Assign each person a number 1-100; each person starts by opening the box labeled with their assigned number. If they don’t immediately lose, they move on to the box labeled with the number corresponding to the name they just saw.

This will succeed as long as the name permutation consists of only a single giant cycle, which happens 1% of the time (quick exercise: prove this if you haven’t seen this result before).

The next question is: can you prove this is optimal?

]]>if we take the 6 posibilities for 3 people

123

132

213

231

312

321

we can agree that we will lose in {123, 132, 213, 321} but ensure that in casas {231, 312} we will win. eace of the people will know, when he opens his number’s box, which from these posibilities is surely wrong, and will ensure that if the other is right, they will not lose.

the same for 100 people. if we agree about the first box each open, we will try just to ensure cases that are not koses cases from begining

]]>But since the entire experiment is completely random, the boxes can all be relabelled at random in each trial without changing the probabilities of each individual player succeeding. Since winning is conditional on all players winning, I don’t think such a strategy will improve the odds.

Ultimately it boils down to picking the correct permutation of 100 objects—or so I conclude at least.

]]>let each begin with another box. if the first opens box 1, if he find 1, they lost, but if he finds 2 then he knows that boxes 2 and 3 there are names 1 and 3, but if in box 3 there is name 3, anyhow they will lose, so he can decide that his name is in box 3, and he will open it.

the same if he finds in box 1 name 3, he will open box 2

now the chance to win is 1/3, much more then 1/6/ ]]>

Alternatively, if the players have the option, each can specify a false name, then when the challenge time arrives, each can specify their real name. Assuming no user specifies a real person’s name as their false name, then the game can be won 100% of the time. ]]>

Then again, I’d probably agree with you, if I hadn’t already seen the answer to this problem. ðŸ˜€

]]>To prove the original problem has probability 1/100!:

Each player must open only one box, and the players win if they all find their own names. Lets say the players choose one box to open beforehand. Since the boxes are random, any choice is equivalent to picking the boxes in order. So player 1 pixes box 1, player 2 box 2, etc. In order for the players to win, player 1 must win AND player 2 must win AND…etc

The odds of player 1 winning are 1/100. After this, player 2s odds are 1/99. Player 3’s odds 1/98, etc.

So the odds of winning with this strategy are

1/100*1/99*1/98*…..*1/2*1/1=1/(100!)

But the odds of guessing the correct permutation of 100 objects is also 1/100!, which is what this problem amounts to. Therefore choosing boxes beforehand is the optimal strategy, for both problems.

Since 100! is greater than the number of known atoms in the universe, the odds of winning are not very good.

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