What is the expected amount of time it takes to first observe a triple for multiple dice rolls? [duplicate]

Question

Assuming we have a fair die, we toss a die multiple times. Also assuming that a triple is defined as when we have three rolls in a roll that result in the same number, and that the rolls are independent, what would be the expected time until a first triple is observed? The answer I have in a book is rather heuristic. It briefly states that if you take 1/36 as the probability of observing a triple at time t>2, then the formula for the time until the first triple should be: $36 = \frac{1}{6}+\frac{5}{6}X$, where $X =$ first time a triple is observed.

I was wondering if there was a more formalistic way of coming up with the answer. Thank you!

Answer 1

Here, $X(t):$number of same outcomes, There will be 4 States $S:{\{0,1,2,3\}}$ But $0$ is reflecting state with probability 1.

from, state $1$, after a roll there is probability $\frac{5}{6}$ that come again in State $1$, and $\frac{1}{6}$ to go to State $2$.

from, state $2$, after a roll there is probability $\frac{5}{6}$ that goes down in State $1$, and $\frac{1}{6}$ to go to State $3$.

Now, Let $e_s:$expected rolls to get triple from state 's'

So, $e_0=1+e_1$ and $e_1=1+\frac{5}{6}e_1+\frac{1}{6}e_2$ and $e_2=1+\frac{5}{6}e_1+\frac{1}{6}e_3$ and $e_3=0$

after solving this you'll get $e_0=43$(Expected time/rolls to get first triple)