Oliver rolls three fair standard six-sided dice. What is the probability that there is at least one pair of dice whose top faces sum to $6$? Express your answer as a common fraction.

I started by saying that there were $6^3$ total possibilities. For the next step, there are $5$ ways to roll a pair of dice and get $6$ when adding them. There is a third dice, so we multiply by $6$ to get $\frac{30}{216}$ but this is wrong! What did I do wrong? Am I on the right track? Or should I do something else to start? (BTW, yes I did simplify the fraction)

  • $\begingroup$ With questions that ask for the probability that at least one event of a given type occurs, it may be easier to compute the probability that no event of that type occurs, and then subtract that from $1$ to obtain the answer. $\endgroup$ – Brian Tung Sep 25 '18 at 23:43
  • $\begingroup$ Why did you multiply by $6$? $\endgroup$ – N. F. Taussig Sep 25 '18 at 23:56
  • $\begingroup$ Because I was being dumb. I figured it out now. $\endgroup$ – Max0815 Sep 25 '18 at 23:57

I suggest you draw a probability tree. On any toss, the probability of a 6 is 5/36. The experiment ends if you get a 6.

  • 1
    $\begingroup$ Ok. Thanks. Solve it :P $\endgroup$ – Max0815 Sep 25 '18 at 23:53
  • $\begingroup$ I am offering a way to solve the problem in the hope you pick up the baton. That's the purpose of this forum. $\endgroup$ – ncmathsadist Sep 26 '18 at 13:09
  • $\begingroup$ Yes, it is. However...idk this is hard choosing which to check... $\endgroup$ – Max0815 Sep 26 '18 at 15:13

Although you figured it out already, here's why your answer didn't work. (At least in the form you left it)

You artificially created an ordering of the dice. You have shown the possibilities of dice 1 and dice 2 summing to $6$ , but you need to also think about dice 1 and dice 3 summing to $6$, as well as dice 2 and dice 3.

So this will give $30+30+30=90$ ways but then you need to remember inclusion/exclusion to remove all the possibilities you over counted.

  • $\begingroup$ Hmmm... a more specific solution. Thanks. Although @WaveX, please take the answer off. I like hints better than answers because I want to learn, not copy. Thanks! $\endgroup$ – Max0815 Sep 26 '18 at 0:29
  • $\begingroup$ I mean can you edit it so that the answer is gone. $\endgroup$ – Max0815 Sep 26 '18 at 0:31

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