If each d6 has a 3/6 chance of producing a +1 and a 1/6 chance of producing a -1, how do I calculate the chances of rolling a total value of at least +1 if I roll 2d6, 3d6, 4d6, 5d6 or 6d6?
>>46608320
I recon this accounts for 2/6 chance of 0?
>>46608338
Yes indeed.
>>46608320
You can probably calculate this in anydice, anon.
>>46608320
http://anydice.com/program/81d2
>>46608320
>how do I calculate the chances of rolling a total value of at least +1
By calculating how many of the possible results fulfill your criteria and dividing by the total number of possibilities.
In the case of 2D6, to achieve at least +1, you need either +1/+1 or +1/0 or 0/+1).
+1/+1 has a chance of (3/6)2=1/4.
+1/0 and 0/+1 each have a chance of 3/6*2/6=1/6. Together they have a chance of 1/3 happening.
1/4 + 1/3 = 7/12
>>46608391
I wish I could. Unfortunately I suck.
>>46608408
Thank you very much.
>>46608320
>>46608391
output 3d{-1, 0, 0, 1, 1, 1}
>>46608320
Bernoulli distribution. Now go look it up. Faggot.
>>46610409
Specifically, you could just use a loop to have it do all the charts the OP said at once:
loop N over {1..6}{
output 3d{-1, 0, 0, 1, 1, 1} named "[N]d420"
}
