Assume we have $n$ red and $n$ green balls in a box. What is the probability that blindfolded, you will pick a red ball on the third pick, if you learn that at least one red ball was picked on the first two picks, assuming the first two balls were not replaced?
My thoughts:
Using Bayes theorem, it is easy to show that if we consecutively pick balls blindfolded without replacement, the probability that the $i$th pick is say, red, is $\dfrac 12 \; (\star)$ for any $i.$ (seems counter-intuitive in the first place!).
Then, let $A$ be the event of picking red in the third pick and $B$ be the event of picking at least one red in the first and second picks. What we want is $P(A|B)$ which by Bayes is $\dfrac{P(B|A)P(A)}{P(B)}.$ Using $(\star),\; P(A) = 1/2.$ Also $P(B)=\dfrac 12 + \dfrac 12 – \dfrac 14=\dfrac 34.$
For $P(B|A),$ I think, it's easier to compute $P(B^c|A)$ where $B^c$ is the complement of $B$ i.e. the event that the first and second picks are green. If I define $D$ to be the event that the first pick is green and $E$ the second pick is green, then I have found no reason to believe that $D|A$ and $E|A$ are independent and so $P(B^c|A) \neq P(D|A) P(E|A).$ This is where I got stuck!
Best Answer
The blindfolded is just supposed to be the same as a random selection. What is the chance the first two picks were red? Without the information that there was at least one red ball in the first two picks, it is $\frac 12 \cdot \frac {n-1}{2n-1}$. The chance of one of each is $\frac n{2n-1}$, again without knowing that there was at least one red in the first two. We need to scale these up based on the new information, so the chance of two red is $\frac 12 \cdot \frac {n-1}{2n-1}\cdot \frac{4n-2}{3n-1}$. Similarly the chance the first two picks are opposite is $\frac n{2n-1}\cdot \frac{4n-2}{3n-1}.$ After the first two picks, either there are $n-1$ of each color and the chance the third is red is $\frac 12$, or there are $n-2$ red ones and $n$ green ones and the chance of the third is red is $\frac {n-2}{2n}$