Rolling a $6$-sided dice indefinitely until lower than the previous throw

You will play a game with a fair 6-sided die.

You will throw the die and as long as the result of the throw is greater than or equal to the previous throw, you will continue throwing. If the throw is lower than the previous one, you will stop and get as many points as the sum of all throws, including the last one.

For example, if you get 2, 5, 5, and 3 as a result of 4 throws, the game will end with 15 points.

What is the expected value of the points you will get at the end of the game?

Consider the expected number of points you will obtain after rolling a particular number.

e.g. Suppose we have just rolled a 6. We need to roll another 6 to keep playing, otherwise we stop. Using conditional expectation, we can compute the number of future points we expect to obtain, $E_6$: $$E_6 = \frac{1}{6}(E_6 + 6) + \frac{5}{6}\times 3.$$

The first term corresponds to rolling a 6, and we are back in the same position as before, just with 6 extra points. The second term gives expected number of points obtained, given that we roll a lower number and stop playing. We solve this to obtain $$E_6 = \frac{21}{5}.$$

Now suppose that, in a new game, we have just rolled a 5. What is the expected number of points from this point, $E_5$? Using the same conditional expectation rules as before: $$E_5 = \frac{1}{6}(E_5+5) + \frac{1}{6}(E_6+6) + \frac{4}{6} \times \frac{5}{2}.$$ Since we know $E_6$ from above, we can now solve for $E_5$.

Repeat this for $E_4, \dots, E_1$, and we know the expected number of future points given the most recent roll number.

Finally, the expected number of points, $E$, will be the conditional sum of these values, i.e.

\begin{equation} E = \sum_{n=1}^6 \frac{1}{6}(E_n + n). \tag{1} \end{equation}

Edit: following an observation by Taner, I thought I'd add a few more lines.

After some generalizing and rearranging, we obtain $$E_n = \frac{1}{5}\left(\sum_{m=n+1}^6 E_m + 21\right),$$ and we can set up a recurrence relation for $E_n$ so we don't have to do this sum every time to compute it. We have

\begin{align} E_{n+1} &= \frac{1}{5}\left(\sum_{m=n+2}^6 E_m + 21\right) \\ &= \frac{1}{5}\left(\sum_{m=n+1}^6 E_m + 21 - E_{n+1}\right) \\ &= E_n - \frac{1}{5}E_{n+1}, \end{align}

which we rearrange to obtain $$E_{n+1} = \frac{5}{6}E_n,$$ which has solution $$E_n = C \left(\frac{5}{6}\right)^n,$$ where $C$ is some constant, for which we can solve by setting $n=6$ and using our value of $E_6$. We obtain (allowing for possible arithmetic errors made by me) $$E_n = \frac{21}{5}\left(\frac{5}{6}\right)^{n-6}.$$

We can substitute this into Equation (1) and use geometric and arithmetic sum formulae to obtain the final answer.

Edit 2: as Alex Zorn pointed out, we don't even need to do this geometric and arithmetic sum stuff in Equation (1). Note that the game is the same before and after rolling a 1, so this tells us straight away that $$E = E_1 = \frac{21}{5}\left(\frac{5}{6}\right)^{-5} = \frac{163296}{15625} \approx 10.45.$$

Here is an answer using generating functions where we demonstrate a variety of relevant techniques. Suppose the die has $n$ sides and we classify by $q$, the last nondecreasing value seen before the decreasing value that ends the game, so $2\le q\le n.$ Using $x$ for the number of points and $z$ for the number of rolls we obtain

$$G_q(z, x) = \left(\prod_{p=1}^{q-1} (1 + x^p z + x^{2p} z^2 + \cdots)\right) \\ \times (x^q z + x^{2q} z^2 + \cdots) (x^{q-1} z + x^{q-2} z + \cdots + x z).$$

