a probability brainteaser from OOT

[DMBFan23]

help me y'all,

I'm either doing something wrong or explaining something wrong.


answer in this thread (http://forumserver.twoplustwo.com/showflat.php?Cat=&Number=1949103&page=0&view=colla psed&sb=5&o=14&fpart=1)

[probman]

After quickly reading through the thread it seems that most people converged to the correct answer of 1/3. The argument you provide it the standard one given for this rather famous problem.

There is usually a second part. Now if the mother instead answers that her eldest child is a girl, does change the probability? Yes, the probability of the younger child being a girl is 1/2 not 1/3 as in the previous case. This is generally referred to as the census takers problem.

[DMBFan23]

did you see the 1% stipulation?

[Gator]

Take an average hand (say Q8u) and figure the probability for having the nuts for that hand. It may not extrapolate perfectly to having the nuts for any hand, but it may be close. So, what's the probability of ending up with a flop that is the nuts for q8u?

[probman]

The 1% stipulation comes from assuming a finite population size, n, with n/2 men and n/2 women. If n is small then the non-replacement effect you mentioned is significant. However, as n grows large this effect becomes insignificant, i.e. much less than 1%.

One should note that the above model for the populations is very inaccurate. A better model is to assume that at every birth there is a 1/2 probability of a boy and a 1/2 probability of a girl. Under this model, the replacement effect is wiped out since we are drawing from a stationary distribution.

[AnyTwoCanLose]

How could it possibly matter what the girl's name is?

The probability is 1/3... right?

[jason_t]

[ QUOTE ]
How could it possibly matter what the girl's name is?

The probability is 1/3... right?

[/ QUOTE ]

A women is more likely to have a girl named Sarah if she has two girls. Conversely, if he she has a child named Sarah....

[Bobby Digital]

Someone please explain how the first answer is 1/3. The woman has two children, on of them IS a girl. This only leaves one other child, which is either a boy or girl. The chances of having a boy or a girl are equally likely. Therefore shoudn't the answer be 1/2.

[jason_t]

[ QUOTE ]

Someone please explain how the first answer is 1/3. The woman has two children, on of them IS a girl. This only leaves one other child, which is either a boy or girl. The chances of having a boy or a girl are equally likely. Therefore shoudn't the answer be 1/2.

[/ QUOTE ]

There are three possibilities, all of which are distinct and equally likely.

G,B
B,G
G,G.

So the answer is 1/3.

[Bobby Digital]

Ok, but if she has at least one girl B,G and G,B are exactly the same. The one that isn't a girl, is a boy.

[jason_t]

[ QUOTE ]
Ok, but if she has at least one girl B,G and G,B are exactly the same. The one that isn't a girl, is a boy.

[/ QUOTE ]

They are not the same. Before more information is received, these possibilities are equally likely:

BB
BG
GB
GG.

Once we are told there is at least one girl, the first possibility is eliminated and three remain, all of which are equally likely.

[MickeyHoldem]

[ QUOTE ]
Now, suppose that 1% of all girls are named Sarah, and that if a mother had two girls that the mother would not name both of them Sarah

You know a woman has exactly 2 children, you ask her if she has a girl named Sarah, she says yes.

What are the chances she has two girls?

[/ QUOTE ]
Clearly without the name stipulation the answer is 1/3.
What does this name stipulation do to the possible outcomes that we must choose from?

the four family options are:

1) BB
2) BG
3) GB
4) GG

Let's assume there are 10,000 of each of these families out there

obviously 1) is out.
for 2) BG the chance that the girl is named Sarah = 1/100 so there are 100 families with BG that have a girl nemed Sarah
same for 3) GB the chance that the girl is named Sarah = 1/100 = 100 families
for 4) GG the chance that she has a girl named Sarah = 1 - (99/100 * 99/100) = 199/10000 = 199 families that have GG will have a girl named Sarah

So there are a total of 399 families that have a girl named Sarah.

We know we must be at a home where there is a Sarah, so we have a 199/399 chance of being at the home of 2 girls.

