"Question for the math guys!" Topic
10 Posts
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otherone  04 Dec 2020 8:07 a.m. PST 
Hi. I was wondering what the percentage chance of rolling at least two "1s" on three dice? Thanks! Rob 
Stryderg  04 Dec 2020 8:44 a.m. PST 
.074 % There are 16 combinations that result in either 2 or 3 1's being rolled out of 216 possible combinations. I brute forced it: went into Excel, wrote out the dice combinations: 1 1 1 1 1 2 1 1 3 etc Then looked for lines where there are 2 or 3 1's. I'm sure are more elegant ways to figure it out, but that's my answer. What did I win? 
Swampster  04 Dec 2020 9:27 a.m. PST 
That'd be 7.4% :) 16 combos is right though. 1,1,1. There are three places where another number can be and 5 numbers other than 1, so 15 other combos with two ones. 
Parzival  04 Dec 2020 9:33 a.m. PST 
There are 6X6X6 or 216 possible combinations of 3 d6. A combination of 1, 1, 1 is therefore only 1 of the possible specific combinations (that it is 1, 1, 1 is only relevant in that there is no ordinal importance to the result, thus making it only 1 possible result). Therefore the odds are simply 1/216, or approx 0.46%. One can also calculate the odds by multiplying the odds of each separate die result, which for each is of course 1 in 6, or 16.66… percent. 16.66… cubed (multiplied by itself 3 times) is approx 0.46%. EDIT: Corrected— forgot to account for the percentage! *face palm* 
MajorB  04 Dec 2020 10:09 a.m. PST 
A combination of 1, 1, 1 is therefore only 1 of the possible specific combinations (that it is 1, 1, 1 is only relevant in that there is no ordinal importance to the result, thus making it only 1 possible result). Except the OP asked for at least 2 1s on any of the 3 dice, not 3 1s. 
pnguyenho  04 Dec 2020 1:01 p.m. PST 
The chance of rolling a 1 on a D6 is 1/6. The chance of not rolling a 1 (ie 26) on a D6 is 5/6. There is only one possible combination of all three 1s, so 1/6 x 1/6 x 1/6 = 1/216. If you label the dice as A, B, and C, in order to roll two 1s only, then the combinations would be: A1, B1, C26 so 1/6 x 1/6 x 5/6 = 5/216 A1, B26, C1 so 1/6 x 5/6 x 1/6 = 5/216 A26, B1, C1 so 5/6 x 1/6 x 1/6 = 5/216 So chance of rolling at least two 1s is 1/216 + 3 x 5/216 = 16/216 (7.4%). 
Stryderg  04 Dec 2020 1:26 p.m. PST 
That'd be 7.4% :) {mumble / grumble} stupid percent sign {mumble / grumble} Thanks for the catch. :) 
Parzival  04 Dec 2020 8:40 p.m. PST 
Ah, missed that detail of 2 out of 3… mea culpa! 
Rudysnelson  04 Dec 2020 9:29 p.m. PST 
1,1,1. 1,1,2. 1,1,3. 1,1,4. 1,1,5. 1,1,6 1,2,1. 1, 3,1. 1,4,1. 1,5,1. 1,6,1 2,1,1. 3,1,1. 4,1,1. 5,1,1. 6,1,1 Those are the combinations for your parameters. Looks like 16 valid rolls. Dice probability makes a great science fair project. My son took his in middle school and won at the regional competition a few years ago. 
otherone  07 Dec 2020 3:03 p.m. PST 
You guys are the best! Thanks! Rob 
