# Probablity & infinity

## Main Question or Discussion Point

If there is an actual infinity of throws of two dice, would all combinations have the same probability (that is each combination would happen an infinite amount of times)?

Related Set Theory, Logic, Probability, Statistics News on Phys.org
mfb
Mentor
They have the same probability for every throw (assuming they are fair).

(that is each combination would happen an infinite amount of times)
That would even happen with loaded dice.
You can make a stronger statement: if you consider the fraction of [result X] in the first N throws, this fraction goes to the same value (1/36 for regular dice) as N->infinity.

They have the same probability for every throw (assuming they are fair).

as N->infinity.
Does N-> Infinity imply a potential infinity instead of actual infinity?

Does N-> Infinity imply a potential infinity instead of actual infinity?
When we discuss heat, we never start with "Oh, fire is caused by phlogiston. But more recent thinkers consider it a chemical process." No. We discard ancient, obsolete ideas, and adopt the new ones.

It's nice that Aristotle talked about actual and potential infinity. If you're discussing the history of philosophical ideas about infinity, it's good information to know.

But since the work of Cantor in the 1880's and the set theorists of the 20th century, we now have much more precise concepts of mathematical infinity.

Let us dispense with the meaningless concepts of "potential" and "actual" infinity. Those terms have no meaning in modern mathematics.

An infinite set of dice rolls would be a function f:N -> {1,2,3,4,5,6} where N is the set of natural numbers. That's how to think about it and talk about it.

Stephen Tashi
An infinite set of dice rolls would be a function f:N -> {1,2,3,4,5,6} where N is the set of natural numbers. That's how to think about it and talk about it.
And for fair dice, each such set of rolls has probability zero.

As to whether a given infinite set of dice rolls happens an infinite number of times, you'd have to say how many infinite sets of rolls you intend to toss for that question to have any meaning.

They have the same probability for every throw (assuming they are fair).

I might not have my terms correct. Craps is based on 7 happening more often (I think the term for that is probability). But if there were an infinite number of throws, won't all combinations happen an equal (that is infinite) number of times?

Office_Shredder
Staff Emeritus
Gold Member
They have the same probability for every throw (assuming they are fair).

I might not have my terms correct. Craps is based on 7 happening more often (I think the term for that is probability). But if there were an infinite number of throws, won't all combinations happen an equal (that is infinite) number of times?
When you think about doing a process an infinite number of times, saying 'each pair of dice comes up an infinite number of times' is not good enough. It's nice to know but if you want to say that all possibilities really come up equally often, then what you are really interested in is the following:

Let Xn, Yn be the results of the two dice on the nth throw, and let
$$S_N = \left( (X_1,Y_1),(X_2,Y_2),....,(X_N,Y_N) \right)$$

To say that, for example the pair (2,2) occurs as often as the pair (4,1), what you are interested in is (my notation here isn't great but should be comprehensive)
$$\# \left( (2,2) \in S_N \right) /N = \# \left( (4,1)\in S_N \right)$$
that is that you want the number of times (2,2) shows up in SN to be equal to the number of times (4,1) shows up. Now this is generally not going to be true, but what is true (and is what we mean when we say that (2,2) and (4,1) have equal probabilities) is that
$$\lim_{N\to \infty} \frac{ \# \left( (2,2) \in S_N \right)}{N} = \lim_{N \to \infty} \frac{ \# \left( (4,1) \in S_N \right)}{N}$$

those limits are just going to be the probability of (1,2) showing up in any roll, or (4,1) showing up in any roll, which is 1/36. If you had a loaded die which came up on a value of 1 with a probability of 9/10 and a 2,3,4,5,6 each with probability 1/20, then we would get that every pair (a,b) still shows up infinitely often as N goes to infinity, but you would not have
$$\lim_{N\to \infty} \frac{ \# \left( (2,2) \in S_N \right)}{N} = \lim_{N \to \infty} \frac{ \# \left( (4,1) \in S_N \right)}{N}$$
because the left hand side would be 81/100 and the right hand side would be 1/400.

mfb
Mentor
They have the same probability for every throw (assuming they are fair).

I might not have my terms correct. Craps is based on 7 happening more often (I think the term for that is probability). But if there were an infinite number of throws, won't all combinations happen an equal (that is infinite) number of times?
"7" (as a sum?) is not one combination, but 6 (out of 36).

If there is an actual infinity of throws of two dice, would all combinations have the same probability (that is each combination would happen an infinite amount of times)?
No.

You are applying concepts from probability on a finite set to an infinite set. This gives no useful result.

• 1 person