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Consider this question: Difficulty with a "trivial" probability question: drawing balls from an urn --- is event probability affected by order in which balls are drawn?

The answer to the issue there is that in some of the sample spaces generated for a possible solution, the elements of the sample space do not all have the same probabilities. Thus, we cannot count the number of elements in an event, and divide that number by the total size of the sample space to determine the probability that an event occurs.

In general, it seems much easier (even if incorrect) to come up with sample spaces where the elements have different probabilities of occurring, as noted in the above question. For instance, a faulty solution of the question raised above could have been:

The sample space consists of two elements: $$\Omega = \{\text{all drawn balls have a different colour}, \text{there are at least two drawn balls with the same colour}\};$$ thus the probability that we draw balls all of different colour is $1/2$."

But, what axiom/rule prevents the generation of such "broken" sample spaces? My guess is that it is not the sample space that is broken (and the sample space is in fact, okay) but it is the assumption that all elements have an equal probability (i.e. there is a uniform probability distribution) that is faulty?

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    $\begingroup$ Kolmogorov's axioms of probability do not require the sample space to have equally likely outcomes. Even under the same "random experiment", you may define different sample spaces some of which have equally likely structure and some not. For ex, consider tossing a pair of fair six-sided die. If the sample space is taken as $(1,1), (1,2), \ldots, (6,6)$ indicating the ordered pairs of numbers observed, they are equally likely. If under the same scenario, taking the sample space to be the sum $2, \ldots, 12$ or to be the max $1, \ldots, 6$, those sample spaces dont have equally likely outcomes. $\endgroup$ – Just_to_Answer Nov 4 '17 at 20:11
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There are probability spaces in which it is not possible to assign a uniform probability distribution over all points in the sample space. For example, consider the sample space, $$ \Omega = \{ 1,2,3,4,\ldots \}, $$ that is, one whose contents are all of the positive integers.

Often, when presenting a question of probability, a non-uniform distribution over some sample space will be explicitly given. The question really is how we approach a problem in which a distribution is not explicitly stated; when do we assume it is uniform, and how do we deal with cases where we do not make that assumption?

There are some "standard" mechanisms of probability in which the apparently symmetry of a set of outcomes is so strong that we tend to assign them a uniform distribution by default. Dealing cards from a deck or drawing balls from an urn tend to fall in this category. So do the toss of an individual coin or the roll of an individual die, although even then we often use the word "fair" in the problem statement (writing a "fair coin" or a "fair die") in order to distinguish the uniform distribution from a class of problems in which the distribution is not uniform (such as a "weighted die").

This only works as long as we believe that relabeling the outcomes would not change the distribution. For example, if there are $r$ red balls and $b$ blue balls in an urn, the balls are identical in every way except color, and we pull out one ball without being able to see what its color is before we choose it, then it seems "obvious" that painting one of the red balls blue and one of the blue balls red should never have an impact on whether the first ball we pull from the urn is red--and it won't as long as we assume the distribution was uniform to begin with.

On the other hand, if the (original) red balls were all smaller than the (original) blue balls, or all heavier, then the assumption of uniform probability would not work so well. Sometimes problem statements will explicitly say how similar the balls are, as the previous paragraph did, but often this is left out of the problem statement and we are expected to assume it because it's conventional to do so. This is OK, I think, as long as we realize that this is what we're doing and don't think that just because there are $n$ listed outcomes they are all equally likely, or even that if the words "balls in an urn" occur in a story that each ball was necessarily equally likely to be taken from the urn.

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  • $\begingroup$ Very well written. Thank you :) $\endgroup$ – user89 Nov 7 '17 at 22:28

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