Someone shows you two boxes and he tells you that one of these boxes contains two times as much as the other one, but he does not tell you which one this is. He lets you choose one of these boxes, and opens it. It turns out to be filled with $10. Now he gives you the opportunity to choose the other box instead of the current one (and skip the $10 of the first box), because the second box could contain twice as much (i.e. $20).

The Question: Should you choose the second box, or should you stick to your first choice to maximize the expected amount of money?

Read the hint below if you're stuck...

A Hint: If you have $10, and you could double this with a chance of 1/2, or half it with a chance of 1/2, one would expect an average of 1/2 * $20 + 1/2 * $5 = $12.5 (so you would expect to gain $2.5)!...

No, there is no reason to change your choice.

ReplyDeleteAn explanation: The hint that is given is misleading! In this puzzle, it is not straightforward to calculate the expectation as suggested in the hint. So, just use your common sense, and note that the chance is and stays 50% that your first choice will be the best one. There is no reason at all to change that choice at the moment you find out that the box contains $10.

I was going to say this one was more opinionated then anything. It's playing odds for money.

ReplyDeleteYou are pretty much betting 5 bucks that says you can get 10 more bucks if you look at it. You are garunteed at least 5 bucks. So, do you feel lucky or not.

Myself, it is such a small amount that I think I would go for the other box.

Oh god. Im still confused...I'd probably give him $10 to leave me alone

ReplyDeleteit is a variation on the monty hall question, which mike will, if he is being kind, post and explain.

ReplyDeleteGerald, I did the Monty Hall problem as my first question/puzzle. I've always loved that puzzle.

ReplyDeleteThis problem is similar, but with only two 'doors' to open, the chances are still 50/50 after revealing what's in the envelope.