In need of some probability/dice experts

using D6, the chance of getting any number is 1 in 6, represented by 1/6

the probability of getting a specific result from 2 D6s is calculated by multiplying 1/6 by 1/6 to derive at 1/36

with a third dice, you just get 1/36 and multiply it by another 1/6 to get 1/216

with every additional dice, just multiply the probability by the previous probability and you will get your answer. this works even when you want to put other types of dice into the mix, like if you want a D10 to be the next dice, just multiply the last probability by 1/10, if you want a D20 just multiply the last probability by 1/20

this is assuming you are looking for a specific result on the dice, you will need to define what result "A" is to appraise if your own calculations are accurate

I always suggest a software called "small roller" to calculate probability without knowing any math. But from what I can see, this app seem to be hard to google. So I decided to upload it on my website until the next moon. It can be found here:

http://lariennalibrary.com/extern/temp/SmallRoller.zip

Enjoy!

That's how I read it first pass through. Winston can use a headbutt, so his die wins on an H independent of the target's defeat requirement. To meet the target, they roll 3 dice and hope to get at least one P and one K amongst the dice.

I should hone up on my combinatorics, but to do this with semi-brute force I'd split the calculation into the 5 die results Winston could have. Then put in the probability from there.

StagC, one quick note on that 75% probability for you. You get there by saying that my chance of getting at least one P on the dice is 100% minus the chance of getting no P. Going backwards you can multiply the "missed" probabilities, but you can't multiply the hit probabilities. A lot of probabilities are calculated that way, but the combinatorics version might look prettier.

If you split your start set, you can multiply the size of the group times its success probability and re-add. In this case, we have Winston's 1/6 of an H times the chance of that winning (100%) and have our first element: 16.7%

Now the P/K die (1/6), needs one P or K amongst C's two dice. Since 4/6 of the die is a P or K, the chance of not getting one is 2/6=1/3. That's the first of C's dice. Same with the second 1/3 of missing. 1/3 * 1/3 = 1/9 of missing. And so 8/9 of at least one P or K, times the 1/6th for this die is 8/54 or 14.8%

etc.

At least in this case, seems simpler to just generate the 216 results in Excel and make a formula to check. All summed up, I get about a 2/3 chance of the two of them winning.

I haven't messed with that dice sim, but I might. Seems useful once you realize how messy branching and multi-role dice can be. I do wonder if it is rough for the designer, how is it for the player... Are they happy enough on guy instinct?

Here's what I would do in Excel:

Create columns of possible die rolls. So you label the columns d1, d2, and d3, for instance.

We'll go with 3d6 example because that's the most complex.

You are going to need a total of 216 (6^3) rows, plus that header row.
In the d1 column, put in H, P, P, P/K, K, and M - each in its own row. (so 6 rows) Repeat that sequence 36 times. (Copy and paste it)
In the d2 column, put in 6 H's, then 6 P's, then 6 P's, then 6 P/K's, then 6 Ks, then 6 Ms - each in its own row (36 total rows). Copy and past that 6 times.
In the d3 column, put in 36 H's, 36 P's, 36 P's, 36 P/K's, 36 K's and finally 36 M's, That's it (that's 6x36 = 216)

You now have all possible combinations for your dice.

The weird parts you will encounter are because of the P/K and the 3rd dice, which may or may not be necessary. (I assume even if you roll 3 P's, that you still win in an encounter that only requires 2)

If you only want to look at 3 dice, it might be easiest to "brute force" it rather than figure out a formula. What I mean by that is to simply do a manual check on what the 3 rolls are and then figure out if it's a "win" or "loss". You'd set up different columns for each scenario that you want. Then just put W or L in it. When you find them all, just add them up (SUMIF) and then divide by 216 for the probability.