When I calculated the answer to this, I got $Cov(X,Y) = 0$. This is because $$Cov(X,Y) = E[XY] - E[X]E[Y]$$ $$= (10)(5)(1/6) - 10(5)(1/6) - 10(5)(1/3) + 10(5)(1/3) - 0 $$ $$ = 0 $$ However, I think there must be a mistake in my calculation because this implies that the two bets are independent. What am I doing wrong?
UPDATE: I know understand why these two bets are independent if there are 36 slots. However, I am not trying to figure out if these bets are independent if there are 38 slots (American Roulette). I think they should be for the same reason the bets were independent with 36 slot roulette, but this is not what my covariance calculation shows.
$$Cov(X,Y) = E[XY] - E[X]E[Y]$$ $$= (10)(10)(6/38) - 10(10)(6/38) - 10(5)(12/38) + 10(5)(14/38)$$ $$ = 2.49 $$
Let's assume that there are $36$ possible outcomes, half of which are red the other half black. Also let's assume (I think this should hold in general) that there are $6$ red numbers in the first dozen numbers.
So lets denote the roulette outcome with $R$ and define $$ X= \begin{cases}1, \quad \text{if $R$ is red},\\ 0, \quad \text{if $R$ is black}, \end{cases} Y = \begin{cases} 1, \quad R \leq 12 \\ 0, \quad R > 12. \end{cases} $$ we have $$ \Bbb P( X = 1) = \frac12, \quad \Bbb P (Y = 1) = \frac 13, \quad \Bbb P(XY = 1) = \frac 6 {36} = \frac 16. $$ Therefore $$ \text{Cov}(5X, 10Y) = 50 \cdot \biggl( \frac 16 - \frac 12 \cdot \frac13 \biggr)= 0. $$ So the covariance is zero, but this does not automatically imply that $X$ and $Y$ are independent. For independence you need to show that $$ \Bbb P(X = x, Y=y) = \Bbb P(X=x) \cdot \Bbb P (Y=y), \quad x,y \in\{0,1\}. $$ An easy calculation shows that $X$ and $Y$ are indeed independent.