# Expectation problem with geometric random variable?

If I have a stream of integers from 0 to 9(each with equal probability and repetition is allowed) that are spewed out until I get the first instance of 4-5 next to each other in this order, how do I find the expected value of the number of integers that will be printed?

I set up a random variable, $$X$$, that is the number of integers printed, and an indicator random variable, $$X_i$$, that is 1 if the $$I^{th}$$ number does not complete the first instance of 4-5, and 0 otherwise. I think that the probability that $$X_i$$ = 1 is calculated as $$1 - (1/10)^2$$, so would the expected value of $$X$$ just be $$1/p$$?

I don't understand your computation. It would appear that the expected value of the $$i^{th}$$ term would be a function of $$i$$, since you would need to ensure that the game wasn't over before the $$i^{th}$$ move. Anyway, here is another approach:
There are three states here...$$\emptyset,4,\{4,5\}$$ according to how much of the $$4,5$$ block occurs at the end of the running string. Letting $$E_S$$ deonte the expected number it will take to finish assuming you are starting in state $$S$$, we note that the answer we want is $$E=E_{\emptyset}$$. Of course $$E_{\{4,5\}}=0$$.
Noting that $$\emptyset$$ can only transition to $$\emptyset$$ (probability .9) or $$4$$ (probability .1) we compute $$E_{\emptyset}=\frac 9{10}\times (E_{\emptyset}+1)+\frac 1{10}\times (E_4+1)\implies E_{\emptyset}=E_4+10$$
Similarly $$E_4=\frac 1{10}\times 1+\frac 1{10}(E_4+1)+\frac 8{10}\times (E_{\emptyset}+1)\implies 9E_4=8E_{\emptyset}+10$$
This system is easy to solve and we get $$\boxed {E_{\emptyset}=100}\quad \&\quad \boxed {E_4=90}$$