FrogReaver
Explorer
It’s just a geometric series. Has a simple formula.
= mean of die roll / [(die size - 1) / die size]
= mean of die roll / [(die size - 1) / die size]
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A down and dirty method is to just round up the normal average. D6 gies from 3.5 to 4. D8 from 4.5 to 5. Etc. Then just add averages together for complex pools. The variance will be off, so remember than the more dice, the more likely the roll will be close to average.Holy Thread Resurrection Batman!
Ok fellow math-masters, I need your help on this, and I can't seem to find a clear concise answer anywhere...
What is the average result for exploding dice when you're rolling multiple dice?
Example, whats the average result when rolling 2d6 vs 4d6 when open ended dice are involved?
What about when different die types are included, like 2d6+1d8+1d12?
Thanks for your input!
If you care, here is why I want to know:
I'm running a Super Hero game using Savage Worlds. I'm having trouble gauging the hero's power level vs opponent's toughness and damage dealing potential. Mainly because you start to stack multiple dice and dice types that can all explode. If I can ballpark a character's average, I can set targets that will provide a more interesting encounter, instead of the opponents getting blasted apart like wet tissue paper by the heroes; or destroying the heroes with a sideways glance by my villain.
2 1.5 2.250000 2.625000 2.437500 2.531250
3 2.0 2.666667 2.888889 2.740741 2.765432
4 2.5 3.125000 3.281250 3.164063 3.173828
5 3.0 3.600000 3.720000 3.624000 3.628800
6 3.5 4.083333 4.180556 4.099537 4.102238
7 4.0 4.571429 4.653061 4.583090 4.584756
8 4.5 5.062500 5.132813 5.071289 5.072388
10 5.5 6.050000 6.105000 6.055500 6.056050
12 6.5 7.041667 7.086806 7.045428 7.045742
16 8.5 9.031250 9.064453 9.033325 9.033455
20 10.5 11.025000 11.051250 11.026313 11.026378
24 12.5 13.020833 13.042535 13.021738 13.021775
30 15.5 16.016667 16.033889 16.017241 16.017260
let P(N) = probability of rolling 20 given you have just rolled NSince this is similar to exploding dice, and this necro-thread has yet to meet a cleric, here's another one for you probability pros:
What is the average result, based on the number you must exceed (target number), of a d20 that you get to re-roll as long as your first roll was higher than the target, and your following roll is higher than your previous roll?
The process goes like this:
Target number is 10. So if you roll a 10 or lower, that's your result.
Roll a 15, and you get to roll again, because you beat the target number.
Roll 8 after that, and your 15 stands. Roll 19, and you get to roll again.
Roll 20 and...just kidding. You're not going to roll a 20 after a 19.
So the result of that process was 19, based on a target number of 10. Which means my question is horribly bad, and should be more like:
What are the odds of getting a 20 result, based on the target number, using the above process?
With exploding dice, I've always done it this way, which builds on what Ovinomancer has posted:A down and dirty method is to just round up the normal average. D6 gies from 3.5 to 4. D8 from 4.5 to 5. Etc. Then just add averages together for complex pools. The variance will be off, so remember than the more dice, the more likely the roll will be close to average.
The real math is to sum (dice average)/(high roll ^ k) as k goes from 0 to infinity. So, for d8, it's 4.5/1 + 4.5/8 + 4.5/64 +4.5/512 ..... The rapidly converges, though, so the first exploder has the largest impact and dominates. A d8 exploder converges to around 5.143, so using 5 is both fast and only slightly wrong.
Odds of rolling a 20, with a target number of 0, are 12.63%? So, 1 in 8?So the answer is 0.1263
P(N-1) = 21/20 P(N)Not sure where your exponents are coming from