Statistics Help

[Lmns Crn]

...I suddenly no longer know what the crap we've been debating about.

[Xeviat]

I don't either LC. I was just trying to figure out the means based on dice pools so I'd know what normal rolls would be.

Now I need to figure out how the exploding dice work into things; in L5R, you get to reroll 10s and add 10 to the new roll. This means the d10s really end up having a range from 1 to 9, and then the last averages to something like 15.6 ... but it goes on forever.

[Elemental_Elf]

Exactly. Every 10 gets an additional roll added to the original 10. So if you roll 5K2 and hit a 9 and a 10, the 10 explodes. You roll an additional die and it comes up 10 as well. So that ten explodes and you roll a 2. So now that original 10 is treated as if it rolled a 22. The total for the entire roll is thus 31.

[Xeviat]

Right E_E, I'm just trying to see how that skews the statistics.

1-9 + (10; 11-19 + (10; 21-29 + ...))

I don't know how far I need to take it to get a viable average.

[sparkletwist]

You've got a 10% chance of rolling a 10 once, a 1% chance of rolling a 10 twice, a 0.1% chance of rolling it three times, and so on. So three times is probably the limit of how far you'd need to go in most cases.

Here's some anydice code to help you with exploding dice:

Code Select Expand

function: explode R:n times N:n plus S:n {
if R > 0 {
   if N = 10 { result: [explode R-1 times d10 plus N+S] }
   result: N+S
}
result: S
}
EXP: [explode 3 times d10 plus 0]

output dEXP


To play with more complicated things, just tweak that last output line. For example, 5k2 would use the previously mentioned syntax of {1,2}@5dEXP. (You just use dEXP instead of d10, basically)

[Ghostman]

The exact average result for a single exploding 10-sided die is 55/9 = 6.111111...

It's calculated by summing up all the possible results of a d10 roll, plus a recursive function call (added to the result of '10', because that one triggers an explosion), and dividing that sum by the number of the possible results:

e(10) = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + e(10)) / 10

Solving the equation above to find e(10), you get:

e(10) = 55/9 = 6.111111...

You can use this method to find the average for any single exploding die. To get the average result for a pool of exploding dice, simply multiply by the number of dice in the pool:

Average[ "XeY" ] = X * e(Y)

^ That's a pool  of Y-sided explosive dice, with X dice in the pool.

This is applicable to straight forward dice pools, not a roll-and-keep system. When you add a rule to keep (or drop) a specified number of the highest (/lowest) rolling dice from the pool, the math gets extremely complicated.