Then, you were conditioning on the 2nd coin being H. Finally, you conditioned on either one being H. You can't just add up conditional probabilities in the normal way, however, if they don't all have the same antecedent condition. Your last case doesn't "include" any of the previous cases, because they all have different conditions. What would be closer to the inclusion relationship you're going for is this:

What is the probability that the second coin is H and the first coin is H? 1/4

What is the probability that the first coin is H and the second coin is H? 1/4

What is the probability that both coins are H and either the first or second coin is H? 1/4

That is, we turn the conditionalizing into conjunctions instead. This makes the third case subsume the previous two, and thus the third case has probability >= each of the previous two. [As it happens, in this example, all three cases turn into the same thing when we do this. Ah well.].

I understand what you mean - maybe I have used the wrong terms or expressions. What I meant to point out is probability is a tricky thing which may easily give wrong answers if used in the wrong context.

Actually, what you claimed about the odds of coin flips landing on head or tails being dependent on whether or not they landed simultaneously had nothing to do with being wrong based on terms and expressions. The logic was nonsensical.

[Guest]

What is the probability that the second coin is H and the first coin is H? 1/4

What is the probability that the first coin is H and the second coin is H? 1/4

What is the probability that both coins are H and either the first or second coin is H? 1/4

It is remarkable you said 1/4, not 1/3 in each of the above!

But you say

the chance of there being two heads if the 1st coin or the 2nd coin is H is 1/3.

Ironic, isn't it? You talk about "re-conditionalized your probability", but aren't you doing just that?

In the first, TT is a member of the set, but in the second, it is (conveniently) not! But there is essential no difference between the two statements "What is the probability that both coins are H and either the first or second coin is H? 1/4" and "the chance of there being two heads if the 1st coin or the 2nd coin is H is 1/3"

You are repeatedly ignoring the fact I am trying to point out - you cannot use permutation and combination together - one in making the set, the other in setting up the conditions. Either use permutation, or combination, not both. If you use permutation to define the set, do not use the word "other". If you use the word "other", use combination in defining the sets.

[Guest]

A lot of the initial confusion in this riddle is that it was worded differently the first few days. Initially, it was implied that a couple was going to have 2 children (future tense), and the question was, if the first child was a girl, what would the probability be for the second to also be a girl.

I do see with the wording now, the actual intent of the OP. It is indeed 1/3 the way it is worded now, just as it is 1/4 the way it was worded originally.

The problem with applying the gamblers fallacy to all probability is that gambling is set up to give better odds to the house, which means that the longer you gamble, the more you will loose. No casino will give straight odds on a 50/50 coin toss, because in the end they will only break even. This is why in roulette black and red are 2:1 payout, yet the odds are not 50/50 (2 greens).

Before the events happen, if the probability is 50/50, 2 consecutive events would yield a 25% probability for each of the 4 possible outcomes. BB / BG / GB / GG

Between events, the probability would be 50% for each of the 2 possible outcomes. GB / GG

After the events, when one of the two is given (but not the order), the probability would be 33 1/3%. BG / GB / GG

[Guest]

A lot of the initial confusion in this riddle is that it was worded differently the first few days. Initially, it was implied that a couple was going to have 2 children (future tense), and the question was, if the first child was a girl, what would the probability be for the second to also be a girl.

Let me assure you that the only time post was edited was to add the solution tag. But am glad you see the reasoning now.

[unreality]

there is a lot of confusion between the two sides of this problem, and i've read all ur arguements, and it isnt about conditionals or conjunctionals or anything

its about wording.

Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Assume safely that the porbability of each gender is 1/2.

this is the problem.

let me requote the important part (middle line)

"They have two kids, one of them is a girl, what is the probability that the other kid is also a girl."

"they have two kids." They already have them. They're not expecting kids... they already have both of them born. The kids could be 20 and 18 years old for all we care.

"One of them is a girl"

okay... so if we have two kids, already born, and we can rule out B/B:

G/B

B/G

G/G

"what is the probability that the other kid is also a girl."

not: "what is the probability that the other kid will be a girl."

In that case it would be 1/2, since the other child does not affect the probability. But it doesnt say that. It said "is also a girl". The child is already born.

it all comes down to two sides:

Side A (answer is 1/2): We're looking at the second child's birth as an independent event.

Side B (answer is 1/3): We're looking at both together.

and I am sorry to say Side B is right (as I was on Side A for quite some time)

Why?

Because the second child has already been born. We dont care what was the probability that it was boy/girl while it was in its mother's womb. We know thats 1/2. The fact is, the gender has already been determined. We're just trying to figure out the probability they are both girls.

Since the options are:

G/B

B/G

G/G

we can safely say, G/G has a 1/3 chance.

There is no fallacy. There is no bla-bla-bla. It all comes down to wording:

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

the "other kid" is already born. The solution doesnt call for the probability that the other kid will be a girl. It calls for the probability that they are (as in right now) BOTH girls.

We're not looking at one girl's chances. In that case there would be two options:

B

G

but we're not. BECAUSE OF THE WORDING, we are looking at both children at once:

G/B

B/G

G/G

It is 1/3.

It pains me to say this (I defended 1/2 for a while until i realized i was wrong) but I'm not being bitter and I'm trying to prove what I now see is right.

its a fine line, based on skale's wording. but the answer is 1/3. I rest my case. I think this topic should be done. The discussion cant go much elsewhere.

[Guest]

Ironic, isn't it? You talk about "re-conditionalized your probability", but aren't you doing just that?

There's nothing wrong with re-conditionalizing of probabilities. You just have to keep in mind that that is in fact what you've done, and make sure not to add things up in the wrong way.

But there is essential no difference between the two statements "What is the probability that both coins are H and either the first or second coin is H? 1/4" and "the chance of there being two heads if the 1st coin or the 2nd coin is H is 1/3"

Those are both true statements. Do you mean to say there is no difference between "The probability that both coins are H and either first or second coin is H" and "The chance of there being two heads if the 1st coin or the 2nd coin is H"? Because those two statements have all the difference in the world; the first of those is a nonconditional probability, and the second is a conditional probability. They're as different as "What is the probability of rolling a 2 and rolling an even number?" (which would be 1/6 on a normal die) and "What is the probability of rolling a 2 given that you've rolled an even number?" (which would be 1/3).

Did you try any of the experiments that I and others have posted? If you haven't, please take the time to try them. You'll find that they won't give the results you're saying they will and I don't think you'll have a convincing argument that the coins are lying.

[Guest]

unreality: well put. Your reasoning is the correct one, and the explanation is most aptly put.

for those who can't agree that the answer is 1/3, READ unreality's post before u post another absurd argument that it is 1/2.

good play, unreality!

[unreality]

thanks

all these posts by dsu and martini are getting way complicated, i was mostly doing that so i could understand it myself