0% chance X will happen per definition means X will never happen.
However, if his 0% chance is some sort of "rounding down" or the like, then of course he is cheating.
And with the infinite fine-pointed dart, the problem is
- such thing does not exist
- in that case, the 0% comes as the lines of the function that assigns the probability of hitting number Y when we split the [0;1] into N disjunct and equally sized intervals, then we look at N->infinite.
Since we cannot have infinite, we look where that function goes to.
Problem is, it never reaches 0%, just gets infinitely close to it. ("for every epsilon radius environment of 0 there is a Z: f(x) <= 0 + |epsilon|, no matter how small we pick epsilon" yadda yadda)
He just doesn't know what he talks about.
But, as far as I know, this is secondary school maths. :)
However, I found most interesting the comment made by David Gordon. He took a unique view on Callahan's statement, not made by any of the other 50 plus commenters.
There is an ambiguity in "There's a 0% probability that X will happen". If "X" means "that a number specified before the toss was hit", then the statement is true. If "X" means "that a number was hit" then the statement does not have a 0% probability. On the contrary, it is certain to occur on the conditions set forward in the example. No interpretation of "There's a 0% probability that X will happen" has been offered in which the statement is true and X happens.David, in other words, blows Callahan's comment out of the water right at the starting gate. They don't call David a genius without good reason.
BTW, David was good friends with both Murray Rothbard and Robert Nozick. Their phone conversations must have been epic. Maybe we can someday coax the NSA to release the recordings.