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I think the answer is $10/216$, but I am not sure. I did it by brute force though, and would like to know the background, as well as knowing if $10/216$ is correct.

Edit: now i think via brute force, the answer is 93/216. if you roll a 1 on any dice there are 3x6 = 18 ways of getting a 6 using any number of dice. similarly for the other numbers.. giving 5 x 18 = 90. now add in the three 6's which adds three more ways... giving 93 total. is the answer 93/216 correct? and if so is there a more elegant was of doing it than manually writing down all the combinations? :)

N. F. Taussig
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nate958
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    First find the probability of no $6$. The number $10/216$ is not right, much too low. – André Nicolas Feb 08 '16 at 23:41
  • You can, by the way, often test these questions: roll 3 dice a bunch of times, see whether the result is closed to $10/216$. – Lee Mosher Feb 08 '16 at 23:46
  • my initial assumption was you had to have a 6 on two dice, i think the problem is more general. meaning probability of a 6 on one roll, using, 1 2 or 3 dice to add up to 6 sorry for confusion – nate958 Feb 08 '16 at 23:50
  • now i think via brute force, the answer is 93/216. if you roll a 1 on any dice there are 3x6 = 18 ways of getting a 6 using any number of dice. similarly for the other numbers.. giving 5 x 18 = 90. now add in the three 6's which adds three more ways... giving 93 total. is the answer 93/216 correct? and if so is there a more elegant was of doing it than manually writing down all the combinations? :) – nate958 Feb 08 '16 at 23:54
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    The probability of not $6$ three times in a row is $(5/6)^3$, so the probability of at least one $6$ is $1-(5/6)^3$. This is $\frac{216-125}{216}$. So instead of your $93$ we have $91$. – André Nicolas Feb 09 '16 at 00:13

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Roll the three dice. Record the result as $(a,b,c)$ where $a$ is the number on the blue die, $b$ the number on the white, and $c$ the number on the red. All $6^3$ such sequences are equally likely.

There are $5^3$ sequences consisting of non-$6$, and therefore $216-125$ sequences with at least one $6$. Thus the probability of at least one $6$ is $\frac{216-125}{216}$.

Note that $216-125$ is $91$, not your $93$. You counted everything correctly, except there is only one sequence $(6,6,6)$, not three.

For this problem, I prefer to work directly with probabilities. The probability of a non-$6$ is $5/6$, so the probability of three of them in a row is $(5/6)^3$, so the probability of at least one $6$ is $1-(5/6)^3$.

André Nicolas
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  • +1 This is the best way, though the direct argument with inclusion-exclusion is not too bad. The number of outcomes with at least one "six" is $(36+36+36)-(6+6+6)+1=91$. –  Feb 09 '16 at 02:17