It's a toss-up ....

The number of sequences of $10$ tosses in which there is no run of $3$ or more consecutive heads or tails can be counted as follows.

We need to look at how many ways we can alternate groups of $1$ or $2$ each of heads or tails. For example, the sequence $HTTHHTHTTH$ can be represented by the ordering $(1,2,2,1,1,2,1)$ . Now this ordering also corresponds to the sequence $THHTTHTHHT$ , so we will have to multiply the number of distinct orderings we find by $2$ to get our final tally.

Now the distinct orderings are all the permutations of any of

(i) 10 1's, resulting in 1 possible permutation,

(ii) 8 1's and 1 2, resulting in $\frac{9!}{8!} = 9$ possible permutations,

(iii) 6 1's and 2 2's, resulting in $\frac{8!}{6!*2!} = 28$ possible permutations,

(iv) 4 1's and 3 2's, resulting in $\frac{7!}{4!*3!} = 35$ possible permutations,

(v) 2 1's and 4 2's, resulting in $\frac{6!}{4!*2!} = 15$ possible permutations,

(vi) 5 2's, resulting in 1 possible permutation.

These add to $89$ permutations, which we then need to multiply by $2$ to get a final tally of $178$ possible sequences of $10$ coin tosses in which there is no run of $3$ or more consecutive heads or tails.

Since there are $2^{10} = 1024$ possible sequences without restrictions, the probability of getting a winning sequence is $\frac{178}{1024} = \frac{89}{512}$ , and the probability of getting a losing sequence is $1 - \frac{89}{512} = \frac{423}{512}$ .

Thus the expected monetary return for one game played is

$5*\dfrac{89}{512} - 1*\dfrac{423}{512} = \dfrac{22}{512} = \dfrac{11}{256}$ .

The expected return from $100$ games played is then

$100 * \dfrac{11}{256} = \dfrac{275}{64}$ .

This means that $a = 275, b = 64$ and $a + b = \boxed{339}$ .

The single game can be modelled as a 3-state Markov Chain: let it be $\textbf{s}=(s_1,...,s_{10})$ , $s_i \in \{H,T\}$ , $i \in [1,10]$ the outcome of a single game. Let's define the states:

$S=\begin{cases} D, & \text{if} \ x_{i+1} \neq x_i, \ 2 \leq i \leq 9 \\ E_2, & \text{if} \ x_{i+1} = x_i, \ 2 \leq i \leq 9 \\ E_3, & \text{if} \ x_i=x_{i+1}=x_{i+2}, \ 2 \leq i \leq 8 \end{cases}$

the set of the possible three reachable states. In other terms, we are in $D$ when the previous outcome is different from the present one, $E_2$ or $E_3$ when the last, respectively, two or three outcomes are the same. Notice that the index $i$ starts from $2$ in $S$ . That's because the first coin toss outcome is not relevant, i.e. you can reach the state $D$ (if the second outcome is different from the first) or $E_2$ (if the second outcome is equal to the first) despite $s_1$ being head or tails.

Hence, our transition matrix is

$P=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 0 & 1 \end{pmatrix}$

and our initial state vector $\textbf{u}$ is

$\displaystyle \textbf{u}=\bigg(\frac{1}{2}, \frac{1}{2}, 0\bigg)$

If we define $\mathbb{P}(g)$ the probability of winning a single game, we have that

$\displaystyle \mathbb{P}(g)=1-\{\textbf{u} \cdot P^8\}_3=\frac{89}{512} \quad \clubsuit$

The expected winning $\mathbb{E}(g)$ after a game is

$\displaystyle \mathbb{E}(g)=5 \cdot \mathbb{P}(g) - (1-\mathbb{P}(g))= \frac{11}{256}$

By linearity of expected value, the expected winning after 100 games is simply

$\displaystyle 100\mathbb{E}(g)=\frac{275}{64}=\frac{p}{q} \\ p+q=\boxed{339}$

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$\clubsuit$ : the notation $\{\cdot\}_k$ means that we are considering the $k$ -th element of the vector.