A much more detailed explanation, with changed values, since the other did not match.
So this is for a war game, where I want to calculate the average number of rounds, it takes a unit to kill another, based what statistics that unit has.
Each round consists of the unit makes a number of attacks, where you roll D6.
For an attack you first roll a D6, to see if it hits, and then if it hits, a D6 is rolled to see if it survives. If it survives the hits, it becomes easier to kill it next round.
To hit you have to roll S or less. If it is more than S, it misses, and that attack is over. The probability to hit the other unit is then (S/6) and to miss it is 1 - (S/6)
Once it hit the other unit might survive. Another D6 is rolled and if it is N or less the other unit dies. If it is more than N the other unit survives, but for the next attack N is increased by 1. Next time it is hit, it has to be (N+1) or less to kill it. So first time it is hit, it is N or less, then (N+1) and then (N+2), until N = 6.
When the unit only has one attack, I can calculate the average number of rounds
First, since the chance to hit is constant, the average number of attempts will be (1/(S/6)).
The second die changes probability, but it can represented as a random variable, that returns the probability of it taking 1, 2, 3 and so on, for the number attacks required to kill it.
If we call this random variable X, and if N starts with N = 4, it will be.
1 = (4/6) 2 = (1-(4/6))(5/6) 3 = (1-(4/5))(1-(5/6))*(6/6)
Since there is only one attack pr round, the average number of rounds can be calculated by (1/(S/6))*E(X)
Where E(X) is the expected number for X
The problem comes, when a unit has more than one attack per round.
Dividing it by 2, for when it has 2 attacks is not correct, and the technique above does not work either. I know it should not work, since the probability for it to hit, is different for when one hit, and two hits.