Dudley’s exercise on finitely additive probabilities Yesterday I found the following exercise in Dudley’s “Real Analysis and Probability”.

In a game, two players, Sam and Joe, each pick a nonnegative integer at random. For each, the probability that the number is in any set $A$ is $\mu(A)$, where $\mu$ is a finitely additive function defined on $2^X$ with $\mu(A) = 0$ for every finite set $A$ and $\mu(X) = 1$, and $X = \mathbb{N}$. The one who gets the larger number wins. A coin is tossed to determine whose number you find out first. It’s heads, so you find out Sam’s and still don’t know Joe’s. Now, who do you think will win?

I was wondering if the following line of reasoning is correct.

The one that will win with probability $1$ is Joe. To prove this, let $m \in \mathbb{N}$ be the number picked by Sam. Then, according to the way in which $\mu$ is defined, $\mu (x > m) = 1$ and $\mu (x \leq m) = 0$. This is so, because the set $\{ x \ | \ x > m \}$ is an infinite subset of $\mathbb{N}$, while $\{ x \ | \ x \leq m \}$ is clearly finite. QED

Is this correct?  
Any feedback is more than welcome as always (even if this exercise looks completely trivial to you), in particular because I feel really uncomfortable with measure theory.  
Thank you for your time and assistance.
 A: 
The one that will win with probability $1$ is Joe. To prove this, let $m \in \mathbb{N}$ be the number picked by Sam. Then, according to the way in which $\mu$ is defined, $\mu (x > m) = 1$ and $\mu (x \leq m) = 0$. This is so, because the set $\{ x \ | \ x > m \}$ is an infinite subset of $\mathbb{N}$, while $\{ x \ | \ x \leq m \}$ is clearly finite. QED
Is this correct?

The problem is that the same reasoning can be applied to show that Sam wins with probability $1$. In particular, let $m \in \mathbb{N}$ be the number picked by Joe. Then according to the way in which $\mu$ is defined, $\mu(x > n) = 1$ and $\mu(x \le m) = 0$. Therefore the probability that Sam picks a higher number is $1$.
You get the idea. Now, this is the whole point of the problem: to show that in an absurd situation, absurd conclusions can be drawn. Here, we can prove that Joe -- or Sam -- wins with probability $1$. But that makes little sense. An alternative interpretation is to consider the fact we find out Sam's answer first to be significant. Then maybe the probability is $1$, but if we found out Joe's number first, then the probability would be $1$ for Sam to win.
You see, the problem is completely symmetric -- Sam and Joe should have the same chance of winning! Yet it seems depending on whose answer we find out first, that person just loses.
So here, simply finding out some information -- even though the actual information, i.e. the actual value of Sam's number, doesn't matter -- seems to change the probability.
This is why we do not allow finitely additive measures to be called "probabilities". They are counterintuitive, and certain desirable properties that we want to be true of probability distributions become false.
A: This is a weird probability distribution, but your reasoning is correct. For any actual value $m$, the set $\{\leq m\}$ is finite and hence has probability zero of being selected. By complement, one of the integers greater than $m$ is almost surely going to be chosen.
A: It is not correct to claim the probability is 1. There are several problems here.  Let $J$ and $S$ be the numbers picked by Joe and Sam.  We are not told the relationship between $J$ and $S$, perhaps $J=S$ always!  Then it is impossible to have $J>S$.  
Another difficulty is that we cannot use the following law of total probability for several reasons: 
$$ P[J>S] = \sum_{i=1}^{\infty} \underbrace{P[J>S|S=i]}_{\mbox{undefined}}\underbrace{P[S=i]}_{0}$$
Problems: 


*

*The "finitely additive" nature of the system does not allow us to sum over a countably infinite number of terms in this way. We indeed have: 
$$ \{J>S\} = \cup_{i=1}^{\infty}\{J>i, S=i\}$$ 
However, this does not allow us to conclude $P[J>S] = \sum_{i=1}^{\infty} P[J>i, S=i]$.  

*The "conditional probability" $P[J>S|S=i]$ is undefined for all $i \in \{1, 2, 3, ...\}$ (because we cannot condition on a probability-0 event). There is no clear mathematical way to work with the information "we are given $S=i$" or to draw conclusions from this information. 

*Standard concepts of “independence” are not helpful: Even if $J=S$ always, $J$ and $S$ are still “independent” according to the standard definition because: 
$$ P[J \leq x, S \leq y] = P[J \leq x]P[S\leq y] =0 \quad \forall x, y \in \mathbb{R}$$

So, in order to “solve” the problem given in your question, you would need to
develop a theory that includes nonstandard definitions of “independence” and 
“conditional probability.” The theory should be consistent and convincing enough 
for people to feel comfortable interpreting the statement of your question within 
the context of your theory. 
