3 ms·

I would assume it's much worse than the (1/2)^(100) approach of picking random each time. If you pick randomly with each of your 50 picks then you've got a 50%

I would assume it's much worse than the (1/2)^(100) approach of picking random each time. If you pick randomly with each of your 50 picks then you've got a 50% chance of winning. If you pick a random cycle and follow it, then you win iff the cycle you pick is your cycle and your number occurs within the first 50 of that cycle. So if the cycle that includes you is small, the first thing is less than 50%, and if the cycle that includes you is large, then the last thing is .. hmm well worth numbering: Say there's a 99 cycle and a 1 cycle. So 0.99 you're in the latter. And assuming that, 50/99 you pick a box that leads you to yourself in 50 picks. And 1/100 you're in the first, and assuming that 1/100 you pick it. So, 1/100 x 1/100 + 0.99 x 50/99 = .0001 + .5 = 0.5001. Hmm, I expected it to be under .5. So .. help?

daxfohl 9y ago
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.
- daxfohl 9y ago
  Yeah I think I forgot to account for not ending up on your cycle in the 99 case (i.e. you're in the 99 but you chose the 1), so it should be .4901 or .4951 or something even in the 99:1 case.

daxfohl 9y ago

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.

daxfohl 9y ago
Yeah I think I forgot to account for not ending up on your cycle in the 99 case (i.e. you're in the 99 but you chose the 1), so it should be .4901 or .4951 or something even in the 99:1 case.