alexjones  02 Jan 2017 10:22 a.m. PST 
I am struggling with the calculations of probability for rolling 3 6 sided die and getting the following results: i) 6 on any of the 3 die rolled. ii) 5 or 6 on any of the 3 die rolled iii) 4 ,5 or 6 on any of the 3 die rolled. Can anyone help please? 
gbowen  02 Jan 2017 10:34 a.m. PST 
There are 3 dice of 6 faces, 6*6*6 possible results = 216. A unique result such as 6, 6, 6 would be 1/216 For one face on 6 the chance is of a 6 on 1 die and any other result on the other 2. 1/6 * 5/6 * 5/6 = 25/216 This can happen with either of the 3 dice so add up the 3 scores. 75/216 = 35% 
JMcCarroll  02 Jan 2017 10:41 a.m. PST 
No genius but… i) 3/6 or 50% of 1 success. ii) 6/6 or 100% of 1 success. iii) 9/6 or 150% of 1 success. hope this helps! 
alexjones  02 Jan 2017 11:01 a.m. PST 
to re phrase to make it a bit clearer i) 6 on at least one of the 3 die rolled. (includes rolling 2 sixes or 3 sixes) ii) 5 or 6 on at least one of the 3 die rolled (includes rolling 2 sixes or 3 sixes or 2 fives or 3 fives) iii) 4 ,5 or 6 on at least one of the 3 die rolled.(includes rolling 2 sixes or 3 sixes or 2 fives or 3 fives or 2 fours or 3 fours). 
MajorB  02 Jan 2017 11:15 a.m. PST 
For one face on 6 the chance is of a 6 on 1 die and any other result on the other 2. 1/6 * 5/6 * 5/6 = 25/216 This can happen with either of the 3 dice so add up the 3 scores. 75/216 = 35% But you also need to add: The possibility of 6 on 2 dice and any other score on the other: 1/6 * 1/6 * 5/6 = 5/216 And this can also happen in 3 ways so: 15/216 Then there is also the possibility of 6 on on all 3 dice: 1/6 * 1/6 * 1/6 = 1/216 75 + 15 + 1 = 91 91/216 = 42% A 5 or 6 on 1 die and any other result on the other 2. 2/6 * 4/6 * 4/6 = 32/216 This can happen with any of the 3 dice so add up the 3 scores. 96/216 Then there is the possibility of 5 or 6 on 2 dice and any other score on the other: 2/6 * 2/6 * 4/6 = 16/216 And this can also happen in 3 ways so: 48/216 Then there is also the possibility of 5 or 6 on on all 3 dice: 2/6 * 2/6 * 2/6 = 8/216 96 + 48 + 8 = 152 152/216 = 70% A 4, 5 or 6 on 1 die and any other result on the other 2. 3/6 * 3/6 * 3/6 = 27/216 This can happen with any of the 3 dice so add up the 3 scores. 81/216 Then there is the possibility of 4, 5 or 6 on 2 dice and any other score on the other: 3/6 * 3/6 * 3/6 = 27/216 And this can also happen in 3 ways so: 81/216 Then there is also the possibility of 4, 5 or 6 on on all 3 dice: 3/6 * 3/6 * 3/6 = 27/216 81 + 81 + 27 = 189 189/216 = 87% 
martin goddard  02 Jan 2017 11:30 a.m. PST 
Good chat. Very interesting may I chip in with some thoughts. It may be simpler to calculate the probability of total failure . Take this away from 1 and what is left is the probability of success! Chance of not gettinga 6 is 5/6. Chance of this happening on 3 dice is 5/6 x 5/6 x 5/6= 125/216 So if that is prob of failure. all othet outcomes woudl give 1 or more successes! So prob is 91/216 of success. A 5 or 6 would be 4/6x 4/6 x 4/6= 64/216 of failure. Thus 152/216 for success. So in closing , find the chance it will not happen and take it way from 1.

DesertScrb  02 Jan 2017 11:38 a.m. PST 

alexjones  02 Jan 2017 12:22 p.m. PST 
thanks for the solutions and especially how you calculated it! very useful. 
wakenney  02 Jan 2017 12:43 p.m. PST 
There are 216 possible combinations of results. Out of those, 91 of the sets will contain a 6 on at least one die. Meaning that the probability of rolling one or more 6s is 91/216 or 42.13%. For 5&6, P = 70.37% For 4,5&6, P = 87.5% Basically, confirmation of everyone else's answers. 
Shaun Travers  02 Jan 2017 1:45 p.m. PST 
I am with Martin. I find it easier to do the chance of it not occurring, and then subtract that from the total possible results. 
(Phil Dutre)  03 Jan 2017 2:34 a.m. PST 
If you're not afraid of a little math: look up binomial distributions. Those describe the number of x successes, on y tries, each with a z probability for success. But with the low numbers of dice you need, it's indeed possible to compute it exhaustively as shown above. 
John Treadaway  16 Jan 2017 7:50 a.m. PST 
If it helps, I found wakenney's answer the easiest to understand…. Other maths too much for my tiny brain… John T Edit: oh and thankfully I have more than one set of percentile dice! 