The first guy can only be helped by the actions of the captain but cannot know anything besides that or what the captain will try to do for them. We have yet to realize what that assist can be and may not waste it until we are certain it kills everything that leads to death for all if that is possible or maximizes their chances if not.

The captain can put the number of the first guy in the right box or do something else that will secure he get to that number before all 50 moves even if it is exactly on the 50th. If someone other than the first guy needs help because there is a way for the first guy to get to its number before 51 moves then the help will go there etc.

So given what i stated above in earlier posts regarding partial derangements its possible there are a number of fixed points in the permutation that goes like e^-1/k! in probability approximately. We cannot ignore their help.

We need a way to choose what we select next after a choice is made that proves not successful, without being able to have anything known about the sequences to us other than that the captain might have found a way to make our worse nightmare not possible to happen or to happen with minimum probability.

If we are #1 and open a box eg 56 and see 67 there, then we know 67 doesnt have 67 in it and it can therefore have 1 in it. The same cannot be said about any other number box because any other box can have still some chance to be a fixed point with not 1 in it.

So it becomes best option then to check 67 after 56 because we remove the fixed point wasted trials. Then at 67 if 1 is not there we must check the number found there for exactly the same reason because the contents of that box are not that number and can be our number still with higher chance than any of the other boxes (conditional probability argument).

Speaking of which what if 1 is already on 1? Then the above logic will never take us to our box. To remove that possibility we must start from our box first.

So we start at 1 and see say 59. Then we go to 59 and see 6. Then we go to 6 and see 9. Then we go to 9 and see 17. Then we go to 17 and see 1? Will it be like that? Yes. Always like that but different number of steps. We will have 1 in our horizon that way because we will never get that way to a fixed point. Why do we get to our number for sure? Because this process never finishes unless it takes us to it. It always takes us to a different number. Because every time we open something and it is not our number it cannot be the end of the process because it will keep changing into something not seen before until we hit our number if we started from its original box eg 1. Basically this process has an end that takes it back to the first box (the box that has 1 in it takes you back to look to box 1 right?). We will be able to get back to our number eventually but the question is with what chance this is in less than 51 moves?

So we must start from our box to see if we got lucky. If we get lucky great. The others must do the same starting from their boxes. If its not there then follow the sequence emerging.

A permutation therefore has what one recognizes as a number of cycles by looking in my link earlier.

The cycles must be disjoint because they start from a number and end to the same number or they would literally take all numbers if 1 went to 2, 2 to 3, 3 to 4 etc. The cycles since disjoint can have at most a period of 100. They are like islands. If we have a period that is longer than 50 then that is the only period longer than 50. So all people would be able starting from their number to either get it right away or after a while that is less than 50 moves.

So the captain can intervene and break that period over 50 (if any) to a smaller one and provided both pieces resulting are less than 50 we now have a solvable situation.

How do you break that? Go at the start of the biggest over 50 cycle eg its smallest number and count how many moves it takes to get back to the same number. Then before its its 50 or exactly there put there that number in that box and the contents of that box to the other box creating 2 cycles that are shorter than 51.

eg 1 5 79 11 23 89 ... 45 67 ...98 1 . Go at 45 and put 1 there (if position of original 1 is very far over 50 steps away).

Thanks for an interesting problem.

That's a cool bit of logic. By eliminating fixed points from your next pick you improve your chances. Then follow what happens if you keep doing that.