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Probability Level 5

Suppose you have a biased coin which comes up heads with probability $\frac{1}{3}$ and tails with probability $\frac{2}{3}$ . You then begin to toss the coin repeatedly, with heads worth 1 point and tails worth 2 points, and add up the points as you go along.

Let $p(n)$ be the probability that at some stage you have accumulated precisely $n$ points. (For example, $p(1) = \frac{1}{3}, p(2) = \frac{2}{3} + (\frac{1}{3})^{2} = \frac{7}{9}$ , etc..)

If $\displaystyle\lim_{n\rightarrow \infty} p(n) = \dfrac{a}{b}$ , where $a$ and $b$ are positive coprime integers. Find $a + b$ .

The answer is 8.

1 solution

Brian Charlesworth
Aug 25, 2014

A quick outline of one solution method....

We have that $p(0) = 1, p(1) = \frac{1}{3}$ , and

$p(n) = (\frac{1}{3})p(n-1) + (\frac{2}{3})p(n-2)$ for $n \ge 2$ .

Assuming that $p(n) = r^{n}$ for non-zero $r$ , we have that

$r^{n} = (\frac{1}{3})r^{n-1} + (\frac{2}{3})r^{n-2} \Longrightarrow r = 1, -\frac{2}{3}$ .

So $p(n) = A*(1)^{n} + B*(-\frac{2}{3})^{n}$ for some constants $A, B$ . Plugging in the initial conditions $p(0) = 1, p(1) = \frac{1}{3}$ , we find that $A = \frac{3}{5}$ and $B = \frac{2}{5}$ .

Thus $p(n) = \frac{3}{5} + (\frac{2}{5})*(-\frac{2}{3})^{n}$ , which as $n \rightarrow \infty$ goes to $\frac{3}{5}$ .

Thus $a = 3, b = 5$ and $a + b = \boxed{8}$ .

Feeling Lucky?

The answer is 8.

1 solution

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