Situation: you're 95% (fail only on a 1) to succeed on two back-to-back d20 rolls, the second of which is a rider-effect dependent upon the success of the first.
You have one and only one reroll available (for example a goblin d20 shirt at a PFS table, or a Luck Blade, etc) which you will expend it necessary.
- What are the odds of achieving the back-to-back combo?
The only way you'll end up with result 1, is when you roll 1 twice. The chance of that is (1/20)^2=0.25%
Yes, but we're potentially rolling three times, so the math is a little more complex than just 1 in 400. And, if the reroll gets used the first time, it's not available the second time. (BTW, refresh your browser, as I've reworded the OP.)
Lets say S=Success, F=Fail, R=Reroll
These are the only ways to win:
SS=(19/20)*(19/20)
SF(RS)=(19/20)*(1/20)*(19/20)
F(RS)S = (1/20)*(19/20)*(19/20)
Add everything up:
(19/20)^2*(1+2/20) = 0.99275
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So, if I understand this correctly:
You roll a d20. If you get a natural one, you fail. If you don't, you roll a second d20. If you get a natural one this time, you fail. You also get a one-off reroll ability.
I'll turn everything into chances out of eight thousand (since there are eight thousand possible combinations when you roll three d20s).
Successes:
2-20 on first d20, 2-20 on second d20, no reroll needed: 7220/8000
Natural one on first d20, use reroll, 2-20 on reroll, 2-20 on second d20: 361/8000
2-20 on first d20, natural one on second d20, 2-20 when rerolling: 361/8000
Total success chance: 7942/8000 = 99.275%
Fails:
2-20 on first d20, natural one on second d20, natural one when rerolling: 19/8000
Natural one on first d20, use reroll, get another natural one: 20/8000
Natural one on first d20, use reroll, 2-20 on reroll, natural one on second d20: 19/8000
Total failure chance: 58/8000 = 0.725%
(Edit: Tsk. Math-ninjad by eleven seconds.)
The chance of having two successes outright is (19/20)^2 = 90.25%.
The chance of having one success and one failure, then succeeding on the reroll is 2*19/20*1/20*19/20 = 9.025%
Total chance of overall success is 99.275%.
This shows the power of rerolls, by having a reroll you decrease your chance of failure over 13 times, from 9.75% to 0.725%.
Thanks, guys. I knew it was somewhere between 399/400 and .95^2.
Is there a ready formula for this sort of thing into which variables could be plugged?
a = hit probability on d20
b = number of sequential rider-effect rolls
c = rerolls available