There are 20 boxes and 20 people in a room. A prize is hidden in only 1 box, and the rest are empty. To win, you must pick the box with the prize. Before the game begins, The organizer tells you that you can choose to go 1st, 2nd, 3rd, 4th, all the way to 21st (last person to pick a box)
What position should you choose?
ex. If you go first, you have a 1/20 chance of picking the box with the prize (5%)
If you go 2nd, there will now be 19 boxes since the 1st person would have already picked one. so 1/19 gives you a 5.2% chance. However, there is the small risk that the 1st person could have already found the prize, in which your chance would be 0%
(I don't know the answer to this)
I remember considering this question while looking at a trivia game where the contestants were seemingly replying randomly...
All the players have an equal probability of $\frac{1}{20}$ to get the prize. One way to show this uses conditional probability:
First player: $\frac{1}{20}$ chances to pick the correct box
Second player: $\frac{19}{20}$ chances that the first player failed $\times $ $\frac{1}{19}$ chances to pick the correct box $=\frac{1}{20}$
Thirst player: $\frac{19}{20} \times \frac{18}{19} \times \frac{1}{18} = \frac{1}{20}$ ...
This way, it looks kind of miraculous, but there is another way to think of the problem which makes the result more natural.
Let's say the players successively choose a box, and then they all open it simultaneously. Then what they are doing is choosing uniformly a random bijection from the set of boxes to the set of players.
Now, one can convince themselves that a uniformly randomly chosen bijection has the same probability to send the winning box to each of the players. (For instance because one way to pick such a bijection is to start by assigning the winning box to a uniformly random player, then randomly assigning the other boxes).