Is the Axiom of Choice inconsistent with Countable Additivity?

Question

Consider a fair lottery among a countably infinite number of people. The Axiom of Countable Additivity says this is impossible to construct: If all people have a positive (and equal) probability of winning, the sum of their probabilities of winning would diverge, while if all people have a $0$ probability of winning, the sum of their probabilities would be $0 \neq 1$.

However, I feel like the Axiom of Choice can be used to construct such a lottery:

1. Have each person select an element from $[0, 1)$. This is possible even with the Axiom of Countable Additivity.

Lemma: With probability $1$ everyone will have different numbers.
Proof:
Let $X_i$ be the number selected by $i$. Let $Y_{ij} = X_i - X_j \mod 1$. In order for two numbers to be equal with nonzero probability, there must be a nonzero probability that at least one $Y_{ij}$ equals $0$. But for all $i$ and $j$, $Y_{ij}$ is distributed uniformly and at random on $[0, 1)$. This means that there is this same nonzero probability that at least one of the $Y_{ij}$ equals $x$ for every $x \in [0, 1)$. Countable additivity would then lead to a divergent sum when considering the probabilities at least one of $Y_{ij} = 1/2$, $Y_{ij} = 1/3$, $Y_{ij} = 1/4$, and so on.

3. Choose a well-order on $[0, 1)$ using the Axiom of Choice.
4. Declare the person with the least number according to this well-order the winner.

What goes wrong with this reasoning? Or is it actually the case that these two axioms are inconsistent with each other?

Answer 1

Presumably your sample space is $[0,1]^\mathbb N$ with probability measure the product of Lebesgue measure on each coordinate. Then "events" such as "person #1 wins the lottery" are not events, i.e. they are not measurable sets for this measure, and as such can not be assigned probabilities.

Answer 2

Edit: In this answer I'm assuming the well-order of the reals is fixed at the outset and not somehow "chosen randomly", though, as commenter on another answer has woken me up to, you phrased the question as if were part of a random process. This would be a whole other can of worms (What distribution does this come from?! Is it independent of the sequence drawn?) but it doesn't affect the upshot of this answer.

Edit 2: Decided I didn't quite agree with how I put my original answer so rewrote it a bit.

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Here's a simpler example, that misses some features of yours but is more direct.

Say we have two players, and a non-Lebesgue-measurable set $E\subseteq [0,1],$ and some measure-zero set $C\subseteq [0,1]\setminus E.$ Draw a value $x\in[0,1]$ uniformly at random (i.e. via the Lebesgue measure on $[0,1]$), and player 1 wins if $x\in E$, it is a draw if $x\in C,$ and otherwise player 2 wins.

Unlike in your case, there's no intuition telling us that player 1 and 2 have equal probability to win, but some other features are there. For instance, since the draw set is measure zero, you would expect the probabilities of player 1 winning and player 2 winning to add up to $1$, as you expect in your case since the event of a tie (subsumed by the event of not having all points of the sequence distinct) is measure zero.

But of course we said we'd pick $x$ according to the Lebesgue measure, then asked what the probability of an event outside the Lebesgue measure was. There is no answer to this question.

Your example is not different in any essential way. In order to say you have 'equal nonzero probabilities for all players' you'd need to show that the events that they win are 1) measurable according to the probability space your sequence is coming from 2) have equal measure and 3) partition almost all of the probability space. If you could do that, there'd be a problem, since as your initial argument (or just common sense) shows, there cannot be infinitely many disjoint measurable sets with the same probability... the probabilities would have to add to infinity. But you can't do that, or at least, per Robert Israel's answer, there's no obvious reason to believe you can pass step 1.

So what can we say about these scenarios? Do they meaningfuly constitute a "lottery"? While we certainly couldn't execute either of these plans in the real world, perhaps it's not too much strain on the mathematical imagination to imagine actually having in hand some 'randomly drawn' $x\in[0,1]$... and if we did, it would certainly either be in $E$, or in $C$, or in neither. We could further imagine doing this over and over, tabulating the results. What would happen?

I don't know of any way of making sense of this, though we seem to have left mathematics and entered philosophy. (But perhaps it's just my ignorance. This answer seems to suggest that you can use outer measure to represent limiting frequencies, but I haven't looked into it, and it seems rather drastic to give up countable additivity.) The abstract probability space formalism is our typical way of making rigorous mathematical sense of the idea of probability, so it seems odd to try to operate outside its jurisdiction on a question like this. Perhaps we've just hit a limit of how much we can extrapolate our intuition into math.

Answer 3

The other answers give the theoretical reasons why this doesn't contradict the axioms of probability. In practice, what goes wrong is as follows. Let's assume you have assigned random reals to each player, that is, each player has an infinite string of digits, and they're all different. You also have chosen a well-ordering on the reals.

Now you want to decide who has won. How do you do that? There is no obvious algorithm, even one that uses infinitely many steps, because comparing approximations for these reals will never stabilise to the actual ordering.

So you haven't really constructed a lottery, fair or otherwise, since there's no way to carry out step 4.