Is there an easier way to solve this probability problem?

Question

Consider the following "game" with two players:

Two players take turns drawing marbles out of a jar, without replacement. The jar starts with some number of red, blue, and white marbles. If a player draws a red marble or a blue marble, they draw again; otherwise the other player draws a marble. The "game" ends when the jar is empty.

I am interested in the random variable $R_{r,b,w}$, which represents the number of red marbles that the first player ends up with, when the jar starts with $r$ red marbles, $b$ blue marbles, and $w$ white marbles. In particular, I am interested in the probability distribution and expected value of $R_{r,b,w}$.

By considering the first marble that the first player draws, I have found the following recursive relationship between the variables: \begin{align} P(R_{r,b,w}=x) &= \frac{w}{r+b+w} P(R_{r,b,w-1}=r-x)\\ &+ \frac{b}{r+b+w} P(R_{r,b-1,w}=x)\\ &+ \frac{r}{r+b+w} P(R_{r-1,b,w}=x-1), \end{align} where $P(R_{a,b,c}=y)=0$ if either $a,b,$ or $c$ is negative, or $y$ is negative.

I've been using this recursion, along with induction, to calculate the distribution for $R_{r,b,w}$, for various values of $r$, $b$, and $w$.

For example, to calculate the distribution for $R_{r,b,2}$, for all possible $r$ and $b$, I first calculated it for the first few values of $r$ ($r=0,1,2$, for instance), made a guess for what it would be for general $r$, then used induction to prove it.

If anyone's curious, I have come up with: \begin{align} P(R_{a,b,1}=x) &= \frac{1}{a+1}\\ P(R_{a,b,2}=x) &= \frac{2(x+1)}{(a+1)(a+2)} \end{align} (The $x$ is understood to be from $0$ to $a$, since the probability is $0$ otherwise, and I didn't want to write the piecewise form.)

I also calculated the expectation of $R_{a,b,1}$ to be $a/2$, while the expectation of $R_{a,b,2}$ is $\frac{2}{3}a$.

1) Is there an easier or more elegant way of calculating the probability distributions of these random variables? It's a lot of (tedious) work calculating the different probabilities.

2) Is there an easier or more elegant way of calculating the expectation of these random variables?

3) Is there an easy way to prove (or disprove) that $R_{r,b,w}$ is independent of $b$ (in other words, that $R_{r,b,w} = R_{r,0,w}$ for all possible $r$ and $w$)?

Answer 1

Here is what I think. It is not a formal proof, but rather a way of thinking.

The outcome of the game has the following general format, if we have an odd number of white marbles.

$Player1: B_1WB_2W...B_{(\frac{w+1}{2})}W \\Player2: B_{(\frac{w+1}{2}+1)}WB_{(\frac{w+1}{2}+2)}W...B_wWB_{(w+1)}$

$W$ is a white marble and $B_i$ is a sequence of red and blue marbles. Note that $B_i$ can also be empty.

In the same way, if we have an even number of white marbles, the general form would be

$Player1: B_1WB_2W...B_{(\frac{w}{2})}WB_{(\frac{w}{2}+1)} \\Player2: B_{(\frac{w}{2}+2)}WB_{(\frac{w}{2}+3)}W...B_{w+1}W$

So, in each case, we have $w+1$ blank spaces $B_i$, to fill in with $r$ red marbles and $b$ blue marbles.

I assume that all the blank spaces $B_i$ are equal and there would be no preference for a marble to choose any of them. This implies that blue marbles have no effect on the distribution of red marbles. Therefore, I would totally ignore them. Having this in mind, we would like to see how many possible arrangement we can have for each case.

If we have $r$ red marbles, we can order them in $r!$ ways. Then take a particular order and find out the number of ways we can put it in blocks $B_1B_2...B_{(w+1)}$. This is a stars and bars problem. So, finally we have

${r!}{{r+w}\choose{r}}$

With a similar reasoning, to find $P(R_{r,b,w}=x)$, we choose $x$ marbles out of $r$ red marbles. Then, we firstly find all the possible permutations of the $x$ marbles and, finally, how to put them, with the same order, in $B_1B_2...B_{(\frac{w+1}{2})}$, for $w=odd$, or in $B_1B_2...B_{(\frac{w}{2}+1)}$, for $w=even$. The final result would be

for $w=even$

$\frac{{x}! {{r}\choose{x}} {{x+\frac{w}{2}}\choose{x}} {(r-x)!} {{r-x+\frac{w}{2}-1}\choose{r-x}}}{{r!}{{r+w}\choose{r}}}=\frac{{{x+\frac{w}{2}}\choose{x}}{{r-x+\frac{w}{2}-1}\choose{r-x}}}{{{r+w}\choose{r}}}$

For $w=odd$, the same process is done, although I have not simplified the fraction.

$\frac{{x}! {{r}\choose{x}} {{x+\frac{w-1}{2}}\choose{x}} {(r-x)!} {{r-x+\frac{w-1}{2}}\choose{r-x}}}{{r!}{{r+w}\choose{r}}}$