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Someone gave me the following riddle and I am really not sure what the answer is.

I have two boxes in front of me. One box contains $x$ Dollar, the other box contains $2x$ Dollar. The rules are as follows: I am allowed to open one box. I can now either keep the money I found in this box, or give this money back and take the money from the second box without knowing how much money I find in there.

The question is now: What is the best strategy to get the most money?

My first intuition was that it makes no difference. I do not get any new information after I opened the first box. I didn't now before how much $x$ was and now I also do not know whether the amount of money $y$ I found is $y=x$ or $y=2x$. So on average I should always get the same amount, namely $1.5x$.

What got me thinking was the following: Let's say I find $y$ Dollar in Box 1. If I change to box 2, I either get $2y$ Dollar or $y/2$ Dollar. Both events can happen with probability $1/2$, so on average I get $5/4 y$ Dollar, if I switch. This result would suggest that switching is better, because without switching I always end up with $y$ Dollars!

At first, this problem looked similar to the Monty Hall problem. But in the Monty Hall problem I actually get new information after I picked the first door, because one "goat door" is opened. In the box problem, however, I do not learn anything new.

So, what is the best strategy?

Merlin1896
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    You need to have some information on the distribution of possible values. "uniform" on the entire positive real line doesn't work (as it is not a probability distribution). If, say, you knew there couldn't be more than $$100$ then you would definitely not switch if your envelope had $$51$ or more. – lulu Nov 01 '17 at 19:23
  • This is known as the Two envelopes problem. It is a paradox exactly because of what you say: it's obvious that switching shouldn't be a better strategy, and here it seems we have a good argument that we should switch! So, what is wrong with that argument? – Bram28 Nov 01 '17 at 19:32

1 Answers1

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The money is already in the boxes.

One box holds $x$ dollars. The other box holds $2x$ dollars.

If you choose the box with $2x$ treasure you stand to loose $x$ dollars if you switch.

If you choose the box with $x$ treasure you stand to gain $x$ dollars if you switch.

Expected gain by switching is $\frac 12 (-x) + \frac 12 x = 0$

Doug M
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