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# One Girl - One Boy

Best Answer Riddari , 13 July 2007 - 07:38 PM

Spoiler for solution
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348 replies to this topic

### #31 Martini

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Posted 22 July 2007 - 04:25 PM

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.
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### #32 skale

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Posted 22 July 2007 - 07:27 PM

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..
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### #33 skale

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Posted 23 July 2007 - 03:01 AM

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!
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### #34 mdsl

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Posted 23 July 2007 - 06:03 PM

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.
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### #35 Martini

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Posted 23 July 2007 - 06:44 PM

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.
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### #36 mdsl

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Posted 23 July 2007 - 06:53 PM

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.
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### #37 Martini

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Posted 23 July 2007 - 07:26 PM

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.
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### #38 mdsl

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Posted 23 July 2007 - 08:49 PM

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.
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### #39 kippi3000

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Posted 25 July 2007 - 04:51 PM

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!!
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### #40 dsu

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Posted 27 July 2007 - 11:57 PM

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].
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