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# One Girl - One Boy

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

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348 replies to this topic

### #11 skale

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Posted 13 July 2007 - 06:25 AM

The answer is 25%. Before they had children the probability for a girl first is 50%. The probability for a girl second is 50%. BUT, to have 2 girls the probability moves to half of half, or 25%.

You might be on the right track but thats not the correct answer.
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### #12 unreality

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Posted 13 July 2007 - 02:45 PM

imtcb, you have fallen into something known as "gambler's fallacy"...

once a first girl is born, THAT DOES NOT CHANGE THE PROBABILITY OF ANOTHER GIRL BEING BORN.

Yes if you look at the whole thing total, it's 25% for two children to be girls

But we're looking at the next child only... and the fact that a girl was born already does not change the 50/50 probability.

One single coin flip is ALWAYS 50/50 (assuming the coin is standard and all that). It doesnt matter what has come before it- will be 50/50 always. But if you're looking at things as a whole, for example if I flip a coin three times the chances of three heads is 1/2 * 1/2 * 1/2 = 1/8. But that doesnt change the fact that the next coin I flip will either be heads or tails, 50/50.

so if we're just looking at whether the next child is a boy or girl, its 1/2 each.
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### #13 skale

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Posted 13 July 2007 - 04:49 PM

imtcb, you have fallen into something known as "gambler's fallacy"...

once a first girl is born, THAT DOES NOT CHANGE THE PROBABILITY OF ANOTHER GIRL BEING BORN.

That is correct BUT,

so if we're just looking at whether the next child is a boy or girl, its 1/2 each.

knowing one of the 2 kids is a girl does alter something else, now am giving away too many hints.. it gives you useful information!
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### #14 mdsl

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Posted 13 July 2007 - 07:12 PM

knowing one of the 2 kids is a girl does alter something else, now am giving away too many hints.. it gives you useful information!

It's 1/2 using conditional probability, so why don't you share your answer already so we can pick it apart.
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### #15 Riddari

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Posted 13 July 2007 - 07:38 PM   Best Answer

Spoiler for solution

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### #16 skale

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Posted 13 July 2007 - 07:42 PM

Spoiler for solution

That sum's it all.. good job Riddari!
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### #17 unreality

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Posted 13 July 2007 - 08:01 PM

ah so we are looking at the thing as a whole... good job riddari!
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### #18 comperr

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Posted 16 July 2007 - 03:21 AM

imtcb - they are imdapendant events: the answer is 1/2
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### #19 Garrek99

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Posted 17 July 2007 - 01:55 AM

If these are the possible combinations and you eliminated the boy-boy combo that means that you are solving this combo problem backwards. What I am trying to say is that since the untouched chances are 50% 50% because:
1st child / 2nd child
Girl / Girl
Girl / Boy
Boy / Girl
Boy / Boy
so for 2nd child the set of possibilities is {girl, boy, girl, boy} which gives 2 out of 4 for girls and gives you 50%.
YOU HAVE TO eliminate the 1st child boy, 2nd child girl because the 1st child is already known to be a girl.

THEREFORE: your only possible outcomes are girl girl and girl boy which gives you 50% 50%.

There is a puzzle about eliminating doors behind which prizes are hidden. I think it is "the price is right" puzzle; where a bit of knowledge helps you increase your chances of picking the right door with the prize but the puzzle you presented isn't it.
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### #20 comperr

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Posted 17 July 2007 - 04:44 AM

the point is:
if whether you had a boy or girl fist does not affect what you have later

the choices therefore are
?/boy(50%)
?/girl (50%)
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