Consider the two wheels of fortune illustrated below.
The first one (left) is constituted by $c$ sectors with the same arc length: $\alpha$ of which are red, $\beta$ of which are blue and $\gamma$ of which are green.
The second one (right) is constituted by three sectors with the same arc length: One is red, one is blue and one is green.
On the first wheel, we perform $2$ turns, and we evaluate the ratio $r$ between the probability "to get at least one red sector and at least one blue sector" and the probability "to get both times a green sector", assuming that the order of the turns matters.
It is easy to show that this ratio is
$$r=\left[\left(\frac{\alpha}{c}\right)\left(\frac{\beta}{c}\right)+\left(\frac{\beta}{c}\right)\left(\frac{\alpha}{c}\right)\right]/\left[\left(\frac{\gamma}{c}\right)\left(\frac{\gamma}{c}\right)\right]= $$ $$ =\frac{2\alpha\beta}{c^2}/\left(\frac{\gamma}{c}\right)^2=\frac{2\alpha\beta}{\gamma^2}.$$
On the second wheel we perform $n$ turns, and we evaluate the ratio $s$ between the probability "to get at least one red sector and at least one blue sector" and the probability "to get all the $n$ times the green sector", in case the order of the turns does not matter.
It is trivial to show that there are $\binom{k+n-1}{n}=\binom{n+2}{n}$ ways in which we can distribute $n$ indistinguishable turns among $k=3$ distinguishable sectors, in case the order of the turn does not matter. It is also easy to see that, among these $\binom{n+2}{n}$ combinations, only one contains all the $n$ turns in one sector, whereas there are $\binom{n}{2}$ combinations which contain at least one turn in one sector and at least one turn in another one (again, these quantities are evaluated assuming that the order of the $n$ turns is irrelevant).
Therefore, the ratio we are looking for is
$$ s=\left[\frac{\binom{n}{2}}{\binom{n+2}{n}}\right]/\left[\frac{1}{\binom{n+2}{n}}\right]=\binom{n}{2}. $$
My question is
How can we build the first wheel, in case the design requires that $r\cdot s=\frac{2\alpha\beta}{\gamma^2}\cdot\binom{n}{2}<1$, and that $n>2$ ?
Thanks for your suggestions!
This cannot be done without the trivial solutions such as setting $\alpha,\beta=0$. This would lead to $s=0$ and $r=0$ so $s\cdot r <1$ trivially.
Since this is probably not what you are looking for note that when $\alpha,\beta$ and $\gamma$ fixed with $\alpha \cdot \beta>0$. $$\lim_{n \rightarrow \infty}\frac{2\alpha\beta}{\gamma^2}\cdot\binom{n}{2}=\lim_{n \rightarrow \infty}\frac{2\alpha\beta}{\gamma^2}\cdot\frac{n!}{2(n-2)!}=\lim_{n \rightarrow \infty}\frac{2\alpha\beta}{\gamma^2}\cdot n \cdot (n-1)=\infty$$
If you were to make $\alpha,\beta$ or $\gamma$ a function of $n$ this could work.