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Consider an experiment of simultaneously spinning 3 spinners. Each spinner is equally divided into 6 numbered sections, 1 through 6. What percent of time will the sum equal 6?

I've tried listing out all possible combinations of the spinner numbers, but that is an infeasible strategy. I am truly stumped on what to do next or how to solve this problem. Any help would be greatly appreciated!
So Otsuru
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    How many ways can three ordered natural numbers sum to $6$? – lulu May 31 '24 at 18:06
  • "but that is an infeasible strategy"... not for just $3$ spinners! There are indeed only three (up to reordering): $1 + 1 + 4$, $1 + 2 + 3$, $2 + 2 + 2$. – K. Jiang May 31 '24 at 18:08
  • Presumably, it is infeasible because you want to calculate the same probability for other spinners and goal numbers. The general solution uses either inclusion-exclusion or generating functions. – Thomas Andrews May 31 '24 at 18:12
  • In thia particular case, you can just use stars-and-bars, and get $\binom 52.$ – Thomas Andrews May 31 '24 at 18:14
  • For the general problem of $~k~$ spinners, each divided into sections representing each element in $~{1,2,\cdots,n},~$ where you want to know the probability that the $~k~$ spinners will sum to $~M,~$ this is a Stars and Bars problem. For Stars and Bars theory, see this article and this article. – user2661923 May 31 '24 at 20:00
  • Re previous comment, for Stars and Bars problems, where there is (potentially) an upper bound on each variable, and/or a lower bound on each variable, see this answer for a blueprint of how to combine Inclusion-Exclusion with Stars-And-Bars to attack this generic type of problem. – user2661923 May 31 '24 at 20:02

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In your case, it's easy to least all possible arrangements

$(1,2,3)$, 6 permutations;

$(1,1,4)$, 3 permutations;

$(2,2,2)$, 1 case.

So the probability is $$ \frac{6+3+1}{6^3}=\frac{5}{108}. $$

van der Wolf
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