What are the chances of getting at least 3 fours when rolling 5 dice?

Question

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

Thanks.

Answer 1

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 $k=C$.

Answer 2

As Andre Nicolas pointed out, for this particular example we can get the answer quickly due to symmetry. But if we want to add the successful events and divide by all possible outcomes (which can also be used in nonsymmetric cases, and suggested in Andre's remark) this is how one can proceed. There are three successful dice rollings: A) We get "exactly" 3 dice with four or higher, B) We get "exactly" 4 dice with four or higher, and C) We get "exactly" 5 dice with four or higher.

Now, we count in how many ways, A , B, and C are possible. Let us focus on A. There are 10 ways to choose 2 (or 3) dice out of five. For those 2 dice in case A we must get a number less than 4 ($3\times3$ cases), and for the remaining three dice we must get a number more than four ($3\times3\times3$ cases). So in total the number of all outcomes for case A is $$10 \times 3 \times 3 \times 3 \times 3\times 3.$$ Now, let us focus on B. There are 5 ways to choose 1 (or 4) dice out of five. For that 1 die in case B we must get a number less than 4 ($3$ cases), and for the remaining four dice we must get a number more than four ($3\times3\times3 \times 3$ cases). So in total the number of all outcomes for case A is $$5 \times 3 \times 3 \times 3 \times 3\times 3.$$ Finally, let us focus on C. In this case for all five dice we must get a number more than four, which gives $$ 3 \times 3 \times 3 \times 3\times 3$$ cases. Adding these successful cases we obtain $16\times 3^5$. So the probability of success rollings is $$\frac{16\times 3^5}{6^5}=\frac{1}{2}.$$