Formula for dice probability where numbers have different values

[omni989]

Hi all

Im hoping someone here can help me out. I'm developing a game and I have a dice system I'm using but I am having trouble mapping the probabilities. I should state that I know very little about such things and my maths is basic but a reasonably explained answer should cause me no problems.

Firstly I need to calculate the probability of rolling X number of successes on multiple D6 where 1-3 are failures, 4 & 5 are one success each and 6 is two successes. I will need to map out the probability of exactly 1 success, 2 success, 3 success etc on 2D6, 3D6, 4D6 up to 6 D6.

In case you are finding that too easy can anyone suggest how I would go about calculate the probabilities with the following variance. There are now two players rolling against each other with their successes cancelling each other out, the winner is the player with successes left. For one player, the attacker, 3 is also a success and cancels upto two 4's that the defender rolls. How can I map how effective the extra number will be to the attacker and how detrimental to the defender? Taking into account that if the attacker does not roll a 3 then the defender gets his 4's but if he does then he loses some of them.

I do hope I have explained that clearly and that someone will offer some solutions.

Thanks in advance

 

[Stephen Tashi]

For 3 dice

Let $$G(x) = (\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{x}{6} + \frac{x}{6} + \frac{x^2}{6})^3$$

$$= (\frac{3}{6} + \frac{2x}{6} + \frac{x^2}{6}) ^3$$

$$= \frac{1}{6^3}( 3 + 2x + x^2)^3$$

If my hand calculation is correct this is:
$$= \frac{1}{6^3}(27 + 54x + 63x^2 + 44x^3 + 21x^4 + 6x^5 + x^6)$$

Probability of zero successes $= \frac{1}{6^3}(27)$
Probability of one success $= \frac{1}{6^3}(54)$ ( = the coefficient of $x$ .)
Probability of two successes $= \frac{1}{6^3}(63)$ ( = the coefficient of $x^2$)
...etc.