How many rolls required to have three consecutive any number

Question

Hi Need Help to solve this problem

How many rolls required for a dice to get any number consecutively $3$ times? for example $111, 222, 333$ ??

I've seen some formula, but most of them looking for certain number instead of any.

please show me the logical structure step by step, so I could understand clearly.

thanks

Answer 1

Presumably, you want the expected number of rolls.

An easy approach is via recursion.

For $k\in\{0,1,2\}$, let $e_k$ be the expected number of rolls to get three consecutive equal values, assuming $k$ is the length of the maximum block of immediately preceding rolls with equal values.

Then we have the system \begin{align*} e_0&=1+e_1\\[4pt] e_1&=1+{\small{\frac{5}{6}}}e_1+{\small{\frac{1}{6}}}e_2\\[4pt] e_2&=1+{\small{\frac{5}{6}}}e_1\\[4pt] \end{align*} of $3$ linear equations in $3$ unknowns.

Solving the system yields $e_0=43$.

To explain the recursion . . .

• The initial roll yields $k=1$ (i.e., $1$ is the length of the maximum block of immediately preceding rolls with equal values).
• After the initial roll, if a roll matches the prior roll, $k$ is incremented by $1$, else $k$ is reset to $1$.
• For each equation, the added constant $1$ on the $\text{RHS}$ counts the roll that just occurred.