I found this problem in a statistics book, and I'm wondering if my solution is correct.
"You and a friend play a game involving 20 balls in an urn, of which 1 is red and 19 are white. The game is considered won if you end up drawing the red ball. Without looking, you draw 14 balls randomly and leave the remaining 6 balls for your friend. What is the probability that you win the game? Would this probability change if after you draw the balls both you and your friend randomly discard all but 1 ball, and it is known that either you or your friend possesses the red ball (i.e., it was not discarded)."
In the first case, the probability of me randomly choosing the red ball and winning is pretty straightforward (70%). However, I'm wondering about whether this probability changes in the second case where I discard 13 of my 14 balls randomly, and my friend does the same with 5 of his 6 balls. My intuition is that I still have a 70% chance of winning, but then again, I had a higher probability of discarding the red ball (so the fact that it is known that the red ball was not discarded may mean that my friend had the higher probability of having it in the first place). Can anyone confirm or refute my reasoning?
The intuitive argument for $70\%$ seems plausible. But we do the problem carefully, in case your initial intuition is not right. Intuitions about probabilities are not entirely reliable.
Let $R$ be the event red is not discarded, and $Y$ the event you end up with the red. We want $\Pr(Y|R)$. This is $\frac{\Pr(Y\cap R}{\Pr(R)}$.
The event $R$ can happen in two ways: (ii) You got the red, and did not discard it or (ii) Your friend got the red, and did not discard it.
The probability of (i) is $\frac{14}{20}\cdot \frac{1}{14}$ and the probability of (ii) is $\frac{6}{20}\cdot \frac{1}{6}$.
While computing $\Pr(R)$, we have already computed $\Pr(Y\cap R)$. It is the probability of (i).
Divide. We get a nice simple answer that is not $70\%$.