Consider this:
Five cards are individually drawn from a shuffled 52-card deck. The corresponding probability space is (Ω, F, ℙ). Define a random variable $X$: Ω → ℝ, which has the same probability mass function $f$, as the random variable representing the value of a single roll of a dice. I.e. $f(x)$ = ℙ{$X = x$} = 1/6, when $x$ ∈ {1,2,...,6}.
As a hint: 48 is divided by 6, and if we have a fixed a set of 48 cards, then at least one of them is included in a set of five cards.
As a fair die has the same chance of showing each face, you are asked to come up with a strategy that takes the set of five card draws from a standard pack and produces a random number from $1$ to $6$ with all numbers equally probable. If we were given five flips of a coin we would note that there are $32$ possible results of flipping a coin five times. We would choose five of those to represent $1$, five to represent $2$ and so on. We would have two left over and if we got one of those we would have to flip again.
There are $52\cdot 51\cdot 50\cdot 49\cdot 48=311875200$ ordered five card draws from a deck. As this is divisible by $6$ you can separate them into six equal groups and always have a result of a number $1$ to $6$. The hint suggests one way that is easier to remember than others.