A box contains 6 number 1, 3 number 2, 2 number 3 and 1 number 4. You have to pay $2 to draw to get the ball A box contains 6 number 1, 3 number 2, 2 number 3 and 1 number 4. We have to pay \$2 to draw to get the number i and will be paid \$i (i = 1, 2, 3, 4).
There are 2 questions here:

*

*Should we play this game?

*Suppose if the we draw the number 4, we can pay $1 to play again, should we play?

I think I've solved the first question, with this table:
\begin{array}{|c|c|c|c|}
\hline
x& 1 & 2 & 3 & 4 \\ \hline
y=r(x) & -1 & 0 & 1 & 2 \\ \hline
P(X=x) & \frac{1}{2} & \frac{1}{4} & \frac{1}{6} & \frac{1}{12} \\ \hline
\end{array}
So $E(Y) = -1 \times \frac{1}{2} + 0 \times \frac{1}{4} + 1 \times \frac{1}{6} + 2 \times \frac{1}{12} \approx -1.667$
So my first answer is we shouldn't play this game, but I don't know how to solve the second question, how should I update the table above? Thank you in advanced.
 A: Your expected value is wrong:
$$-1\cdot\frac12+0\cdot\frac14+1\cdot\frac16+2\cdot\frac1{12}=-\frac16\;.$$
It doesn’t make a difference for the first question, since it’s still negative. But if the expected value had indeed been less than $-1$, the second question would have been easy to answer, since in that case it would still be less than $0$ even if we pay only $\$1$.
As has been mentioned in the comments, it makes a difference whether the $4$ gets replaced. To me, “play again” sounds like we play exactly the same game again, so I’ll assume the $4$ gets replaced.
That raises another question: If we draw the $4$ again, do we get to play yet again, or does it stop there? It’s perhaps instructive to solve the problem for both cases.
If the answer is no, so there’s at most one replay, you can use the value you calculated for a single game to find the value of drawing a $4$. If we draw a $4$, we get $\$4$ as before, but now we get to play a game for $\$1$ that has an expected value of $-\frac16$ when it costs $\$2$, and thus an expected value of $\frac56$ if it only costs $\$1$. So drawing a $4$ is now worth an additional $\frac56$. Replacing the value of $2$ by $2\frac56$ in your table leads to an additional contribution to the expected value of $\frac1{12}\cdot\frac56=\frac5{72}$, so the expected value is now $-\frac16+\frac5{72}=-\frac7{72}$, still negative. So if you can only get a single replay, you still shouldn’t play.
Now consider the case where you can keep replaying as long as you keep drawing the $4$. This is a bit trickier. The idea is to introduce a variable for the value of the game, derive an equation for it and solve the equation.
So say the value of the initial game, i.e. when you pay $\$2$ for it, is $x$. If we draw the $4$, in addition to the $4$ points we get to play a game that’s worth $x+1$ (the same as the initial game, but $\$1$ more since we have to pay $\$1$ less for it). So now the expected value of the game is $x=-\frac16+\frac1{12}(x+1)$. Solving for $x$ yields $x=-\frac1{11}$. So you should still not play the game.