This is

$$G_q(z, x) = \left(\prod_{p=1}^{q-1} \frac{1}{1-x^p z}\right) \frac{x^q z}{1-x^q z} (x^{q-1}+\cdots+x) z \\ = z^2 \left(\prod_{p=1}^{q} \frac{1}{1-x^p z}\right) (x^{2q-1}+\cdots+x^{q+1}).$$

Introducing $$G(z, x) = \sum_{q=2}^n G_q(z, x)$$

the desired quantity is given by

$$\left. \frac{\partial}{\partial x} G(z, x) \right|_{x=1, \; z=1/n.}$$

Differentiate for the expected number of points:

$$z^2 \left(\prod_{p=1}^{q} \frac{1}{1-x^p z}\right) \left(\sum_{p=1}^q \frac{p x^{p-1} z}{1-x^pz} \right) (x^{2q-1}+\cdots+x^{q+1}) \\ + z^2 \left(\prod_{p=1}^{q} \frac{1}{1-x^p z}\right) ((2q-1) x^{2q-2}+\cdots+(q+1)x^{q}).$$

Set $x=1$ to find

$$z^2 \left(\prod_{p=1}^{q} \frac{1}{1-z}\right) \left(\sum_{p=1}^q \frac{pz}{1-z}\right) (q-1) \\ + z^2 \left(\prod_{p=1}^{q} \frac{1}{1-z}\right) \left(\frac{1}{2} (2q-1)(2q) - \frac{1}{2} q(q+1)\right).$$

Now

$$ \frac{1}{n^2} \prod_{p=1}^q \frac{1}{1-1/n} = \frac{1}{n^2} \prod_{p=1}^q \frac{n}{n-1} = \frac{n^{q-2}}{(n-1)^q}$$

and we obtain

$$\frac{n^{q-2}}{(n-1)^q} \left((q-1) \frac{1/n}{1-1/n} \frac{1}{2} q(q+1) + \frac{3}{2} q (q-1)\right).$$

Summing the contributions we get

$$\sum_{q=2}^n \frac{n^{q-2}}{(n-1)^q} \left(\frac{1}{2} \frac{1}{n-1} (q-1) q (q+1) + \frac{3}{2} q (q-1)\right).$$

The first of these is

$$\frac{3}{(n-1)^3} [w^{n-2}] \frac{1}{1-w} \sum_{q\ge 0} \frac{n^q}{(n-1)^q} w^q {q+3\choose 3} \\ = \frac{3}{(n-1)^3} [w^{n-2}] \frac{1}{1-w} \frac{1}{(1-wn/(n-1))^4}.$$

The second is

$$\frac{3}{(n-1)^2} [w^{n-2}] \frac{1}{1-w} \sum_{q\ge 0} \frac{n^q}{(n-1)^q} w^q {q+2\choose 2} \\ = \frac{3}{(n-1)^2} [w^{n-2}] \frac{1}{1-w} \frac{1}{(1-wn/(n-1))^3}.$$

Adding we find

$$\frac{3}{(n-1)^3} [w^{n-2}] \frac{1}{1-w} \frac{1+(n-1)(1-wn/(n-1))}{(1-wn/(n-1))^4} \\ = \frac{3}{(n-1)^3} [w^{n-2}] \frac{1}{1-w} \frac{1+(n-1)-wn}{(1-wn/(n-1))^4} \\ = \frac{3n}{(n-1)^3} [w^{n-2}] \frac{1}{(1-wn/(n-1))^4}.$$

Extracting the coefficient we obtain

$$\frac{3n}{(n-1)^3} {n-2+3\choose 3} \frac{n^{n-2}}{(n-1)^{n-2}} = \frac{1}{2} (n+1) n (n-1) \frac{n^{n-1}}{(n-1)^{n+1}}.$$

The result is the following closed form:

$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{2} (n+1) \frac{n^n}{(n-1)^n}.}$$

This will produce

$${\frac {163296}{15625}}$$

for $n=6,$ confirming the accepted answer.

Observe that the above uses pen-and-paper type manipulations only.