How do I calculate the average expected value of the dice bowl game? I am trying to better understand the "dice bowl game" from Goldratt's The Goal.
In the book, five kids (named Andy, Ben, Chuck, Dave and Evan) each start with an empty bowl, and a 6-sided die. There is a box of matches next to Andy.
For 10 rounds, each kid rolls the die, and tries to take that number of matches from the previous bowl (or matchbox), but they cannot take more matches than the bowl has. Andy will always take the number he rolls since the matchbox is always full. (There is no choice involved - the kids take the number they roll, limited by the number of matches in the bowl they take from.)
An example round would look like this:
A B C D E
6         (Andy rolls 6, takes from matchbox)
2 4       (Ben rolls 4, takes 4 from Andy)
2 0 4     (Chuck rolls 5, takes 4 from Ben)
2 0 1 3   (Dave rolls 3, takes 3 from Chuck)
2 0 1 2 1 (Evan rolls 1, takes 1 from Dave)

The second round would look something like this:
A B C D E
2 0 1 2 1 (Starting position from the previous round)
4 0 1 2 1 (Andy rolls 2, takes 2 from matchbox)
1 3 1 2 1 (Ben rolls 3, takes 3 from Andy)
1 0 4 2 1 (Chuck rolls 5, takes 3 from Ben)
1 0 1 5 1 (Dave rolls 3, takes 3 from Chuck)
1 0 1 1 5 (Evan rolls 4, takes 4 from Dave)

Note that Chuck rolled a 5 but could only take 4 from Ben, since Ben's bowl only contained 4.
I'd like to understand how I can calculate the expected value of Evan's bowl after 10 rounds.
The idea here is to gain a deeper understanding of variability on the throughput of a factory process.
I've simulated the problem, but I'd like to understand how to solve it without simulation.
I have high school math, and have dabbled in discrete math and probability, but my understanding is fairly elementary.
UPDATE: So this was not clear from the initial statement of the question, but the bowls don't empty between rounds. Players will also try to take the number they roll, limited only by the matches available.

Let $R_i, i = 1, 2, \ldots n$ be the die value of the $i$-th kid, where $n$ is the total number of kids participating in the game. Assuming the die are fair, we have $R_i$ are i.i.d. discrete uniform random variables on $\{1, 2, 3, 4, 5, 6\}$. Let $T_i$ be the number of matches taken from the previous bowl (or matchbox) by the $i$-th kid. Then by the rules of the game,
$$T_1=R_1, T_i=\min\{T_{i−1}, R_i\},i = 2, 3, \ldots n $$
The number of matches remain in the $i$-th bowl after being taken by the next kid is 
$$T_i − T_{i+1},i = 1, 2, \ldots n−1$$ 
and since there is no kid taking from the last bowl, all $T_n$ matches remain in the $n$-th bowl. Now you are interested in $E[T_n]$, the expected number of matches that the last kid have (in $1$ round). If you play $m$ rounds, you can just multiply by $m$ as the expectation is linear.
If we further investigate into those $T_i$, it is not hard to check that
$$ T_2 = \min\{T_1, R_2\} = \min\{R_1, R_2\}$$
$$ T_3 = \min\{T_2, R_3\} = \min\{\min\{R_1, R_2\}, R_3\} = \min\{R_1, R_2, R_3\}$$
and so on. So inductively we have
$$ T_i = \min\{R_1, R_2, \ldots, R_i\}, i = 1, 2, \ldots n $$
i.e. the running minimum of the previous rolls. You can verify this with the example you made, where the roll result $R_i$ are $(6, 4, 5, 3, 1)$ and the number of matches taken $T_i$ are $(6, 4, 4, 3, 1)$.
With all the above assumption, we can compute the survival function of $T_n$:
$$ \begin{align} \Pr\{T_n > t\} &= \Pr\{\min\{R_1, R_2, \ldots R_n\} > t\} \\
&= \Pr\left\{\bigcap_{i=1}^n R_i > t\right\} \\
&= \prod_{i=1}^n \Pr\{R_i > t\} \\
&= \begin{cases} 
1 & \text{when} & t < 1 \\
\displaystyle \left(1 - \frac {\lfloor t \rfloor} {6}\right)^n & \text{when} & 1 \leq t < 6 \\
0 & \text{when} & t \geq 6
\end{cases}
\end{align}$$
Since $T_n$ is a positive, discrete random variable, we can make use of the survival function to compute the expectation:
$$ E[T_n] = \sum_{t=0}^5 \Pr\{T_n > t\} 
= \sum_{t=0}^5\left(1 - \frac {t} {6}\right)^n
= \frac {1^n + 2^n + 3^n + 4^n + 5^n + 6^n} {6^n} $$
and multiply by the number of rounds $m$.
From the nature of minimum, when the number of players increase, the number of matches expected decrease and approaching $1$.
