Discussion
Two identical random walkers have started walking randomly from the origin on a 1 d lane. They are taking steps simultaneously and of equal length l. They travelled for 10 steps. The probability of meeting together at 4th step is ......
Answer: C
The keyed probability is \(\left\{\frac{10!}{4!6!}\left[\frac{1}{2}\right]^{10}\right\}^{2}\).
Each walker takes 10 independent steps, each left or right with probability 1/2. The probability that one walker takes 4 steps one way and 6 the other is binomial:
\[ \begin{aligned} p &= \frac{10!}{4!\,6!}\Big(\frac{1}{2}\Big)^{10} \\ &= \frac{210}{1024} \end{aligned} \]
The walkers are independent, so the probability that both make that same choice, and so stand together on the same site after 10 steps, is the product \(p \times p = p^{2}\).
A single factor \(p\) is the probability for one walker only, and a cube would apply to three walkers. Squaring only the binomial coefficient while squaring the whole bracket again, as in the last option, double-counts the factors.
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