Probability Problem Solving Red, yellow and blue counters are placed on a board as shown,
and they ‘race’ to the finish (F) by moving up one square at a
time. The moves are determined by picking a bead at random
from a bag containing one red bead, two yellow beads and
three blue beads. After the colour of the bead which has been
drawn is noted, the bead is returned to the bag before the next
bead is picked. The race is over as soon as one of the counters
lands on the square marked F. Find the probability of winning
for each of the counters.

So far, I have only been able to figure out the basics of probability of red>probability of yellow>probability of blue. My estimate is that:
Probability of red winning = 4/9
Probability of yellow winning 1/4
Probability of blue winning = 11/36
What do you guys think?
 A: 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
$$\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{2}{2\cdot 3}+\frac{3}{2^2\cdot 3}\right)\cdot\frac{1}{6}=\frac{4}{9}.$$
A: Here's a graph of all the possible sequences with their probabilities, created using a small Python script and GraphViz.

We can easily determine the winning probabilities by summing the values in the terminal nodes. 
Thus for red to win we get $\frac16 + \frac1{18} + \frac1{12} + 2\cdot\frac1{36} + \frac1{24} + 3\cdot \frac1{72} = \frac49 = \frac{16}{36}$
For yellow, we get $\frac19 + 2\cdot \frac1{18} + 3\cdot \frac1{36} =\frac{11}{36}$
For blue, we get $\frac18 + 3\cdot \frac1{24} = \frac14 = \frac{9}{36}$
