The Axiom of Determinacy (AD) fails.
What that means: Partition the set ωω into two sets S and T, and think of this partition as a game (S, T) with two players. To play, player 1 picks a natural number a0, then player 2 picks b0 (as a function of a0), then player 1 picks a1 (as a function of b0), then player 2 picks b1 (as a function of a0 and a1), and so on until an and bn are selected for all n ∈ ω. Then the sequence a0, b0, a1, b1, … is either in S (in which case player 1 wins), or in T (in which case player 2 wins).
The game (S, T) is determined if either player 1 or player 2 has a winning strategy, i.e., if there are functions fn: nω → ω where choosing an = fn( b0, …, bn–1 ) guaranteed player 1 victory, or similarly for player 2. (We can't have both.) AD is just the statement that every such game is determined, which is false in ZFC. As with most of the weird examples, the undetermined game is constructed with a well-ordering of R.
What makes this so unintuitive to me is that both AC and AD are generalizations of statements that are easily seen for finite objects. (Any finite game, or even any game with finite depth, is determined, by an easy induction on the depth.)
There are apparently many set theorists that agree with this assessment, since they try to rescue AD as relativized to L(R). That the relative consistency strength of this statement is equivalent to that of large cardinals is considered good evidence that those large cardinals are, in fact, consistent. More precisely, ZF + AD is consistent iff ZFC + "there are infinitely many Woodin cardinals" is consistent, and ADL(R) is outright provable in ZFC + "there is a measurable cardinal which is greater than infinitely many Woodin cardinals".