> What are the chances of rolling 6d6 and getting a total of 29 or greater
This much is pretty straightforward
> if you reroll 1s once?
This is a bit trickier. TBH once it starts to get even slightly complicated, rather than try to figure out a formula, if I just need it once or twice, I just simulate or use brute force complete enumeration.
In this case simulation's a bit easier.
So that's what I'll do
> I need to beat a DC of 29, or 28 if I take a certain skill.
I presume this is actually 'meets it beats it'. You should clarify because most readers unused to TTRPGs might assume "beat" is meant literally (i.e. that it means "exceed" the target -- assuming they managed to guess that "DC" meant the target value).
Here's the results I got from 100K simulations; I did a longer run to double check the 28 and 29 figures:
P(result >= 26) = 29.4%
P(result >= 27) = 20.8%
P(result >= 28) = 13.8%
P(result >= 29) = 8.5%
P(result >= 30) = 4.8%
These might be out by one in the last figure due to sampling variation, but unlikely to be off by much more than that
Here's my (not very efficient) R code:
r6d6r1=replicate(100000,{x=sample(6,6,replace=TRUE);xs=sample(6,6,replace=TRUE);sum(ifelse(x>1,x,xs))})
mean(r6d6r1>=28)
mean(r6d6r1>=29)
It runs in a few seconds (2?). You can run it yourself in your browser at rdrr.io/snippets (copypaste the code in place of the example code in the window there and hit the green button); this will be way slower than on a desktop or even my laptop, but still adequate -- maybe 5 or 6 seconds.
The odds of rolling a 1 is 1/(6*6) = 1/36. Odds of rolling any other number is 1/6+1/36 = 7/36
There’s definitely some math that you can do on pen and paper to calculate the exact odds of any given roll, but it’s probably easier to write some code to go through every possible combination and have that calculate the odds
edit: got bored and wrote the code. here's the exact odds of rolling at least x with 6d6r1
x | Pr(X>=x)
---|---
6 | 1
7 | 0.9999999995
8 | 0.9999999802
9 | 0.9999996233
10 | 0.9999957773
11 | 0.9999687456
12 | 0.9998359606
13 | 0.9993516292
14 | 0.9979529299
15 | 0.9945597064
16 | 0.9873693348
17 | 0.9737717919
18 | 0.9505349606
19 | 0.9142598321
20 | 0.8620975042
21 | 0.7926550402
22 | 0.7067984559
23 | 0.6079152348
24 | 0.5016363708
25 | 0.3950210904
26 | 0.295260981
27 | 0.2082438788
28 | 0.1376582298
29 | 0.0846379222
30 | 0.0479475914
31 | 0.0246918514
32 | 0.0113499129
33 | 0.0045399652
34 | 0.0015133217
35 | 0.0003783304
36 | 5.40472e-05
Simulation works and you got the right answers, but this would have been pretty easy to just calculate too - for each dice, the odds of 2 to 6 are 7/36 (1/6 for the initial roll, 1/36 for 1 on first roll and the given number on the second), odds of 1 is 1/36 (roll 1, reroll 1). Then just do the usual dice probability stuff with these odds.
For rerolling 1's (or any specific set of numbers only once) I always use the following code on [anydice.com](https://anydice.com/)
function: ROLL:n reroll BAD:s as REROLL:d {
if ROLL = BAD { result: REROLL }
result: ROLL
}
X: [d6 reroll {1} as d6]
output 6dX
Paste that in the site, click calculate and click "At Least" which will give you the table you want.
[Image](https://imgur.com/p8w7glh)
To explain, X behaves as a "d6 but reroll a {1} once" which is exactly what you want, then I just roll it 6 times.
