Say I know of a man with 3 sons.
I belive that the odds that a son has green eyes is 1/4, and 3/4 for brown.
Given that I know one son has green eyes, the probability of at least two having green eyes is: $\frac{p(\text{2 or more have green eyes})}{p(\text{one or more has green eyes})} =\frac{1-(3/4)^3-3(1/4)*(3/4)^2}{1-(3/4)^3}$
Given that I know that the first son has green eyes, the probability of two or more having green eyes is: $1-(3/4)^2$
The same second probability applies if I condition on the second son having green eyes, or condition on the third.
Intuitively, however, it would seem that the first probability should be a weighted average of these 3 equal probabilities, and therefore equal to them. But I calculated them, and that does not seem to be the case.
Could someone please shed some light on the matter?
"The father with three sons of green or brown eyes" problem can be modeled by the following set of elementary events:
$$\Omega=\{(g,g,g),(g,g,b),(g,b,g),(g,b,b),(b,g,g),(b,g,b),(b,b,g),(b,b,b)\}$$ where the $i^{\text{th}}$ components of the triplets denote the eye color of the $i^{\text{th}}$ son. What abou the probabilities assigned to these elementary events? If we assume that the eye color of the brothers is independent and that color green is of probability $\frac14$ and then color brown is of probability $\frac34$ then we have the following probabilities:
$$P:\left\{\frac1{4^3},\frac3{4^3},\frac3{4^3},\frac9{4^3},\frac3{4^3},\frac9{4^3},\frac9{4^3},\frac{27}{4^3}\right\}$$ in the order of the elementary events listed above.
1
a conditional probability of the following form:
$$P(\{(g,g,g),(g,g,b),(g,b,g),(b,g,g)\}\mid \{(g,b,b),(b,g,b),(b,b,g),(g,g,g),(g,g,b),(g,b,g),(b,g,g)\})=$$ $$=\frac{P(\{(g,g,g),(g,g,b),(g,b,g),(b,g,g)\})}{P(\{(g,b,b),(b,g,b),(b,b,g),(g,g,g),(g,g,b),(g,b,g),(b,g,g)\})}=$$ $$=\frac{\frac1{4^3}+\frac3{4^3}+\frac3{4^3}+\frac3{4^3}}{1-\frac{27}{4^3}}=\frac{10}{37}.$$
That is, the first result is OK.
2
is a conditional probability of the following form
$$P(\{(g,g,g),(g,g,b),(g,b,g),(b,g,g)\}\mid \{(g,g,g),(g,g,b),(g,b,g),(g,b,b)\})=$$ $$=\frac{P(\{(g,g,g),(g,g,b),(g,b,g),(b,g,g)\}\cap\{(g,g,g),(g,g,b),(g,b,g),(g,b,b)\})}{P(\{(g,g,g),(g,g,b),(g,b,g),(g,b,b)\})}=$$ $$=\frac{P((g,g,g),(g,g,b),(g,b,g))}{P(\{(g,g,g),(g,g,b),(g,b,g),(g,b,b)\})}=$$ $$=\frac{\frac1{4^3}+\frac3{4^3}+\frac3{4^3}}{\frac1{4^3}+\frac3{4^3}+\frac3{4^3}+\frac9{4^3}}=\frac7{16}.$$
That is, the second result is OK as well.