Probability of Scoring Exactly 30 Points in Coin Tosses

To find the probability that the man scores exactly $30$ points, we can break the problem down into recursive steps by scaling the points.

Since each head ($H$) awards $10$ points and each tail ($T$) awards $5$ points, all possible cumulative scores are multiples of $5$. We can define $1$ unit of score as $5$ points.

Thus, a head ($H$) corresponds to an increase of $2$ units, and a tail ($T$) corresponds to an increase of $1$ unit. Scoring $30$ points corresponds to reaching a target of: $N = \frac{30}{5} = 6 \text{ units}$

Let $P_k$ denote the probability that the man scores exactly $5k$ points (or $k$ units) at some stage during the repeated coin tosses.

1. Recurrence Relation

To reach a score of $k$ units:

• The man could have reached $k-1$ units and then thrown a Tail (probability $\frac{1}{2}$).
• Or, he could have reached $k-2$ units and then thrown a Head (probability $\frac{1}{2}$).

This gives us the linear recurrence relation for $k \ge 2$: $P_k = \frac{1}{2} P_{k-1} + \frac{1}{2} P_{k-2}$

2. Initial Conditions

• At the start, the score is $0$ units, so $P_0 = 1$.
• To score $1$ unit ($5$ points), the first toss must be a Tail: $P_1 = \frac{1}{2}$

3. Solving the Recurrence Relation

The characteristic equation for the recurrence relation $P_k - \frac{1}{2}P_{k-1} - \frac{1}{2}P_{k-2} = 0$ is: $r^2 - \frac{1}{2}r - \frac{1}{2} = 0 \implies (r - 1)\left(r + \frac{1}{2}\right) = 0$

The roots are $r_1 = 1$ and $r_2 = -\frac{1}{2}$. Thus, the general solution is: $P_k = A(1)^k + B\left(-\frac{1}{2}\right)^k$

Using the initial conditions:

• For $k = 0$: $A + B = 1$
• For $k = 1$: $A - \frac{1}{2}B = \frac{1}{2}$

Subtracting the second equation from the first: $\frac{3}{2}B = \frac{1}{2} \implies B = \frac{1}{3}$ $A = 1 - \frac{1}{3} = \frac{2}{3}$

Therefore, the formula for $P_k$ is: $P_k = \frac{2}{3} + \frac{1}{3}\left(-\frac{1}{2}\right)^k$

4. Calculating $P_6$ (Probability of Scoring 30 Points)

For $k = 6$: $P_6 = \frac{2}{3} + \frac{1}{3}\left(-\frac{1}{2}\right)^6$ $P_6 = \frac{2}{3} + \frac{1}{3} \cdot \frac{1}{64}$ $P_6 = \frac{2}{3} + \frac{1}{192}$ $P_6 = \frac{128 + 1}{192} = \frac{129}{192}$

Simplifying the fraction by dividing both numerator and denominator by $3$: $P_6 = \frac{43}{64}$

Here, $m = 43$ and $n = 64$. Since $43$ is prime and does not divide $64$, we have $\gcd(43, 64) = 1$.

Finally, we calculate $m + n$: $m + n = 43 + 64 = 107$