[Math] Rolling three dice – sum of two results equals third one

combinatoricsprobability

Is my answer correct?

Three six-sided dice are thrown, what is the probability that the sum of two of them equals the third one?

If order does not matter, there are 9 such possibilities:

  1. (1, 1, 2)
  2. (1, 2, 3)
  3. (1, 3, 4)
  4. (1, 4, 5)
  5. (1, 5, 6)
  6. (2, 2, 4)
  7. (2, 3, 5)
  8. (2, 4, 6)
  9. (3, 3, 6)

The total number of possibilities is $\frac{6^3}{3!}=6^2=36$. Therefore, the probability is $\frac{9}{36}$.

Best Answer

If you imagine there are three distinguishable dices (either different colors, or you throw them one after the other), then there are $6^3=216$ possible cases.

Now your enumeration counts down to :

$(1,1,2)$ : $3$ cases

$(1, 2, 3)$ : $6$ cases

$(1, 3, 4)$ : $6$ cases

$(1, 4, 5)$ : $6$ cases

$(1, 5, 6)$ : $6$ cases

$(2, 2, 4)$ : $3$ cases

$(2, 3, 5)$ : $6$ cases

$(2, 4, 6)$ : $6$ cases

$(3, 3, 6)$ : $3$ cases

for a total of $45$ cases, and a probability of $\frac{45}{216}=\frac{5}{24}$.

Hope I did it the right way :-(

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