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

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!

• Set up a markov chain like any of the other similar problems on this site? – Neil G May 2 '15 at 21:44
• – Neil G May 2 '15 at 21:45
• What do you mean by "three rolls in a roll"? You say you have one die, and when you roll it ("a roll"), you only get one result, not three. Can you clarify your question to explain more precisely what you're asking? [edit: a thought occurs -- Do you maybe mean 'three rolls in a row'?] – Glen_b May 3 '15 at 1:44

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)