I like to add a simpler argument for the average waiting time of 42 throws from the last blog entry.

We throw until a 6 and once more. On average, we take 7 throws for this. The extra number after the 6 is either a 6 or not.  So we need to repeat the experiment 6 times for an average time of 42 throws.

This argument can easily be extended to three successive throws of 6. We throw until we get two times a 6 and once more. So we are at

$$6 \cdot (42+1) = 258$$

That was the result of the previous blog entry. The average number of throws for k times 6 thus satisfies

$$N_{k+1} = 6 (N_k + 1), \quad N_0=0$$

This is an inhomogeneous difference equation. The solution is

$$N_k = \dfrac{6}{5} \left(6^k-1\right)$$

Clearly, these arguments require a precise imagination of the virtual experiment necessary to get the goal. We required just one result: If we have to repeat a random event with a chance p for success, the average waiting time till success is 1/p.

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