On an infinitely sided die, is it possible to roll and get the same number twice?
In music class, our teacher was saying that every sound we make is uniquely different and has never been heard before. So I kind of thought of a different problem, with the dice. Is there some type of way to calculate the chance? It can’t be zero or can it?
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$\begingroup$First of all, such a die does not exist.
Thus speaking of probability assignment to each event takes us to defining $\frac {1}{\infty} =0$
On the other hand, if probability of some event is zero, it does not mean that it does not happen. Since these events are independent, the next face could as well be the same as the previous one.
$\endgroup$ 4 $\begingroup$I like this question a lot.
So lets say the die is a sphere, with a continuously changing and unique hue on every point of its surface.
The probability of getting a hue in a given range can be calculated by dividing the solid angle that contains that hue by $4\pi$ (the solid angle of the whole sphere). In this case we are interested in an infinitesimal solid angle to capture one colour only.
Let's call the 'area' $\delta$ and we will take $\delta \rightarrow 0$ in the end. Let us call the total number of ways the sphere could land (or exact colours) we could obtain $C$ and we will take $C \rightarrow \infty$ in the end.
The number of ways two dice could fall is the product of each individual possibility; $C^2$ (infinite number of possibilities, like in the music example). The number of ways they could fall and have hues in the same area is;
$$ C * \left(C\frac{\delta}{4\pi}\right) $$That is, the number of ways the first die can land times the number of ways the second die can land times the fraction of positions of the second die that would be the same hue.
So now the probability of getting the same hue on the two dice is the number of same hue outcomes divided by the total possible outcomes;$$ \left(C * \left(C\frac{\delta}{4\pi}\right)\right)/(C^2)= \frac{\delta}{4\pi} $$
Directly proportional to the area of the die that we consider to be the same hue. As we take $\delta \rightarrow 0$ to require an exact colour match this probability also goes to $0$.
$\endgroup$ 3 $\begingroup$It depends on how you model the die. For instance, following your comment to another question, you might assume that the die is a sphere, and that rolling it means sampling a single point on its surface using a uniform distribution.
If you do so, the probability of landing on a given region of that sphere is given by
$$ \dfrac{\mbox{area of the region}}{\mbox{area of the whole surface}} $$
Here one should define "area" precisely (and possibly cope with the fact that there might be regions without an area). Let's assume, say, the Lebesgue measure, which feels quite natural.
So, rolling a point on the upper "hemishpere" is $1/2$.
The probability of rolling a single point $x$ is zero, since the area of the set $\{x\}$ is zero.
If we roll twice such die, we are sampling from two independent random variables. The probability of landing on the same point on these rolls is zero.
It's not obvious to visualize this since we are dealing with two 2D surfaces, so two dice make a 4D surface, and the region where they are equal is a 2D subregion of the 4D surface. Let's consider instead an example with lower dimension.
Assume the die is a circle instead. We roll a point on the circumference. Cut the circumference, and straighten it, so that it becomes an interval $[0,1)$. We can assume that rolling the die gives us a point $0\leq x < 1$ with uniform probablity.
In this scenario, rolling twice the die means choosing a (uniformly) random point in the square $S = [0,1)\times [0,1)$, each coordinate of the point $p=(x,y)$ being the result of the individual roll.
The case where the two rolls are equal is the event $E=\{(x,y)\ |\ 0\leq x=y< 1\}$, i.e. the diagonal of the square $S$.
What is the probability of $E$? Well, it is
$$ \dfrac{\mbox{area of }E}{\mbox{area of } S} $$
Since $E$ is only a line in a square, a one-dimensional object in a two-dimensional space, it has area zero, so the probability of $E$ is zero as well.
$\endgroup$ $\begingroup$If you roll a two sided dice twice then the chance of getting the same number twice is $\frac{1}{2}$. If you roll a three sided dice twice then the change is $\frac{1}{3}$ and so if you roll a dice with n sides twice the change of getting the same number is $\frac{1}{n}$.
Now as the number n gets larger $\frac{1}{n}$ gets smaller. If you take the limit of this to infinity (basically asking what number does this function get close to as n gets larger without bound) $\lim_{n \to \infty}\frac{1}{n} = 0$. You can see that you would roll the same number twice with probability zero.
However "With probability zero" is an odd thing. If you consider all the possible combinations of dice rolls there are clearly it is possible to happen. However all the other possibilities are so much more numerous as to crowd them out. This sort of suggests that talking about probabilities with infinite things might not be a sensible thing to do.
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