We have two urns. You can't tell them apart from the outside, but one has three \$1 chips and seven \$10 chips, and the other has eight \$1 chips and two \$10 chips. You randomly draw a chip from one of the urns and it happens to be a $10 chip. Without replacing this draw, I offer you a chance to draw and keep a chip from either urn. Should you draw from the same urn or the opposite urn, and what is the expected value of the chip you draw? Why?
I did by casework,
say we had drawn from 1st urn , then E(2nd time 1st urn)=\$7, E(2nd time 2nd urn)=\$7.3
say we had drawn from 2nd urn , then E(2nd time 1st urn)=\$6.4, E(2nd time 2nd urn)=\$8
so finally E(2nd time same urn)=0.5*7+0.5*8 = 7.5
E(2nd time other urn)=0.5*7.3+0.5*6.4 = 6.85
so same urn is favourable. Now this answer is incorrect. But I cannot find my mistake. Help!
Best Answer
Let $X$ be the event the first chip drawn was worth $\$1$. Let $A$ be the event the first chip was drawn from the first urn, what I'll hereafter refer to as "urn $A$." Similarly, $B$ will be the event $A^c$, the event of drawing from urn $B$. Let $Y$ be the random variable corresponding to the value of the second drawn chip, that is $Y=10$ is the event where we draw a $\$10$ chip and $Y=1$ is the event where we draw a $\$1$ chip.
We are tasked with calculating $E[Y\mid X]$ given the strategy of not switching as well as $E[Y\mid X]$ given the strategy of always switching.
Now... while true, $\Pr(A)=0.5$, this shouldn't appear in your calculations where it did. What is more relevant is $\Pr(A\mid X) = \frac{3}{11}$. This calculation is able to be conveniently and immediately performed thanks to the symmetry of the problem where each of the $11$ chips worth $\$1$ were equally likely to have been drawn, three of which originated from urn $A$.
If you insist on calculating $\Pr(A\mid X)$ the long way, approach with Bayes' Theorem, $\Pr(A\mid X) = \dfrac{\Pr(X\mid A)\Pr(A)}{\Pr(X)} = \dfrac{\frac{3}{10}\cdot\frac{1}{2}}{\frac{3}{10}\cdot\frac{1}{2}+\frac{8}{10}\cdot\frac{1}{2}} = \frac{3}{11}$
Now... $\Pr(Y=10\mid X) = \frac{\Pr(Y=10\cap X)}{\Pr(X)}=\frac{\Pr(X)\Pr(A\mid X)\Pr(Y=10\mid X,A) + \Pr(X)\Pr(B\mid X)\Pr(Y=10\mid X,B)}{\Pr(X)}$
The $\Pr(X)$ can cancel top and bottom, leaving you with:
$\Pr(A\mid X)\Pr(Y=10\mid X,A) + \Pr(B\mid X)\Pr(Y=10\mid X,B)$
I already covered how to calculate $\Pr(A\mid X)$. As for $\Pr(Y=10\mid X,A)$, that is as simple as looking at how many chips are still in the specified urn of the appropriate value after noting one of the $\$1$'s was taken out and is $\frac{7}{9}$
So, we have $\Pr(Y=10\mid X) = \frac{3}{11}\cdot\frac{7}{9} + \frac{8}{11}\cdot\frac{2}{9}$
We can similarly calculate $\Pr(Y=1\mid X)$. We calculate then $E[Y\mid X] = 10\Pr(Y=10\mid X)+1\Pr(Y=1\mid X) = \frac{48}{11}\approx 4.36$ for not switching.
For switching it is similar except $\Pr(Y=10\mid X,A)$ will have to do with the contents of urn $B$ instead of the remaining contents of urn $A$ yielding $\frac{334}{55}\approx 6.07$ for switching.
This agrees with our intuition... if we had drawn a $\$1$ chip, we should have thought it more likely that we were previously drawing from the urn with a larger proportion of $\$1$ chips and we should want to switch to improve our expected result.