I think my math for the above case is wrong, but I've got seven empty beer cans in front of me so. But even if it's right, you can somewhat rationalize. If th

I think my math for the above case is wrong, but I've got seven empty beer cans in front of me so. But even if it's right, you can somewhat rationalize. If there are two cycles each of len 50 then it's 50% chance. And if there are two cycles where one is 99 and the other is 1, then you've got very marginally better than 50% chance. And if it's one cycle of 100 you've got 50%. But if it's more than two cycles, say three of 33 or 34, then you can quickly see that you've only got a 33% chance of finding yourself. And plenty of outcomes like that exist. So it seems like 1 cycle (or apparently 2 with 99:1) is an upper bound. And the remainder only serve to lower the odds. And given they're plenty likely (I guess), your odds are even worse than randomly guessing with each pick.