# B What are the odds of getting a number on a hypothetical infinity sided dye?

1. Mar 24, 2016

### KarminValso1724

Let's say for example, there was a dye in which any number with any amount of digits could be scored. You also had an equal chance of scoring every number. Which means that you have the same chance of rolling a 1 as you do 5 billion. If you rolled that dye, how many digits would that number likely be. Considering this, the percentage of numbers that have less than gogol digits compared to an infinite amount of numbers would be infinitesimal . For example, there are a certain amount of numbers that have gogol digits or less. But out of an infinity of numbers, less than 1 in, let's say a quintillion have gogol digits or less. So you have much less than 1 in a quintillion chance of scoring a number that has less than a gogol digits. And that number can be expanded infinitely into something much larger than 1 in a quintillion or gogol digits. Bit a number has to be rolled, and it must represent a finite amount, so how large would that number likely be?

2. Mar 24, 2016

### PeroK

There is no way, practically or theoretically, of selecting a whole number at random, where there is no upper limit on that number.

3. Mar 24, 2016

### mathman

An infinite number of disjoint events with equal probability is impossible, since the total probability has to be 1.

4. Mar 24, 2016

### pwsnafu

Its die, not dye.

Not possible. If there is an equal chance, the die has a finite number of faces.

5. Mar 24, 2016

### WWGD

In a sense that can be made precise, a die with number of sides approaching infinity approaches the shape of a ball.

6. Mar 24, 2016

### FactChecker

The expected value would be infinite. For any proposed finite value, x > 0, you can show that the expected value is larger than x.

7. Apr 8, 2016

### Zafa Pi

With the exception of the spelling correction the previous replies assume that a probability is countably additive. There are many finitely additive probabilities that give probability 0 to each individual number, but certain subsets have probability 1 , including the entire set. For example take any ultrafilter (google it) on the integers and give any member of it probability 1 and any non-member probability 0.

8. Apr 9, 2016

### pwsnafu

Those are quasi-probability. The axioms of probability as commonly understood require countable additivity.

9. Apr 9, 2016

### Zafa Pi

They are more commonly called finitely additive measures (probabilities). Quasi-probabilities are generally finitely additive, but often have other properties as well and are found in quantum mechanics.
I am a bit surprised that you didn't think I was aware of the usual definition of a probability.
Everything in my post was correct and I was just adding a little pazzazz.

10. Apr 15, 2016

### zinq

I have discovered ways that, theoretically, a random integer can be selected so that all integers are treated with perfect symmetry.

But of course, this does not translate into any kind of real-valued countably additive measure on the integers for the usual reason: There is no real number which, when added to itself countably many times, sums to 1.

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P.S. "In a sense that can be made precise, a die with number of sides approaching infinity approaches the shape of a ball" is not true. There are more than countably infinitely many ways to have a (compact convex) polyhedron in R3 with countably many faces.

For instance, just take a circle in the xy-plane and mark one point P, and then also points at these fractions of the way around the circle from P: 1/2, 3/4, 7/8, 15/16, .... Connect these successively by lines to get an infinite-sided polygon. Finally pick the point in 3-space that is 1 unit above the center of the circle and use this to create a cone on the polygon. This is the convex hull of all the points we used, so it is convex.

(There are also ways to do this so that each of the infinitely many faces of the polyhedron is just a finite-sided polygon.)

Last edited: Apr 15, 2016
11. Apr 15, 2016

### WWGD

EDIT: I may have been too snarky here:this isthe notion of convergence of a sequence of metric spaces , as used in Metric Geometry, which I am referring to

Specifically, the claimed statement is: there is a sequence of maps $f_n: X_n \rightarrow S^1$ from compact metric spaces and a sequence $s_n$ so that each $f_n$ is an $s_n$-isometry. This is statement 7.5.8 in Gromov's "Metric Geometry". A similar argument can be made for $S^2$ and, AFAIK for $S^n$. If you find something flawed with this statement please let me know.

Last edited: Apr 15, 2016
12. Apr 15, 2016

### zinq

WWGD, I'd really like to see a quote that makes that claim. (And maybe you're speaking of the book by Burago, Burago, and Ivanov?)

13. Apr 15, 2016

### WWGD

Yes, it is the BBIvanov book. Unfortunately I have it in a box somewhere, but I did look up the quote in Google books. I will look up the book; I don't have access to a university library where I stand a resonable chance to find it, but if I do find it, I will state the actual quote.

EDIT: I found this: http://math.stackexchange.com/quest...dorff-convergence-to-a-circle/1116840#1116840

Last edited: Apr 15, 2016
14. Apr 16, 2016

### Hornbein

Die with an infinite number of faces? The probability that any given face is rolled is zero. Proof: any number greater than zero leads to a contradiction.

Get used to it.

It IS possible to choose an integer at random with for example P[X=n] = 1/2^n.