The game of hand cricket. (If you could add a more suitable title, I'd be very grateful to you)
My friend gave me this question:

In the game of hand-cricket (hand baseball, if you like it :D) two players $X$ - the batsman and $Y$ - the bowler, randomly choose numbers from $1$ to $6$ (simultaneously). Now, if both put the same number, then player $X$ is declared out and the game ends (for $X$). If both put different numbers, then the score of $X$ is incremented by the number he chose. Before starting, the score of $X$ is $0$. If the game goes on like this, what is the most probable score of $X$?

For example, if $X$ and $Y$ start by choosing $1$ and $6$, then the score of $X$ is 1. If $X$ puts $3$ and $Y$ puts $1$, then the score of $X$ becomes $4$. If $X$ and $Y$ both put $5$, then $X$ is out and the final score is $4$.
The question was just to write a program to find the average score - nothing big at all. But I wondered if there should be an expected score.
The Question:
Now, my question is why should there be a 'most probable score' at all? To investigate this, I just wrote a small program and it turns out that the average score is about $12$. Then, I changed the parameter $6$ to various numbers and I got various answers. If the parameter is increased, I get a greater average score and on reducing it, I get a lower average score. This seems quite sensible as by increasing the parameter, we are effectively reducing the chances of 'collision'. But is there any mathematical way to estimate the average score given a parameter $K$ such that the numbers chosen are between $1$ and $K$?
Please note that I am not familiar with 'advanced' probability and such stuff, so if possible, give the answer in an intuitive way! Maybe I am asking too much. In that case, you may just tell that it is way above my current 'level'.
 A: Let $a$ be the expected score. Let's start the game. With probability $\frac{1}{6}$, the game will be over in one round, with score $0$. With probability $\frac{5}{6}$, we will get a score with expected value $\frac{7}{2}$, and get to play again, with expected additional score $a$. Thus
$$a=\frac{1}{6}(0)+\frac{5}{6}\left(\frac{7}{2}+a\right).$$ 
Solve for $a$. We get $a=\frac{35}{2}$.
A: Firstly, I assume that instead of "most probable" score, we're interested in the expected (average) score, as that is what you talk about in the rest of your question.
The probability of the batsman (or batter) getting "out" is $\frac16$. So, in expectation, it will take $6$ rounds before the batsman gets out, i.e., the batsman gets (in expectation) $5$ rounds of "not-out" batting. In each round, the expected score is the average of the numbers $1$ to $6$ (as all of them are equally likely), which is $\frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$. The expected final score is therefore $3.5 \times 5 = 17.5$.
To make the above argument rigorous: let $X_i$ be the score in the $i$th round, let $N$ be the (random) number of rounds played, and note that the total score is $X = X_1 + X_2 + \dots + X_{N-1}$. As the random numbers $X_i$ are independent and identically distributed, by Wald's equation, 
$$E[X] = E[X_1] E[N-1] = \frac{21}{6}(6-1) = \frac{35}{2} = 17.5.$$
A: Here is a more tedious approach:
Let $O$ denote an out and $\overline{O}$ a 'non-out'. Then the events are $\overline{O}^k O$. The probability of $O$ is $x=\frac{1}{6}$, and the 'payout' for the event $\overline{O}^k O$ is $\frac{7}{2}k$.
You want to compute $\sum_{k=0}^\infty \frac{7}{2}k (1-x)^k x = \frac{7}{2} x (1-x) \sum_{k=1}^\infty k (1-x)^k = \frac{7}{2} x (1-x) \frac{1}{x^2} = \frac{7}{2} \frac{1-x}{x}$,
which has value $\frac{35}{2}$.
Running the following printed a value of $17.47003$.
import random

def die():
    return random.randint(1,6)

def run():
    payout = 0.0
    while True:
        if die()==1:
            return payout
        payout = payout + die()

n = 100000
total = 0.0
for i in range(n):
    total = total + run()

print("average=%s" % (total/n))

