1
$\begingroup$

My understanding of some statistics:

For a given experiment a finite number of samples can be taken, defining the sample size. However this experiment may have an infinity large population size.

Given that in an infinitely large population size any possible outcome will occur an infinite number of times, how can we truly say one outcome is more likely than another. Mathematically lets say from our sample we've found that there is a 0.1 chance of drawing a blue ball, and a 0.0000001 chance of drawing a red ball. If the balls are always replaced these odds do no change. However since the ball can be drawn any number of times this will happen:

blue ball count: 0.1 * x red ball count: 0.0000001 *x

where x is any number. Clearly the odds do not change with increasing x. But if x is infinitely large then:

blue ball count: 0.1 * infinity = infinity red ball count: 0.0000001 8 infinity = infinity

So how can we confidently state any odds when the population size is infinity?

P.S. This is just out of interest so I would appreciate as much detail, links etc. as possible

$\endgroup$
1
  • 1
    $\begingroup$ This is one reason the concept of measure had to be invented. Problems like this one show that probability is a number that is associated with certain kinds of subsets of a "population" rather than with individual members of the population. An important consequence of this generalization was the realization that in uncountably infinite populations, not every set can always have a probability associated with it (in a consistent way). In particular, probability generally is not a "ratio of the number of favorable cases to all cases." $\endgroup$
    – whuber
    Apr 14 '14 at 16:55
2
$\begingroup$

Practically speaking, there cannot be an infinite population of balls - there isn't room in the universe.

However, your multiplying infinity by .1 doesn't relate to the problem.

A sample is random if any combination of objects is equally likely. This is possible even with an infinite population. For example, if the balls are numbered $1 - \aleph_0$ then what is required is that, in a sample of size 3 (for example) the probability of picking

1, 2, 3

is the same as

100020101, 2912091, 209101100021

$\endgroup$
4
  • 1
    $\begingroup$ That is quite clever. $\endgroup$ Apr 12 '15 at 6:57
  • 1
    $\begingroup$ It seems like you can also create transitive relationships. For instance, is it correct to say that the chance of drawing 1,2 in a row is greater than the chance of drawing 1,2,3? $\endgroup$ Apr 12 '15 at 6:59
  • $\begingroup$ Yes, because one contains the other as a subset. $\endgroup$
    – Peter Flom
    Apr 12 '15 at 11:00
  • $\begingroup$ And is it equally correct to say that the chance of either occurring is 0? Since we are picking an integer out of all integers? $\endgroup$ Apr 13 '15 at 21:37
0
$\begingroup$

I think your confusion stems from seeing probability as based on a number as opposed to a relative frequency. If you actually drew an infinite sample with replacement, then determining the probability via counting is pointless, as you have shown.

However, you can use the frequentist notion of probabilty (see 1, 2) to get an answer:

Let R = Draw a Red Ball, B=Draw a blue ball, then the relative frequency definition of probability is:

$P(R) \equiv \lim\limits_{n_{R+B}\rightarrow \infty} \frac{n_{R}}{n_{R+B}}$

What this definition says is that its the relative fequency of the occurance of red vs. blue draws that defines the probability. Your infinite sample size experiment merely takes the relative frequency notion of probability to its logical extreme, and will result in the calculation of the exact relative frequency of blue balls to red balls, even though both are infinite.

The actual numbers you are using are confusing, because they don't add to 1. However, if you state that there is a 0.1 chance of getting a red ball, then you would expect to get 1 red ball out of every 10, or, in the limit, your proportion of red balls will approach 0.1.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.