Probability question involving throwing a die until we get a certain result

You roll a dice until you get the number $1$ four times. What is the probability that you get exactly $10$ numbers under the value of four?

My attempt: First, I tried to calculate only the event that there are $10$ rolls. So, we want to choose two numbers $(2,3)$ to appear $6$ times, and of course the number $1$ four times. The probability to get those numbers: $(\frac{2}{6})^6(\frac{1}{6})^4$ . On top of that, we would like to arrange them, so we get $\binom {9}{3}$ to arrange the number $1$ (I chose $\binom {9}{3}$ here because we know there is $1$ at the end). For numbers $2$ and $3$, we have $\binom {2}{1}$ possibilities for choosing one of the numbers, and $$\binom {6}{6}\binom{0}{0} + \binom {6}{5}\binom{1}{1} + \binom {6}{4}\binom{2}{2} + \binom {6}{3}\binom{3}{3} + \binom {6}{2}\binom{4}{4} + \binom {6}{1}\binom{5}{5} + \binom {6}{0}\binom{6}{6}$$ ways to arrange them. So, the probability for that event is :

$$(\frac {2}{6})^6(\frac {1}{6})^4\binom {9}{3}\binom {2}{1} \left[\binom {6}{6}\binom{0}{0} + \binom {6}{5}\binom{1}{1} + \binom {6}{4}\binom{2}{2} + \binom {6}{3}\binom{3}{3} + \binom {6}{2}\binom{4}{4} + \binom {6}{1}\binom{5}{5} + \binom {6}{0}\binom{6}{6}\right] \approx 0.01138$$

That is, as mentioned above, only one event of the probability question. There is an infinite sum of events, where in those events we get a number that is bigger or equal to four. So, we have $\frac 3 6$ possibilities for choosing one of those numbers. The probability of this event is $$\sum_{n=0}^\infty (\frac {1}{2})^n = \frac {1}{\frac {1}{2}} = 2 $$

So I multiplied the result above with $2$, and my final answer is $\approx 0.02276$.

I'm not certain with my final answer. In fact, I'm pretty sure I'm mistaken. I think my logic here is not right. Any help would be welcomed.

To summarize the discussion in the comments:

The only relevant tosses are $\{1,2,3\}$. Getting anything higher than $3$ just wastes a turn. If, say, the toss sequence was $\{6,1,2,5,1,4,3,6,1,1\}$ we would only be concerned with the subsequence $\{1,2,1,3,1,1\}$, simply ignoring the higher tosses.

Thus the problem is equivalent to the following: Given a fair three sided die, what is the probability that the fourth $1$ is observed on the tenth toss?

Since each of the faces on our (virtual) three sided die are equally probable, this is simply $$\frac 13\times \binom {9}3\times \left(\frac 13\right)^3\times \left(\frac 23\right)^6\approx .09104$$

Sanity check: considering the six sided die, we expect it to to take six tosses to see a $1$, hence $24$ tosses to see four $1's$. Of the $20$ non-$1's$ in that expected sequence we expect $\frac 25$ of them to be either $2$ or $3$, hence $8$. A straight forward computation shows that the above calculation does indeed yield $8$ for this expectation.

Note: of course you can do the calculation using the six sided die but in that case you need to consider all possible sequence lengths (well, all those of length $≥10$ at least). Nothing theoretically wrong with that, but it forces you to compute a rather messy summation. To be sure, that infinite sum converges quite quickly so you only need to consider the first thirty terms or so. Worth doing if no other reason than to confirm the result.

I think the answer is a little more involved.

Let $L$ be the index when the $4$th one appears. Then the probability in question is (note the denominator is $5$. not $6$!) $\sum_{k=10}^\infty P[L=k] \binom{k-4}{6}({2 \over 5})^6 ({3 \over 5})^{k-10}$.

To compute $P[L=k]$, note that the $k$th index is always $1$, and the other three can appear anywhere else. Hence $P[L=k] = {1 \over 6} \binom{k-1}{3} ({1 \over 6})^3 ({5 \over 6})^{k-4}$.