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-   -   Super Duper Extra Hard Brainteaser (http://archives2.twoplustwo.com/showthread.php?t=215071)

partygirluk 03-17-2005 08:12 PM

Re: Super Duper Extra Hard Brainteaser
 
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


I will bet money that I am right.

[/ QUOTE ]

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

[/ QUOTE ]
I'll take it.

[/ QUOTE ]

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

Patrick del Poker Grande 03-17-2005 08:14 PM

Re: Super Duper Extra Hard Brainteaser
 
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


I will bet money that I am right.

[/ QUOTE ]

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

[/ QUOTE ]
I'll take it.

[/ QUOTE ]

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

[/ QUOTE ]
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 03-17-2005 08:16 PM

Re: Super Duper Extra Hard Brainteaser
 
[ QUOTE ]
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....

[/ QUOTE ]

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 03-17-2005 08:21 PM

Re: Super Duper Extra Hard Brainteaser
 
[ QUOTE ]
[ QUOTE ]


I will bet money that I am right.

[/ QUOTE ]

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

[/ QUOTE ]

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 03-17-2005 08:21 PM

Re: Super Duper Extra Hard Brainteaser
 
[ QUOTE ]
[ QUOTE ]
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....

[/ QUOTE ]

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.

[/ QUOTE ]
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 03-17-2005 08:23 PM

Re: Super Duper Extra Hard Brainteaser
 
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 03-17-2005 08:24 PM

Re: Super Duper Extra Hard Brainteaser
 
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


I will bet money that I am right.

[/ QUOTE ]

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

[/ QUOTE ]
I'll take it.

[/ QUOTE ]

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 03-17-2005 08:24 PM

Re: Super Duper Extra Hard Brainteaser
 
[ QUOTE ]

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


[/ QUOTE ]

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 03-17-2005 08:26 PM

Re: Super Duper Extra Hard Brainteaser
 
[ QUOTE ]
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.

[/ QUOTE ]

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 03-17-2005 08:28 PM

Re: Super Duper Extra Hard Brainteaser
 
you are allowing both daughters to be named sarah, at the least.


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