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When rolling 3d6 what is the chance to get 1;1(snake-eyes)?

Chance for getting 1;1 with one pair of dice = 1/36
Number of dice combination (1.+2. , 2.+3. , 1.+3.) times the probability = 3/36 or 18/216

We now have counted the result 1;1;1 (probability with three dice = 1/216) three times (once for every combination) which means we must substract it two times ---> 18/216 - 2*(1/216) = 16/216 = 2/27



Meaning the old ghosthelm raises the chances of perrils from 1/36 to 2/27 (this doesn't take into account that you would have fewer 6;6).
Did I make any mistakes?

Old ghosthelm gave a 4+ save versus perils.

Did you mean the runes of witnessing? Where you rolled 3d6 and take the two lowest? Because yes it increase your chance to roll a 1:1 peril but decreased your overall chance to peril (due to 6:6 being so rare).

"We now have counted the result 1;1;1 (probability with three dice = 1/216) three times (once for every combination) which means we must substract it two times ---> 18/216 - 2*(1/216) = 16/216 = 2/27 "


I'm not sure where this part is coming from - your odds of rolling 1,1,1 on 3d6 are 1/216 - there is one set of results those 3 dice can produce that is 111.

For what it's worth, the Mathhammer standard of calculating average isn't really useful anyway. Standard deviation is far more interesting.

 daedalus wrote:
For what it's worth, the Mathhammer standard of calculating average isn't really useful anyway. Standard deviation is far more interesting.

Err...

Standard deviation is not a specific operation that you use to calculate something (so comparing it to division doesn't get you anywhere even though the normal method most Dakka posters use to calculate expected results is very bad, yes) and it doesn't have anything to do with the question at hand, which is largely "what is the chance of rolling 11 on 2d6 vs the chance of rolling 111 on 3d6"

This is what you want to answer this question: http://en.wikipedia.org/wiki/Binomial_distribution

If you put the numbers (3 trials, 2 successes required, 1/6 probability per trial) into a calculator you get about a 7.5% chance to roll at least two 1s on 3D6. So your 2/27 number is correct, but you had a somewhat odd method of getting to it.

 Peregrine wrote:
This is what you want to answer this question: http://en.wikipedia.org/wiki/Binomial_distribution

If you put the numbers (3 trials, 2 successes required, 1/6 probability per trial) into a calculator you get about a 7.5% chance to roll at least two 1s on 3D6. So your 2/27 number is correct, but you had a somewhat odd method of getting to it.

It's pretty much the same aproach just manualized.

3choose2 is the same as looking how many possible combinations there are.