Today's Sunday Times had an interesting article that in the early morning (I've heard of that time of day somewhere) Radio 4 poses a puzzle. Anyway there was an interesting one they posed.
Assuming the ratio of boys and girls are 50/50 who have more sisters: boys or girls. (Given the answer I'm assuming they mean count the number of people who have one or more sisters.)
https://www.bbc.co.uk/programmes/p063yhf0
https://www.thetimes.co.uk/edition/news/helen-willetts-blows-away-john-humphrys-in-today-mind-games-zfs86rkp9
https://www.thetimes.co.uk/article/test-yourself-with-today-brain-teasers-gmqhljkfq
Also, girls have more brothers than boys.
BBB
BBG
BGG
GGG
Of the 6 boys listed, only three have sisters. But of the 6 girls listed, five have sisters.
Both the boys and the girls have the same number of sisters.
Take a family of N children.
There is a probability 1 / 2N that they are all boys, none of which would have sisters,
There is a probability (N)C(1) / 2N that there is 1 girl and N-1 boys;
each of the boys has a sister but the girl has not, so this is a fraction of (N)C(1) / 2N x (N-1) boys with sisters
There is a probability (N)C(2) / 2N that there are 2 girls and N-2 boys;
each boy has at least one sister, so this is a fraction of (N)C(2) / 2N x (N-2) boys with sisters,
and each girl has a sister, so this is a fraction of (N)C(2) / 2N x 2 girls with sisters
. . .
There is a probability (N)C(N-1) / 2N that there are N-1 girls and one boy;
the boy has at least one sister, so this is a fraction of (N)C(N-1) / 2N x 1 boys with sisters,
and each girl has at least one sister, so this is a fraction of (N)C(N-1) / 2N x (N-1) girls with sisters
There is a probability 1 / 2N that they are all girls;
all of them have sisters, so this is a fraction of 1 / 2N x N girls with sisters
The sum of the fractions of boys with sisters is 1 / 2N x ( (N)C(1) x (N-1) + (N)C(2) x (N-2) + ... + (N)C(N-1) x 1)
and the sum of the fractions of girls with sisters is 1 / 2N x ( (N)C(2) x 2 + ... + (N)C(N-1) x (N-1) + N)
Each term (N)C(N-K) in the girls' value can be changed to (N)C(K), and the last N can be changed to (N)C(N-1) x 1, so the two values are identical regardless of the size of the family.
Wizard - if this is right, don't make any plans for next June 16-20, since that is when I have my next Vegas trip penciled in.
GGB
GBG
BGG
BBG
BGB
GBB
BBB
There's 8 combinations of 3 siblings.
I count 9 boys have a sister and 9 girls have a sister.
Quote: rudeboyoiGGG
GGB
GBG
BGG
BBG
BGB
GBB
BBB
There's 8 combinations of 3 siblings.
I count 9 boys have a sister and 9 girls have a sister.
Well I don’t think that generally works:
Quote: unJonQuote: rudeboyoiGGG
GGB
GBG
BGG
BBG
BGB
GBB
BBB
There's 8 combinations of 3 siblings.
I count 9 boys have a sister and 9 girls have a sister.
Well I don’t think that generally works:If at least some families have only two kids, then it will tip to boys having more sisters because of: BB, BG, GB and GG
The question is, are there more boys or girls that have sisters?
In BB, nobody has sisters.
In BG, the boy has a sister.
In GB, the boy has a sister.
In GG, both girls have sisters.
Two of each.
Quote: ThatDonGuyQuote: unJonQuote: rudeboyoiGGG
GGB
GBG
BGG
BBG
BGB
GBB
BBB
There's 8 combinations of 3 siblings.
I count 9 boys have a sister and 9 girls have a sister.
Well I don’t think that generally works:If at least some families have only two kids, then it will tip to boys having more sisters because of: BB, BG, GB and GG
The question is, are there more boys or girls that have sisters?
In BB, nobody has sisters.
In BG, the boy has a sister.
In GB, the boy has a sister.
In GG, both girls have sisters.
Two of each.
Good point.
Sigh....Quote: DJTeddyBearVery counter-intuitive. But the answer is...
Take this very small sample of 4 families, 3 kids each:
BBB
BBG
BGG
GGG
Of the 6 boys listed, only three have sisters. But of the 6 girls listed, five have sisters.
I can’t believe I screwed this up. 🤬
If a hen and a half can lay an egg and a half in a day and a half, then how many eggs can six hens lay in six days?
Quote: unJonHere’s one from my childhood:
If a hen and a half can lay an egg and a half in a day and a half, then how many eggs can six hens lay in six days?
