We toss a symmetrical cube until we get the first sequence of two identical results in two consecutive tosses. Calculate the expected value of the number of tosses made
I don't have an idea for this.
I just know that
$1\cdot 0\%+2\cdot \left(\frac{1}{6}\right)^2+...$
Because it's 0 percent that you get two consecutive numbers when rolling a dice once, and if twice it's pretty clear from where this did came form. But I don't know the probability for 3, I'd guess that it's $\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)$ because in the first run we only have 5 options so that the last two would be two consecutive of the same number.
But what about for 4? Basically $\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\left(\frac{4}{6}\right)$ because your first two have to be different so you can't pick any (that's why 5/6), but then for the second toss you only can have 4 out of 6, because the next shouldn't be the same one and the former shouldn't be same one
And so on
Is this the right way?
With a Markov chain: start in state $\emptyset$. The first toss doesn't matter, so w.p. $1$ you get to state $S_1$. Then, if you toss the same number, you are done. Otherwise you are back in state the same. The set of equations is just $$ m_{1,S} = (1-p) m_{1,S} + 1 $$