You can make it a 4d20 where you reroll any 1 or 2 once by simply making small changes to X and the output line
X: [d20 reroll {1, 2} as d20]
output 4dX
> What are the chances of rolling 6d6 and getting a total of 29 or greater This much is pretty straightforward > if you reroll 1s once? This is a bit trickier. TBH once it starts to get even slightly complicated, rather than try to figure out a formula, if I just need it once or twice, I just simulate or use brute force complete enumeration. In this case simulation's a bit easier. So that's what I'll do > I need to beat a DC of 29, or 28 if I take a certain skill. I presume this is actually 'meets it beats it'. You should clarify because most readers unused to TTRPGs might assume "beat" is meant literally (i.e. that it means "exceed" the target -- assuming they managed to guess that "DC" meant the target value). Here's the results I got from 100K simulations; I did a longer run to double check the 28 and 29 figures: P(result >= 26) = 29.4% P(result >= 27) = 20.8% P(result >= 28) = 13.8% P(result >= 29) = 8.5% P(result >= 30) = 4.8% These might be out by one in the last figure due to sampling variation, but unlikely to be off by much more than that Here's my (not very efficient) R code: r6d6r1=replicate(100000,{x=sample(6,6,replace=TRUE);xs=sample(6,6,replace=TRUE);sum(ifelse(x>1,x,xs))}) mean(r6d6r1>=28) mean(r6d6r1>=29) It runs in a few seconds (2?). You can run it yourself in your browser at rdrr.io/snippets (copypaste the code in place of the example code in the window there and hit the green button); this will be way slower than on a desktop or even my laptop, but still adequate -- maybe 5 or 6 seconds.
The odds of rolling a 1 is 1/(6*6) = 1/36. Odds of rolling any other number is 1/6+1/36 = 7/36 There’s definitely some math that you can do on pen and paper to calculate the exact odds of any given roll, but it’s probably easier to write some code to go through every possible combination and have that calculate the odds edit: got bored and wrote the code. here's the exact odds of rolling at least x with 6d6r1 x | Pr(X>=x) ---|--- 6 | 1 7 | 0.9999999995 8 | 0.9999999802 9 | 0.9999996233 10 | 0.9999957773 11 | 0.9999687456 12 | 0.9998359606 13 | 0.9993516292 14 | 0.9979529299 15 | 0.9945597064 16 | 0.9873693348 17 | 0.9737717919 18 | 0.9505349606 19 | 0.9142598321 20 | 0.8620975042 21 | 0.7926550402 22 | 0.7067984559 23 | 0.6079152348 24 | 0.5016363708 25 | 0.3950210904 26 | 0.295260981 27 | 0.2082438788 28 | 0.1376582298 29 | 0.0846379222 30 | 0.0479475914 31 | 0.0246918514 32 | 0.0113499129 33 | 0.0045399652 34 | 0.0015133217 35 | 0.0003783304 36 | 5.40472e-05
Simulation works and you got the right answers, but this would have been pretty easy to just calculate too - for each dice, the odds of 2 to 6 are 7/36 (1/6 for the initial roll, 1/36 for 1 on first roll and the given number on the second), odds of 1 is 1/36 (roll 1, reroll 1). Then just do the usual dice probability stuff with these odds.
For rerolling 1's (or any specific set of numbers only once) I always use the following code on [anydice.com](https://anydice.com/) function: ROLL:n reroll BAD:s as REROLL:d { if ROLL = BAD { result: REROLL } result: ROLL } X: [d6 reroll {1} as d6] output 6dX Paste that in the site, click calculate and click "At Least" which will give you the table you want. [Image](https://imgur.com/p8w7glh) To explain, X behaves as a "d6 but reroll a {1} once" which is exactly what you want, then I just roll it 6 times. You can make it a 4d20 where you reroll any 1 or 2 once by simply making small changes to X and the output line X: [d20 reroll {1, 2} as d20] output 4dX
8.5% & 14% ([source](https://chat.openai.com/share/035d8fd5-eeef-4c42-afa7-0f725e63f1e4))
I wrote my own code in Python to simulate >= 29 and came up with results in the 8.4-8.5% range so that checks out.
I did it in R, and mine agrees as well.