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Starting with the number 0, Casey performs an infinite sequence of moves as follows: he chooses a number from {1, 2} at random (each with probability $\frac{1}{2}$) and adds it to the current number. Let $p_m$ be the probability that Casey ever reaches the number m. Find $p_{20}$ − $p_{15}$. Answer: $\frac{11}{2^{20}}$

This problem appeared in Harvard-MIT math tournament. The solution for this is posted at this website, http://hmmt.mit.edu/static/archive/february/solutions/2015/combo.pdf. The solution given in this problme does not reach my stone head. But I solved it in another way that is very primitive. Could someone let me know how this recurrence relation is obtained in this solution.

Satish Ramanathan
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In order for Casey to fail to reach $n$, he must reach $n-1$ and then pick $2$. The probability that he reaches $n-1$ is $p_{n-1}$, and the probability that he picks $2$ on his next step is $\frac12$, so the probability that he fails to reach $n$ is $\frac12p_{n-1}$. The probability of the complementary event that he does reach $n$ must therefore be $1-\frac12p_{n-1}$, so we have the recurrence $p_n=1-\frac12p_{n-1}$.

The rest of the posted answer is simply one way of solving this recurrence to get a closed form. Subtracting $\frac23$ from both sides, we have

$$p_n-\frac23=1-\frac12p_{n-1}-\frac23=\frac13-\frac12p_{n-1}=-\frac12\left(p_{n-1}-\frac23\right)\;.\tag{1}$$

Now let $x_n=p_n-\frac23$; $(1)$ shows that $x_n$ satisfies the recurrence $x_n=-\frac12x_{n-1}$. Thus, $x_1=-\frac12x_0$, $x_2=\left(-\frac12\right)^2x_0$, and in general $x_n=\left(-\frac12\right)^nx_0$: at each step you’re simply multiplying by $-\frac12$. Moreover, $x_0=p_0-\frac23=1-\frac23=\frac13$, so

$$x_n=\left(-\frac12\right)^n\cdot\frac13=\frac{(-1)^n}{3\cdot2^n}\;.$$

And $p_n=x_n+\frac23$, so we have finally

$$p_n=\frac23+\frac{(-1)^n}{3\cdot2^n}\;.$$

By the way, that $-\frac23$ doesn’t really appear by magic; this answer explains where it came from.

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Brian M. Scott
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  • Is that all it is? Some wording such as yours in the solution would have been very helpful. I have always found such solution to these contests are so cryptic and eat away a lot of meat puts students like me completely baffled. Thanks anyway. You must have been a great teacher during your times. – Satish Ramanathan Apr 01 '15 at 04:28
  • @satish: You’re welcome, and thank you. (Sometimes I find published solutions pretty cryptic. But it’s good practice for reading published papers, which tend to leave a lot to the reader.) – Brian M. Scott Apr 01 '15 at 04:31