Alfonzo is a kangaroo. Each second, he takes a 2-yard hop forward with probability 60% or a 1-yard hop backward with probability 40%.

Question

In 15 seconds, I believe that, in expectation, Alfonzo has hopped 15[(2*0.6) + (-1 * 0.4)] = 15*0.8 = 12 yards forward.

How long, in seconds, before Alfonzo takes a hop backwards?

Answer 1

The probability of Alfonzo taking a hop backwards for any given second is 0.4, so the chances of it taking a given time t is $0.4^{t}$.

I'm not sure if the average distance travelled in 15 seconds is relevant to your question, sorry if I've missed something.

Answer 2

The expected value is the sum of the product of each potential value times its probability. In this case, the probability of taking n seconds before hoping backwards is $0.4 * 0.6^{n-1}$ because you would go forward (probability 60%) for n-1 seconds before going backwards (40%).

$E[x]=\sum_{n=1}^{\infty}{\left(n*0.4*0.6^{n-1}\right)}=\frac{0.4}{(1-0.6)^2}=2.5$