Yes. That means the game cannot go on for infinite time.
No! "The game will end in finite time with probability one" does not mean the game cannot go on for infinite time.
Let me give another example. Randomly select a real number between 0 and 1. The probability of you picking that number p is 0. If it were more than 0, then by the Archimedean property of the real numbers, there is a integer n such that np > 1; that is, there is a greater then 100% chance of picking one of n numbers. But we know there are more then n real numbers between 0 and 1, so the odds of picking any one number is 0, even though you're guaranteed to pick some number.
But, we can see scenarios where it clearly does not complete in finite time.
A set of zero probability.
I am not sure where you're getting that. It seems to me that each game is independent. The number of games remaining to be completed does not depend upon the number already played.
Right, each game is independent. It's like we're flipping coins, and every time someone gets heads without the other person getting heads, they win. There's a 50% chance a round of someone winning, so the number of games that extend to round n+1 is half that of the number of games that extend to round n. The probability of a game continuing to n+1 rounds is (1/2)^n, so the probability of any game continuing to infinity is lim as n goes to infinity of (1/2)^n, which equals 0. Same thing here, with slightly more complicated numbers.
There are examples of games going to infinity, but the probability of any or all of them happening is zero.