Super Duper Extra Hard Brainteaser

[partygirluk]

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I will bet money that I am right.

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I will bet up to $100 that my answer of 1/2 is correct.

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I'll take it.

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I'm tempted to take it as I am pretty sure the real answer is 199/399

[Patrick del Poker Grande]

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I will bet money that I am right.

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I will bet up to $100 that my answer of 1/2 is correct.

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I'll take it.

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I'm tempted to take it as I am pretty sure the real answer is 199/399

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I am willing to entertain doubts that it is not 1/3, even that it's 199/399, but I am quite quite sure it is not 1/2.

[sam h]

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Think about it like this, it might change your mind: knowing that a women has two girls increases the likelihood that she has a child named Sarah. Conversely, knowing she has a child named Sarah....

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I'm don't buy it yet but could be pursuaded if you explained more.

This might be where we disagree: You say that there are equal probabilities of the four outcomes but I'm not so sure.

B, G(s)
G(s), B
G, G(s)
G(s), G

I just don't see it. If I took a random sample of the population that had two kids, one of which was a girl named Sarah, I could not justify expecting 75% of the firstborn children to be girls.

[DMBFan23]

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I will bet money that I am right.

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I will bet up to $100 that my answer of 1/2 is correct.

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gaming mouse,

I hate to be a dick, but even if your logic is correct that the answer must be closer to 1/2 than 1/3, the answer CANNOT be exaclty 1/2 due to the stipulation that both kids cannot be sarah.

[Patrick del Poker Grande]

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Think about it like this, it might change your mind: knowing that a women has two girls increases the likelihood that she has a child named Sarah. Conversely, knowing she has a child named Sarah....

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I'm don't buy it yet but could be pursuaded if you explained more.

This might be where we disagree: You say that there are equal probabilities of the four outcomes but I'm not so sure.

B, G(s)
G(s), B
G, G(s)
G(s), G

I just don't see it. If I took a random sample of the population that had two kids, one of which was a girl named Sarah, I could not justify expecting 75% of the firstborn children to be girls.

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That's because you're giving equal weight to G,G(s) and G(s),G to other combinations like B,G(s), i.e. that G,G(s) has a 25% likelihood, as does G(s),G, when really it's both of them put together that have a 25% likelihood.

[partygirluk]

OK right. Lets assume 1,000,000 families.

250,000 will be 2 boys
250,000 willl be 2 girls
500,000 will be one of each.

Of the 500,000 with 1 boy, 1 girl, there will be 5,000 instances where the girl is Sarah.

Of the 250,000 instances with 2 girls, there will be 4975 instances with a girl named Sarah.

Therefore there is a 4975/9975 (199/399) chance that the mother has 2 girls.

[gaming_mouse]

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I will bet money that I am right.

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I will bet up to $100 that my answer of 1/2 is correct.

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I'll take it.

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All calcs assume exactly 2 children.

P(2 girls | girl named sara) = P(2 girls & 1 named Sara)/P(girl names sara)

P(girl named sara) = P(1st child sara OR 2nd child sara) = 2*P(1st child sara) = 2*.5*.01

P(2 girls and one named sara) = 2*P(1st child sara, 2nd child girl) = 2*.5*.01*.5

Plugging back in, we get .5. Where is the mistake?

[jason_t]

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I will bet up to $100 that my answer of 1/2 is correct.


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Not sure who'd take that bet; I won't as I believe the answer is 1/2 also. My math confirms it, as does my simulation.

I'll make the same bet.

[sam h]

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That's because you're giving equal weight to G,G(s) and G(s),G to other combinations like B,G(s), i.e. that G,G(s) has a 25% likelihood, as does G(s),G, when really it's both of them put together that have a 25% likelihood.

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That's basically what I've been thinking. I think it might be that, among the population that has two kids and one girl named Sarah, G, G(s) + G(s), G still equals 33%.

[DMBFan23]

you are allowing both daughters to be named sarah, at the least.