# Waiting time for successive occurrences of a result, when rolling a die

In consecutive throws of an ordinary die, which of the following two possibilities is more likely to happen first:

a) Two successive occurrences of 5 or

b) Three successive appearances of numbers divisible by 3?

I thought that "time to event" random variable follows a geometric distribution. In the first case the probability of the event would be $p_1=1/36$ so the waiting time would be $E(X_1)=\frac{1-p_1}{p_1}=35$, plus 2 (for the successive occurrences of 5) = 37.

Accordingly the probability of the second event would be $p_2=\frac{2^3}{6^3}$ and the waiting time for the second event would be $E(X_2)=\frac{1-p_2}{p_2}=26$, plus 3 = 29.

As I was not sure about the validity of the above considerations, I decided to try some monte carlo simulations (the code can be found here: http://ideone.com/TbLdDe). According to the results, the second event will happen first, indeed. But the expected values are larger than the ones calculated before (About 42 and 39, accordingly)

What is the right way to calculate the waiting time?

• "I thought that "time to event" random variable follows a geometric distribution." -- you have overgeneralized. Some 'number of trials to event A' variables have a geometric distribution, but many other distributions occur (e.g. 'number of trials to the second success in Bernoulli trials' is not geometric, but negative binomial). When you seek several of the same event in a row you generally don't have a geometric. – Glen_b Mar 27 '14 at 9:26