What are the chances of getting at least three or more "fours" or higher number when rolling a fair six-sided die five times?

Actually, this is a problem I am curious about the answer and I can't find it anywhere. I would be glad if explained.

The question can be generalized to getting at least $A$ times a number $B$ or higher when rolling $C$ dice of $D$ sides. I would like to know the function that returns this probability given $A$, $B$, $C$ and $D$ parameters.

Examples of success: $(1,4,4,3,4)$, $(2,6,4,3,5)$, $(6,1,5,5,4)$;

Examples of failure: $(3,4,4,1,1)$, $(6,5,1,2,1)$, $(5,1,4,3,3)$.


  • 3
    $\begingroup$ Use the binomial distribution. $\endgroup$ – nomen Jun 5 '14 at 0:07
  • $\begingroup$ You got 3 numbers higher than or equal to four, and that is valid. $\endgroup$ – Fagner Jun 5 '14 at 0:14

Because of the special choice of numbers, we can find a simple expression for the probability.

Call a number $\ge 4$ a success. The probability of success when tossing a single die is $\frac{3}{6}$, that is, $\frac{1}{2}$.

By symmetry, the probability of $3$ or more successes in $5$ trials is the same as the probability of $3$ or more failures. But the events "$3$ or more successes" and "$3$ or more failures" are disjoint, and together they encompass all possibilities. It follows that the required probability is $\frac{1}{2}$.

Remark: For the general problem, there will not be a simple expression. You can easily determine the probability $p$ of success on any one trial. The probability of $k$ successes in $n$ trials is $\binom{n}{k}p^k(1-p)^{n-k}$. Add up these expressions over the desired range. So in the case of your parameters, we have $n=C$, and we sum from $k=A$ to $k=C$.


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