Someone's bound to post this eventually so it might as well be me, I've altered some things to hopefully make it slightly less google-able.
Suppose you're on a cruise ship, and you're given the choice of three boxes: inside one box is a super-deluxe pogo stick; the others contain radishes. You choose a box, say No. 1, and the host, who knows what's inside the boxes, opens another box, say No. 3, which has a radish. He then says to you, "Do you want swap your box for box No. 2?" Is it to your advantage to switch your or keep your original box? Or is there no difference?
Also you definitely want the super-deluxe pogo stick instead of a radish.
you want to switch, this is the famous monty hall problem. Look up how to do it online but i will give you the basics. at the beginning they each have a 1/3 chance, and then the host shows you one that does not have it. Because he intentionally did this then there is still a 1/3 chance for the one you picked and a 2/3 chance for the one you did not pick
oh sorry :P
next problem: there are two envelopes and one has twice as much money in it than the other does you open one envelope and it has 50 dollars in it. do you want to keep the fifty dollars or take the amount of money that is in the second envelope and why?
If you get to see the amount of money the point to this paradox is lost. You swap and see whether you were right to swap and the game stops there.
The classical problem is you get the envelope and then without seeing what's inside are asked whether you want to swap. Some say your expected return by swapping is greater then by sticking so you swap. You are then asked whether you want to swap again and the same logic apples so you swap. Basically you might conclude that it is beneficial to swap indefinitely. Clearly it is not so we have a paradox.
Maths is funny sometimes. It is 50 50. You shouldn't hurt your brain by overanalysing.
In the Monty Hall problem there are 3 doors. Two have no prizes, one has some valuable prize. You choose one, and the host removes an empty door from the two unchosen ones. You then have the option to swap, and you should always swap.
I understood it pretty quickly after reading about it and thinking it through, but apparently lots of people can't fathom it at all.
The one in this thread is simpler, because even though the chance of gaining or losing is 50-50, the potential loss is $25 but the potential gain is $50. It's more like a psychological test: if you can't get it, you just aren't much of a risk-taker.
For the Robertson Hotel doesn’t merely have hundreds of rooms — it has an infinite number of them. Whenever a new guest arrives, the manager shifts the occupant of room 1 to room 2, room 2 to room 3, and so on. That frees up room 1 for the newcomer, and accommodates everyone else as well (though inconveniencing them by the move).
Now suppose infinitely many new guests arrive, sweaty and short-tempered. No problem.
How does he do it?
Later that night, an endless convoy of buses rumbles up to reception. There are infinitely many buses, and worse still, each one is loaded with an infinity of crabby people demanding that the hotel live up to its motto, “There’s always room at the Hilbert Hotel.”
The manager has faced this challenge before and takes it in stride.
First he does the doubling trick. That reassigns the current guests to the even-numbered rooms and clears out all the odd-numbered ones — a good start, because he now has an infinite number of rooms available.
But is that enough?
Note: learend this in a cantor lecture, you can look it up on google but don't ruin the fun
The Monty Hall Problem - you should swap, its intuitive and easy to prove.
I have no idea x how you can say that the two envelope problem is easier. Surely you are not suggesting that it is mathematically provable that you should swap. Swapping is just the same as picking the other one first a sticking. It is for sure of no benefit to swap, the difficulty is in proving it. Very tricky.