There are four ‘day and a half’ periods in 6 days.
6 hens times 4 periods = 24 eggs.
A certain debilitating and deadly virus has swept the nation over the course of a month. 10% of the population currently has the virus. Responding to the epidemic, a company has created a new test that is 90% accurate. (Meaning the test is correct in saying you do have the virus or do not have the virus 90% of the time.) The country has instituted mandatory testing for everyone and you show up at the doctor’s office for the test. The results are instant and the test, unfortunately, comes back positive for the virus. You are distraught.
What is the probability you actually have the virus?
Quote: ThatDonGuy
Both the boys and the girls have the same number of sisters.
Take a family of N children.
There is a probability 1 / 2N that they are all boys, none of which would have sisters,
There is a probability (N)C(1) / 2N that there is 1 girl and N-1 boys;
each of the boys has a sister but the girl has not, so this is a fraction of (N)C(1) / 2N x (N-1) boys with sisters
There is a probability (N)C(2) / 2N that there are 2 girls and N-2 boys;
each boy has at least one sister, so this is a fraction of (N)C(2) / 2N x (N-2) boys with sisters,
and each girl has a sister, so this is a fraction of (N)C(2) / 2N x 2 girls with sisters
. . .
There is a probability (N)C(N-1) / 2N that there are N-1 girls and one boy;
the boy has at least one sister, so this is a fraction of (N)C(N-1) / 2N x 1 boys with sisters,
and each girl has at least one sister, so this is a fraction of (N)C(N-1) / 2N x (N-1) girls with sisters
There is a probability 1 / 2N that they are all girls;
all of them have sisters, so this is a fraction of 1 / 2N x N girls with sisters
The sum of the fractions of boys with sisters is 1 / 2N x ( (N)C(1) x (N-1) + (N)C(2) x (N-2) + ... + (N)C(N-1) x 1)
and the sum of the fractions of girls with sisters is 1 / 2N x ( (N)C(2) x 2 + ... + (N)C(N-1) x (N-1) + N)
Each term (N)C(N-K) in the girls' value can be changed to (N)C(K), and the last N can be changed to (N)C(N-1) x 1, so the two values are identical regardless of the size of the family.
I agree! I was lazy and just worked out all possible cases of two- and three-child families. I'll work through your proof later.
DJTB, the reason you're wrong is you need to triple-count for the GBB and BGG families, since there are three possible birth order positions for the singleton.
Quote:Wizard - if this is right, don't make any plans for next June 16-20, since that is when I have my next Vegas trip penciled in.
You're right, so worthy of an audience with me. Actually, I owe you enough beers to supply a fraternity pledge party, so would have seen you either way. However, Charlie owes you one for this problem.
Quote: unJonThat’s right. This one should be easy for the probability people:
A certain debilitating and deadly virus has swept the nation over the course of a month. 10% of the population currently has the virus. Responding to the epidemic, a company has created a new test that is 90% accurate. (Meaning the test is correct in saying you do have the virus or do not have the virus 90% of the time.) The country has instituted mandatory testing for everyone and you show up at the doctor’s office for the test. The results are instant and the test, unfortunately, comes back positive for the virus. You are distraught.
What is the probability you actually have the virus?
1/2
There are four possibilities:
You don't have the virus, and the test says you don't: probability = 0.9 x 0.9 = 0.81
You don't have the virus, but the test says you do: probability = 0.9 x 0.1 = 0.09
You have the virus, and the test says you do: 0.1 x 0.9 = 0.09
You have the virus, but the test says you don't: 0.1 x 0.1 = 0.01
The probability that you really have the virus = 0.09 / (0.09 + 0.09) = 0.5
However, I have seen this problem before, and I don't think this should be considered the correct answer as I challenge the assertion that the probability that you have the virus is 10% based solely on the fact that 10% of the population has it.
Quote: unJonWhat is the probability you actually have the virus?
I get 50%.
Classic Bayesian problem. Probability (virus given positive result) = Probability (virus and positive result) / Probability (positive result) =
.1 * .9 / [.1 * .9 + .9 * .1] = 0.5
Quote: WizardYou're right, so worthy of an audience with me. Actually, I owe you enough beers to supply a fraternity pledge party, so would have seen you either way. However, Charlie owes you one for this problem.
Tell you what - I might be able to make it out for the next Spring Fling, in which case, you can buy "a round for the group" and we'll call it even.
Quote: ThatDonGuyTell you what - I might be able to make it out for the next Spring Fling, in which case, you can buy "a round for the group" and we'll call it even.
Deal, contingent on you being there.