A Coin Toss Problem

Let $X_n$ , $n \in N_0$ be the number of heads minus the number of tails obtained so far.

Then, $[X_n]_{n\in N_0}$ is a simple symmetric random walk, and we stop tossing the coin when $X$ hits 3 for the first time. This will happen during the first 10 tosses, if and only if $M_{10} > 3$ where $M_n$ denotes the (running) maximum of X .

According to the reflection principle ,

$\begin{aligned} P[M_{10} > 3] & =P[X_{10} > 3] + P[X_{10} > 4] \\ & = 2(P[X_{10} = 4] + P[X_{10} = 6] + P[X_{10} = 10])\\ &= 2^{-9} [\binom{10}{3} + \binom{10}{2} + \binom{10}{1} + \binom{10}{0}]\\ &= \frac{11}{32} \end{aligned}$ .

Hence our answer being $32 - 11 = 21$ .

NOTE : For those who don't know about Random Walks and the Reflection Principle , I would suggest you to read up those topics .