Calculating probability of values for multiple dice rolls where sum is equal or greater than 4 Assuming we have 6 sided dice and play game of rolling dice each turn until we get sum value of at least 4.
For example we know that game will always finish with value less or equal than 9 because on 1. turn max value we can get without finishing the game is 3, then we can get highest value  of 6, all values on 2. turn will result in finishing game.
My question is, what is the probability that game will finish with value 5, 6 or 7?
What's the correct approach to solve this probabilities? My difficulty is that each next turn is conditional on probability of previous turns.

You know that the final sum will be in the range of $4$ through $9$. Determine all possible rolls that yield the final result. Add up the probabilities as each event will be disjoint.
Example: to get to 4, you have the following roll patterns:
$$4 \\ 3, 1 \\ 2, 2 \\ 2, 1, 1 \\ 1, 3\\ 1,2,1 \\1, 1, 2 \\ 1, 1, 1, 1$$
This gives:
$$P(4) = P(5) = P(6) = \dfrac{1}{6}+3\cdot \dfrac{1}{6^2}+3\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4} = \dfrac{344}{1296} \\ P(7) = 3\cdot \dfrac{1}{6^2}+3\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4} = \dfrac{127}{1296} \\ P(8) = 2\cdot \dfrac{1}{6^2}+3\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4}=\dfrac{91}{1296} \\ P(9) = \dfrac{1}{6^2}+2\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4}=\dfrac{49}{1296}$$
As a semi-verification, let's add it all up to make sure it adds to $1$:
$$\dfrac{3\cdot 343+127+91+49}{1296} = 1$$
This appears to be the correct probability for each result.
Another way to think about this: Let's consider the game in rounds. It can end on round $1$ with a roll of $4,5,6$. So that is a $\dfrac{1}{6}$ probability for each.
It can end on round $2$ with a first roll of $1,2,3$ for round $1$ to reach any of $4,5,6,7$ (so $3\dfrac{1}{6^2}$ probability of each), a first round roll of $2,3$ can reach $8$ for a probability of $2\cdot \dfrac{1}{6^2}$, or $3$ can reach $9$ for a probability of $\dfrac{1}{6^2}$.
It can end on round $3$ with the first two rolls as $(1,1),(1,2),(2,1)$ and reach $4,5,6,7,8$ for a probability of $3\cdot \dfrac{1}{6^3}$. And the first two rolls being $(1,2),(2,1)$ can reach $9$ with probability $2\cdot \dfrac{1}{6^3}$.
And it can reach any result on the fourth round by rolling three consecutive $1$'s followed by any roll of the $d6$ for a probability of $\dfrac{1}{6^4}$ each.