[jason1990]

Food for thought:

In this setup, there are 40,000 girls (10,000 each from groups (2) and (3), and 20,000 from group (4)). But there are only 399 Sarahs. The problem states that 1% of all girls are named Sarah. Where's the 400th Sarah?

[mostsmooth]

[ QUOTE ]
Food for thought:

In this setup, there are 40,000 girls (10,000 each from groups (2) and (3), and 20,000 from group (4)). But there are only 399 Sarahs. The problem states that 1% of all girls are named Sarah. Where's the 400th Sarah?

[/ QUOTE ]
i think the equation should be amended from 1-(99/100*99/100) to 1-(99/100*98/99)
that would give you 200 sarahs i believe rather than 199
somebody chime in please
/images/graemlins/cool.gif

[MickeyHoldem]

Good point... I just got done reading the ENTIRE thread in OOT... eeek.... I should have stayed away!!!

But here's some more food for thought... Suppose that 4% of the mother's that had a boy first, and then a girl, named that girl Sarah! All other mothers named their girls something else. You could now say "Now, suppose that 1% of all girls are named Sarah, and that if a mother had two girls that the mother would not name both of them Sarah" The answer here is obviously not 1/2

We have to assume that all mothers name their girls Sarah with the same probability... but we can't because the GG family can't have 2 girls named sarah, so the probabilty changes for the second girl.

I declare the problem has bad form! /images/graemlins/grin.gif

[Nomad84]

[ QUOTE ]
Food for thought:

In this setup, there are 40,000 girls (10,000 each from groups (2) and (3), and 20,000 from group (4)). But there are only 399 Sarahs. The problem states that 1% of all girls are named Sarah. Where's the 400th Sarah?

[/ QUOTE ]

When there are 2 girls, if the first is named sarah, there is a 0% chance of the other girl being named sarah. She is exempt from the 1% rule. If the 2 girls could both be named sarah, this would happen 0.01*0.01=0.0001 or 1 time in 10000. So in the 10000 pairs of girls, once you would have both named sarah. This is not allowed, therefore you are exactly one sarah short of what would otherwise be expected.

FWIW, I got 199/399 as well. Same math as the first couple of math verifications in the post in OOT.

[mostsmooth]

does my amended formula resolve this?
(ps i have to education, especially in math)

[Nomad84]

[ QUOTE ]
[ QUOTE ]
Food for thought:

In this setup, there are 40,000 girls (10,000 each from groups (2) and (3), and 20,000 from group (4)). But there are only 399 Sarahs. The problem states that 1% of all girls are named Sarah. Where's the 400th Sarah?

[/ QUOTE ]
i think the equation should be amended from 1-(99/100*99/100) to 1-(99/100*98/99)
that would give you 200 sarahs i believe rather than 199
somebody chime in please
/images/graemlins/cool.gif

[/ QUOTE ]

I am assuming that for each girl, there is a 1% chance that she is named Sarah. Your equation says that there are 100 girls, and 1 is named Sarah. I do not believe that this was the intent of the OP.

[Mano]

I think this is how you would do it:

P(child is a girl named Sarah) = (.01)*(.5) = .005
P(child is a girl not named Sarah) = (.99)*(.5) = .495

We also know how to calculate conditional probability:
P(A|B) = P(A n B)/P(B)

(or in equivalent english the probability of A given B is equal to the probability of the intersection of A and B divided by the probability of B).

P(A|B) = P(having two girls given exactly one is named Sarah)
P(B) = P(exactly one of two children is named Sarah)
P(A n B) = P(having two girls with exactly one named Sarah)

P(A n B) = P(first child named sarah)*P(second child girl not named Sarah) + P(first child girl not named Sarah)*P(second child named sarah)

P(A n B) = (.005)*(.495) + (.495)*(.005) = .00495

P(B) = 1 - P(neither Child Named Sarah) - P(both children named Sarah) = 1 - (.995)*(.995) - (.005)*(.005) = 0.00995

So P(A|B) = .00495/0.00995 = .4975