Expected value of the recursive algorithm 
Sara and Macey want to play the game of Truth or Dare. They use the following recursive algorithm to decide who goes first:  

$\textbf{Algorithm} \hspace{2mm} GoesFirst(k):$
    $// \hspace{2mm}k ≥ 1$, the die is fair, and all rolls are independent
    Shelly rolls the die, let $s$ be the result;
    Macey rolls the die, let $m$ be the result;
    $  
\textbf{if } s > m \\
\textbf{then} \text{ Sara goes first} \\ 
\quad \quad return \hspace{2mm} k \\
\textbf{end if } \\
\textbf{if } s < n \\
\textbf{then} \text{ Macey goes first} \\ 
\quad \quad return \hspace{2mm} k \\
\textbf{end if } \\
\textbf{if } s = m \\
\textbf{then} \text{ GoesFirst(k + 1)} \\ 
\textbf{end if } \\
$

The ladies run algorithm $GoesFirst(1)$, i.e., with $k = 1$. Define the random variable $X$ to be the value of the output of this algorithm.
  What is the expected value $\mathbb{E}(X)$ of the random variable $X$?
(a) $3/2$
  (b) $5/4$
  (c)  $5/6$
(d) $6/5 \hspace{2mm} \text{ This is the answer} $ 

Why is this the answer? I know the formula for the expected value is $\mathbb{E}(X) = \sum_{}^{} k \cdot Pr(X = k)$.  
From class I learned that the Expected Value of a Geometric Distribution is $1/p$ where $p$ is the probability of success.  
Is $5/6$ is the probability of success? (ie. $5/6$ times we will get an answer as to who will go first).  
If so, then we repeat this algorithm until we reach a success, which I assume follows the Geometric Distribution:  
$\frac{1}{5/6} = 6/5$ which is the answer. 
I tried doing the sum but realized it was pointless.     
$\mathbb{E}(X) = 1 \cdot 5/6 + 2 \cdot 1/6$ ... 
I think my first solution is correct. If anyone can formalize my thinking that would be great.
Thanks.
 A: The algorithm returns the number of the first roll of the die on which they get a different number.  As you say, the probability of success (the probability that they roll different numbers) is $5/6.$  The expected number of trials until the first success is $1/p$ as you know.  The answer is $6/5$.  
It isn't "pointless" to compute the sum.  That's how the expected value of $1/p$ is arrived at in the first place.  However the second term should be $$2 \cdot 1/6\cdot 5/6$$ It's $2$ times the probability that they get the same number on the first roll, times the probability that they get different numbers on the second.
A: There are $36$ possible outcomes of $(s,m)$ drawings, of which $6$ make a tie, with probability $\dfrac16$.
Then we have the recurrence
$$p_{k+1}=\frac{p_k}6$$
because $k$ is incremented only in case of a tie, and we indeed have a geometric law with parameter $1-p=\dfrac56$. The expectation is known to be the inverse of the parameter.

By direction computation, the expectation is
$$E=\frac{1p+2p^2+3p^3+\cdots}{p+p^2+p^3+\cdots}=1+\frac{1p^2+2p^3+\cdots}{p+p^2+p^3+\cdots}=1+p\frac{1p+2p^2+\cdots}{p+p^2+p^3+\cdots}=1+pE$$ and we draw
$$E=\frac65.$$
A: Your output X follows a 
Negative binomial distribution. X is the number of trial before the first success. 
Here the success S is when the 2 dices roll different number. Let's say we have rolled the first dice what is the probability that the second dice gives a different number? The probability of success is $p=P(B=1)=\frac 56$  
Then we repeat this experiment until success and X=k means $B_1=0$ and $B_2=0$ and $\dots$  $B_k=1$.
The expectation of the negative binomial is $E(X)=\frac 1p$. A good explanation is available in Wikipedia. A formal proof is the following:
$E(X)=E(X|B_1=0)P(B_1=0) + E(X|B_1=1)P(B_1=1) = (E(X)+1)(1-p) + p$
$E(X)=\frac 1p $
