# 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?

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

1. $(0, 0, 0)$ (the starting state)
2. $(0, 1, 0)$
3. $(0, 0, 1)$
4. $(0, 1, 1)$
5. $(0, 0, 2)$
6. $(0, 1, 2)$
7. $(1, *, *)$ (win state for $R$)
8. $(*, 2, *)$ (win state for $Y$)
9. $(*, *, 3)$ (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).

• @RobertZ Thanks, yes, I noticed this myself too. Of course there are $6$ beads in total in the bag, not $9$. Commented Feb 13, 2018 at 17:18
• Well done. Great answer (+1)! Commented Feb 13, 2018 at 17:20
• @ploosu2 This is a really cool way of solving it that I never expected. Commented Feb 13, 2018 at 19:35

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 $\frac18 + 3\cdot \frac1{24} = \frac14 = \frac{9}{36}$