Let's build an RPG together.

Oh, wait, yeah, I think I see what mistake I made...

Still odd that your chance of rolling doubles on two dice would be teh same as your chance of rolling any given number on a single dice. Doesn't sound quite right...

I think it's more like you multiply your chance of rolling any given result on a dice by your chance again, no? Like 25% x 25% = 6.25% for d4s, 16.67% x 16.67% = 2.78% for d6s and so no?

Yeah, looks like I'm starting to get closer...

I think you'll find that you didn't list all of the combinations in your brute force analysis. With 2d4, there are 16 combinations (4 for each dice), four of which are doubles.

Think of out this way. Given one roll, what is the chance that the second roll is the same as the first. It's 1/X for a dX.

Best way to consider rolling doubles:

You have a 1 in 1 chance of rolling any number on the first dice.
You have a 1 in x chance of rolling that same number on the second dice of x sides.

Which means your total chance is 1 in x to roll a double.

If you roll a 1 on your first die you chances of getting another one is 25%, same if you roll a 2, 3, or 4.

The 6.25% is your chance of rolling two of a specific number. Rolling two 1s is 6.25%, two 2s is 6.25%, two 3s.... Etc.

Overall, though, it adds up to 25%.

You have 1 of X chance to roll a number on one die, but then you have 1 out of X * X chance of rolling the same result on your second die, no? otherwise, you'd roll double at the same rate you'd roll any given number. Add a 3rd die, and your chances increase a bit in where your chances are now 2 out of X * X since you need to match one of two possible results rolled? add a 4th die and it's now 4 out of X * X?

Urgh, headache... I hate maths... XD

Now I'm confused...

kalajel: you're right, but X * X is the chance of a SPECIFIC number (so 2 1's, specifically). But two of the same is just a 1 in X (because the probability of anything on the first dice is 1 of 1.

But that's the same chance of rolling any given result on 1 die. It does not makes sense that you have the same chance of rolling doubles. You are aiming for a specific result on your second dice, so it would be closer to the, "I aim to roll two 1s" analogy, no?

Yeah, I think this is more right...

What are we trying to answer? If we're just asking for the odds of rolling the same number on two dice, it is in the case of 2d6, 1/6.

If we want a specific result, such as "double 1s" on 2d6, it is 1/36.

The first example, we're just looking for any double, which is a result of {1,1}, {2,2}, {3,3}, {4,4}, {5,5}, {6,6}.

The second example, we're looking for only one result out of the many combinations, {1,1} out of {n,n}.

Condider 2d2. There are four results:

1,1 (double)
1,2
2,1
2,2 (double)

So lets say your goal is to find the odds of rolling a 1 on the first dice: your chances are 1 in X, where X is the number of sides.
Now your goal is to get another 1, which means you need to roll another 1 in X, bringing the total to 1 in X * X.

You already showed that part.

Instead, you're not concerned with what the roll of the first dice is, as a double can happen on any of the dice. So your chances of rolling a number on the first dice is 1 in 1 (guaranteed).
Now that you have that result, you want a specific result on the second die, a 1 in X chance. That means the total chance is 1 in 1 * X, or 1 in X.

If you look at the original, 1 in X * X, the chances of getting a specific double, there are X number of possible doubles in that collection, so it's 1 in (X * X) / X (chances of a specific combination / the number of combinations you'll accept), which is 1 in X.

Yeah, I think I get it now. I think I was overthinking the whole thing...

so 2d4 you have 4 x 4 =16 possible results, 4 x 3 = 12 of them won't be doubles, so 16 - 12 = 4 of them will be doubles, so 25% chance of rolling doubles.

On 3d4, you have 4 x 4 x 4 = 64 results, 4 x 3 x 2 = 24 of those won't be doubles or triples, so 64 - 24 = 40 of htem will be doubles or triples, so 62.5% chances or rolling at least a double...

and so on...

You got it :)

Can we rename the thread "Let's learn to teach math together?" because we're all not that great at it :p

I like it. Are you thinking like you get so many notice points before you have to deal with the ramifications. Like maybe an extremely trustworthy guy gets extra notice points whereas a shifty looking guy would get lower notice. I don't know...

Alright, let's mix and match for fun...

1d12 + 1d8 + 1d6...

total number of possible results: 12 x 8 x 6 = 576

number of results which won't result in doubles or triples: 12 x 7 x 5 = 420

number of results which will result in doubles or triples: 576 - 420 = 156

% chance of rolling at least a double on 1d12 + 1d8 + 1d6: 156 ÷ 576 x 100 = 27.08333333 or about 27.083%...

Makes sense?

Oh you... I think I was just overthinking it... It's a bad habit I have...