# Design of a peculiar Wheel of Fortune

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 $n>2$ ?

• I have a problem with your second computation : if you take order into account (we are talking probabilities, right ?), then I find $s=3^n-2^{n+1}+1$, which is clearly different from your answer. – Nicolas FRANCOIS Aug 7 '18 at 11:23
• First a note that in your final block you write $c^2$ instead of $\gamma^2$. I don't quite see the problem yet. You want to have a design that fixes the problem. But do you want this for general $\alpha, \beta, \gamma$ and $n$? If you fix $\alpha,\beta=1$ we can find an expression for $\gamma$ in terms of $n$ that satisfies it for all $n >2$. Or are you looking for a fixed combination of $\alpha,\beta,\gamma$ such that this holds for all $n>2$ such that $\gamma$ does not depend on $n$? – Jan Aug 7 '18 at 11:26
• What @NicolasFRANCOIS is trying to say is that while the amount of distinct events may be smaller, the probability of them is not the same. Back to the dice example: If you throw 2 dice there are 11 different possible outcome. However the probability of throwing a sum of 2 is 1/36 while the probability of throwing a 7 is 1/6. – Jan Aug 7 '18 at 11:53
• No. The probability to get the green sector $n$ times if you turn this wheel $n$ times is $3^{-n}\ne\binom{n+2}n^{-1}$. Not distinguishing between the turns doesn't just scale the numerator and the denominator; it leads to different ratios. Of course you're right that these ratios also satisfy the probability axioms; and you may even be able to construct some physical system in which they occur as probabilities. I'm just saying that the wheel you describe is not that physical system, and talking about these ratios as probabilities in the context of this wheel is meaningless and misleading. – joriki Aug 7 '18 at 19:05
• @joriki I see your point, that is in agreement with the comments of other users. Thanks for pointing this out. I will think better about it, especially about the physical system satisfying this peculiar definition of probability. Any suggestion, maybe? Thanks again! – user559615 Aug 7 '18 at 19:18

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.
• Well, he didn't ask for the ratio to be lower than $1$ for ALL $n$, did he ? So maybe you can take $c=n$, in which case all you have to do is take $\alpha>1$ or $\beta>1$... – Nicolas FRANCOIS Aug 7 '18 at 11:57
• If you are asking it as a function of $n$ you can simply set $\alpha,\beta$ to 1 and easily find $\gamma$ as a function of $n$ that would satisfy that constraint. – Jan Aug 7 '18 at 11:58