10 faced die success chance

Question

If a ten faced dice is rolled and anything above 6 is considered a success and rolling 10 gets a reroll but rolling 1 removes a success however rolling 1 while rolling 10's extra does not remove a success what is the expected number of successes when 10 dice are rolled?

Answer 1

The difficulty in this problem is what happens if we get a $10$, as we get another roll. Lets assume we do roll a $10$, then we will have some amount of expected successes $\eta$ given that we already got one $10$. In this case we will have a $\frac{3}{10}$ chance of getting just a simple $+1$ success, a $\frac6{10}$ chance of getting a score of $0$, and a $\frac1{10}$ chance of getting a $10$ in which case we expect $1 + \eta$ points (as we get one from rolling the $10$ and then we reroll we $\eta$ expected points). This means we may write:

$$\eta = \frac{3}{10} + \frac{1}{10}\left(1+\eta\right)$$

Now we may solve for $\eta$, and we will find $$\frac{9}{10}\eta = \frac{4}{10} \iff 9\eta = 4 \iff \eta = \frac49$$

Let us now finish up with writing our original probability in terms of $\eta$ and then we may substitute in. We have $\frac1{10}$ chance for a $-1$ with rolling a $1$, we have a $\frac{3}{10}$ chance for a $+1$ with rolling a $1$, and then if we roll $10$ we get $1$ point for rolling $10$ and then expect another $\eta$points, giving us:

$$-\frac{1}{10} + \frac{3}{10} + \frac{1}{10}\left(1+\eta\right)$$

Now we simplify and find:

$$\frac{3}{10} + \frac\eta{10} = \frac3{10}+\frac9{40} = \frac{12+9}{40} = \frac{21}{40}$$ as our expected number of success per roll, we then multiply by $10$ for the $10$ dice, thus we get a final answer of: $$\frac{21}4$$