I am having trouble understanding how you get $91/216$ as the answer to this question.

say a die is rolled three times

what is the probability that at least one roll is 6?

  • 1
    $\begingroup$ Well, what's the probability of getting no 6s ? "None" is so much simpler to calculate than "at least"! $\endgroup$
    – DJohnM
    Mar 10 '13 at 1:59
  • $\begingroup$ not getting a 6 would be 5/6. The probability of not rolling a six 3 times is (5/6)(5/6)(5/6) = (5/6)^3. I think I am getting this. Just a bit confusing. $\endgroup$
    – Josh
    Mar 10 '13 at 2:02


$1 - \left(\dfrac{5}{6}\right)^3\; \;$ to obtain the probability that at least one 6 is rolled in three tosses.

That is, we find the probability that at least one $6$ is rolled by subtracting from $1$ the probability that it does not appear on any of three rolls:

The probability that $6$ does not appear given one roll is $5/6.\;$ The probability that a $6$ does not appear on three roles of the die is given by $$\left(\frac 56\right)^3 = \dfrac{125}{216}$$

  • We use $1$ to represent certainty: that is 100 percent probability. It must be the case that

    P(Not rolling at least one 6 on three rolls) + P(rolling at least one 6 on three rolls)] =1

    $\implies 1 - $ (probability of not rolling at least one 6) = (probability of rolling at least one 6).

So the probability of obtaining a $6$ on at least one roll is:

$$1 - \dfrac{125}{216} \quad = \quad \dfrac{216}{216} - \dfrac{125}{216} \quad = \quad\dfrac{91}{216}$$


Rolling a single die, the probability that it does not show a 6 is $5/6$. So rolling three dices, the probability for no 6 is $(5/6)^3$, and therefore the probability for at least one 6 is $$1-\left(\frac{5}{6}\right)^3 = \frac{91}{216}.$$

  • $\begingroup$ Why the 1 - though? $\endgroup$
    – Josh
    Mar 10 '13 at 2:02
  • $\begingroup$ It turns the probability for "no 6" into the probability of the complement "NOT no 6", which is the same as "at least one 6". $\endgroup$
    – azimut
    Mar 10 '13 at 2:04
  • $\begingroup$ @Josh The possible cases are 0, 1, 2, 3 sixes. If you ask What is the probability that I roll 0 or 1 or 2 or 3 sixes?, the answer is 1, since one of these cases must happen. Now in your scenario, the only case you do not want is 0 sixes, since for the other three possibilities the statement at least one six rolled is always true. Therefore you need to remove the possibility of getting 0 sixes from the probability above (which is 1, as said). $\endgroup$
    – Karlo Grba
    Jul 5 '16 at 10:16

There are two answers already that express the probability as $$1-\left(\frac56\right)^3 = \frac{91}{216},$$

I'd like to point out that a more complicated, but more direct calculation gets to the same place. Let's let 6 represent a die that comes up a 6, and X a die that comes up with something else. Then we might distinguish eight cases for how the dice can come up:





We can easily calculate the probabilities for each of these eight cases. Each die has a $\frac16$ probability of showing a 6, and a $\frac56$ probability of showing something else, which we represented with X. To get the probability for a combination like 6X6 we multiply the three probabilities for the three dice; in this case $\frac16\cdot\frac56\cdot\frac16 = \frac5{216}$. This yields the following probabilities:

$$\begin{array}{|r|ll|} \hline \mathtt{666} & \frac16\cdot\frac16\cdot\frac16 & = \frac{1}{216} \\ \hline \mathtt{66X} & \frac16\cdot\frac16\cdot\frac56 & = \frac{5}{216} \\ \mathtt{6X6} & \frac16\cdot\frac56\cdot\frac16 & = \frac{5}{216} \\ \mathtt{X66} & \frac56\cdot\frac16\cdot\frac16 & = \frac{5}{216} \\ \hline \mathtt{6XX} & \frac16\cdot\frac56\cdot\frac56 & = \frac{25}{216} \\ \mathtt{X6X} & \frac56\cdot\frac16\cdot\frac56 & = \frac{25}{216} \\ \mathtt{XX6} & \frac56\cdot\frac56\cdot\frac16 & = \frac{25}{216} \\ \hline \mathtt{XXX} & \frac56\cdot\frac56\cdot\frac56 & = \frac{125}{216} \\ \hline \end{array} $$

The cases that we want are those that have at least one 6, which are the first seven lines of the table, and the sum of the probabilities for these lines is $$\frac{1}{216}+\frac{5}{216}+\frac{5}{216}+\frac{5}{216}+ \frac{25}{216}+\frac{25}{216}+\frac{25}{216} = \color{red}{\frac{91}{216}}$$ just as everyone else said.

Since the first 7 lines together with the 8th line account for all possible throws of the dice, together they add up to a probability of $\frac{216}{216} = 1$, and that leads to the easier way to get to the correct answer: instead of calculating and adding the first 7 lines, just calculate the 8th line, $\frac{125}{216}$ and subtract it from 1.


probability of getting at least a six on rolling = 1- probability of getting no sixes.

the probability of getting no sixes on one rolling is $\left(\frac{5}{6}\right) $

so, for 3 times rolling , we have that, the probability of getting no sixes is $\left(\frac{5}{6}\right)^3 = \frac{125}{216} $

so, the probability of getting at least one six is $1- \left(\frac{5}{6}\right)^3 = \frac{91}{216}$ i.e your answer.
I think this could be helpful for you


For rolling 3 dice without getting a six, to get the probability, pretend your 6-sided dice are 5-sided dice. That can help you figure out the probability. Therefore, the probability is 125/216.


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