There is a 6 sided die which is thrown repeatedly while summing the die values. If the sum exceeds 11, the sum is reset to 0, and throwing continues. If we get 11, we stop throwing. For example, the series of die throws 6, 6 (sum = 12 exceeds 11 --- reset sum to 0), 5, 6 (sum = 11, stop throwing) requires 4 die throws to get a sum of 11.
On average, how many die throws does it take to get a sum of 11?
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You should try setting this situation up as a Markov chain.
Your states can correspond to the total score achieved. Eg $x_8$ is "score $= 8$."
Your starting state is $x_0$.
The transition probabilities are given by the result of the die throw. For example, from $x_4$ you could go to $x_5$ with probability $\frac 16$, to $x_6$ with probability $\frac 16$, to $x_7$ with probability $\frac 16$ etc.
For larger scores there would be the chance of returning to zero score. For example, from $x_7$ you could go to $x_8$ with probability $\frac 16$, to $x_9$ with probability $\frac 16$, to $x_{10}$ with probability $\frac 16$, to $x_{11}$ with probability $\frac 16$ and to $x_0$ with probability $\frac 26$.
The expected time to reach a particular state is a straightforward procedure once you have the transition matrix set up.
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Q: A dice is thrown and the result is summed to the previous throw until the sum equals or exceeds $11$. Call each attempt to reach $11$ a run of dice. How many times can we expect to roll the dice until we roll a run which sums to exactly $11$?
The minimum length of a run is $2$: $(5, 6)$ or $(6,5)$. The maximum length of a run is $11$: Rolling $1$ eleven times in a row.
The amount of total runs possible which might sum to $11$ is $6^{11} + 6^{10} + \dots + 6^2$, exactly $435356460$ runs. This does not exclude permutations of runs whose sums are equivalent.
There is exactly $1$ run of length $11$ whose sum is $11$, the trivial $11$-tuple snake eyes. We must find the total number of runs of each length from $11$ down to $2$ (not excluding permutations whose sums are equal) and divide that by $435356460$ to find the probability that a run will sum to $11$. The next step will be calculating the average length of a run, which is $$\frac{\text{the length of each run multiplied by the number of runs of that length which sum to $11$, the sum of these numbers}}{\text{the total number of runs which sum to 11}}$$
This number, multiplied by the probability that a run will sum to $11$, will be the average number of times a dice must be thrown until a run which sums to $11$ is rolled.
Let $E(s)$ be the expected number of throws with current sum $s$.
$E(11) = 0$ by definition, and $E(s) = E(0)$ for $s>11$.
Then:
$$E(s) = 1 + \frac{1}{6}\sum_{k=1}^{6}E(s+k)$$
This yields a system of equations. Solve for $E(0)$.
For example:
$$E(7) = 1 + \frac{1}{6}(E(8) + E(9) + E(10) + E(11) + E(12) + E(13))$$
can be framed as
$$-6E(7) + E(8) + E(9) + E(10) + E(11) + E(12) + E(13) = -6$$
and then simplified to
$$-6E(7) + E(8) + E(9) + E(10) + E(11) + E(0) + E(0) = -6$$
or
$$2E(0) - 6E(7) + E(8) + E(9) + E(10) + E(11)= -6$$
which would correspond to the $E(7)$ row in the transition matrix.
The final matrix equation would be as follows:
$$\begin{bmatrix} -6 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & -6 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -6 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -6 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -6 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & -6 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & -6 & 1 & 1 & 1 & 1 & 1 \\ 2 & 0 & 0 & 0 & 0 & 0 & 0 & -6 & 1 & 1 & 1 & 1 \\ 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -6 & 1 & 1 & 1 \\ 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -6 & 1 & 1 \\ 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -6 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} E(0) \\ E(1) \\ E(2) \\ E(3) \\ E(4) \\ E(5) \\ E(6) \\ E(7) \\ E(8) \\ E(9)\\ E(10) \\ E(11) \end{bmatrix} = \begin{bmatrix} -6 \\ -6 \\ -6 \\ -6 \\ -6 \\ -6 \\ -6 \\ -6 \\ -6 \\ -6 \\ -6 \\ 0 \end{bmatrix}$$
You can use Gaussian elimination to solve the rest.