Need help with transforming a binomial distribution This week, Bob heads to the roulette table. When betting on red, if the spin lands on red then you win the value of the bet, otherwise you win nothing. Bob will use the Martingale strategy: starting with a bet of $\$1$, he will double his bet after each loss, and return his bet to $\$1$ after each win. Bob will bet on red every time.
(a) Let $Y$ be the net winnings from playing 3 spins using Strategy B,
i) [2] What is the range and pmf of $Y$ ?
The probably the roulette wheel will be red is $\frac{18}{38}$.
So I found the $R(X)=0,1,2,3$
and the $p(x)=\binom3x\cdot (9/19)^x\cdot(10/19)^x$
I am completely slumped with finding the range and pmf of $Y$.
I think the range of $R(Y)$ might $=\{-6,-2,0,1,2,3\}$, (based on the money factored) but I might be wrong.
If anyone can help me answer this question that would be fantastic!
 A: You should be able to do this with a tree diagram, for instance, listing all 8 possibilities.
Another way to help compute the winnings would be to code Bob's outcomes, as W for win, L for loss, to give sequences like $WLL$ (for a win on the first throw, and losses on the second and third throw). Then notice, by the Martingale strategy, substrings of type $L^n W$ ($n \ge 0$) always win a dollar (regardless of "n"). Finally, if Bob finishes his gambling with a string of $n$ losses, $L^n$ amounts to the loss $-(2^n-1)$.
This gives, for the range:
$WWW = 3$; $WWL = 1$; $WLW = W + LW = 1+1 = 2$; $WLL = -2$; $LWW = LW + W = 2$; $LWL = LW + L = 1-1 = 0$; $LLW = 1$; $LLL = -7$ (poor Bob!)
The tricky part is these win/loss amounts are not commutative with respect to the string ($WLL$ does not equal $LWL$).
For the pmf of your discrete RV, $Y$, the probabilities are commutative. You'd just have your odds calculation as above: $(9/19)^{\#wins} (10/19)^{\# losses}$--so eg $P(WLW) = (9/19)^2 (10/19)^1$. At any value in the range of $Y$, you'd add up all the probabilities (for instance $P(Y=-7)=(10/19)^3$).
