Probability Problem Solving
You are almost correct. Just swap the probabilities of Blue and Yellow (and therefore the probability of Yellow is LESS than the probability of Blue).
Blue wins if and only if the sequence of picked beads is one of the following: $BBB$, $YBBB$, $BYBB$, or $BBYB$. So the probability that Blue wins is $$\left(1+\frac{3}{3}\right)\cdot\frac{1}{2^3}=\frac{1}{4}.$$
Yellow wins if and only if the sequence of picked beads is one of the following: $YY$, $YBY$, $BYY$, $YBBY$, $BYBY$, or $BBYY$. So the probability that Yellow wins is $$\left(1+\frac{2}{2}+\frac{3}{2^2}\right)\cdot \frac{1}{3^2}=\frac{11}{36}.$$
Finally Red wins with probability $$1-\frac{1}{4}-\frac{11}{36}=\frac{4}{9}.$$ As a double-check, we have that Red wins if and only if the sequence of picked beads is one of the following: $R$, $YR$, $BR$, $YBR$, $BYR$, $BBR$, $YBBR$, $BYBR$, or $BBYR$. So the probability that Red wins is $$\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{2}{2\cdot 3}+\frac{3}{2^2\cdot 3}\right)\cdot\frac{1}{6}=\frac{4}{9}.$$
This can be modeled as the Markov chain with the states storing the amounts of steps each color has taken. Write these as tuples $(R, Y, B)$. We don't need all possible states, but rather have three states $(1,*, *), (*,2,*)$ and $(*, *, 3)$ that will be absorbing (once they are reached, the race is over and the winner is the one "whose state" we reached. All the states are
- $(0, 0, 0)$ (the starting state)
- $(0, 1, 0)$
- $(0, 0, 1)$
- $(0, 1, 1)$
- $(0, 0, 2)$
- $(0, 1, 2)$
- $(1, *, *)$ (win state for $R$)
- $(*, 2, *)$ (win state for $Y$)
- $(*, *, 3)$ (win state for $B$)
The transition probabilities are $\frac{1}{6}$ for incrementing the $R$-value, $\frac{2}{6}$ for incrementing the $Y$-value and $\frac{3}{6}$ for incrementing the $B$-value. The absorbing states have self-transition with probability $1$.
The question can then be solved with the theory of finite absorbing Markov chains by forming the Fundamental Matrix and calculating the matrix of Absorption Probabilities. For reference see Markov Chains and the section 11.2 on absorbing chains.
EDIT:
The transition matrix is (the states indexed in the given order)
$$\left[ \begin{array}{cccccc|ccc} 0 & \frac{2}{6}& \frac{2}{6} & 0 & 0 & 0 & \frac{1}{6} & 0 & 0 \\ 0 & 0 & 0 & \frac{3}{6} & 0 & 0 & \frac{1}{6} & \frac{2}{6} & 0 \\ 0 & 0 & 0 & \frac{2}{6} & \frac{3}{6} & 0 & \frac{1}{6} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{3}{6} & \frac{1}{6} & \frac{2}{6} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{2}{6} & \frac{1}{6} & 0 & \frac{3}{6} \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{2}{6} & \frac{3}{6} \\ \hline \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] $$
The blocks are $Q$ on the upper left, $R$ on the upper right. Calculate the fundamental matrix
$$N=(I-Q)^{-1}$$
And then the probabilities of the chain being absorbed to a particular absorbing state are given by the matrix
$$B=NR.$$
The chain starts at the state indexed $1$ so the first row of $B$ will give the probabilities.
Calculating (with computer) I get:
$$N = \frac{1}{12} \begin{pmatrix} 12 & 4 & 6 & 4 & 3 & 3 \\ 0 & 12 & 0 & 6 & 0 & 3 \\ 0 & 0 & 12 & 4 & 6 & 4 \\ 0 & 0 & 0 & 12 & 0 & 6 \\ 0 & 0 & 0 & 0 & 12 & 4 \\ 0 & 0 & 0 & 0 & 0 & 12 \\ \end{pmatrix} $$
And $$ R = \begin{pmatrix} \frac{1}{6} & 0 & 0 \\ \frac{1}{6} & \frac{2}{6} & 0 \\ \frac{1}{6} & 0 & 0 \\ \frac{1}{6} & \frac{2}{6} & 0 \\ \frac{1}{6} & 0 & \frac{3}{6} \\ \frac{1}{6} & \frac{2}{6} & \frac{3}{6} \\ \end{pmatrix} $$
So the first row of $B$ is
$$ \begin{pmatrix} \frac{4}{9} & \frac{11}{36} & \frac{1}{4} \end{pmatrix} $$
(EDIT2: I had $9$ in the denominator before where there should be $6$. Now it's fixed).