First, let's begin by looking at the classic Monty Hall Problem

You died and went to the purgatory, where the devil himself shows you three doors. He tells you that behind two doors are express hand baskets to the lake of fire, and behind one door is an escalator to the heavenly realms. Then God himself showed up and vouched for the devil being honest here. The devil asked you to pick a door at random. After you picked your door. The devil opens one of the remaining doors and shows you that behind that door, there's a hand basket waiting to be filled. The devil then offers you the chance to change your mind and switch your pick. Do you take his offer?

The solution is really simple: Yes.

Why? This is an exercise in conditional probability. When you made the first choice, there is 2/3 chance that you would choose a door that goes to hell, and 1/3 chance you choose the door that goes to heaven. Since the devil opens the other door "after" you made the choice, it doesn't change those statistics. So that means that the door you chose has a 2/3 probability being the one going to hell. That also means that the other door has a 2/3 probability of going to heaven. Quite simple, eh?

Now, let's do a modified Monty Hall Problem that I shall call the Knight's Choice.

A Knight, after fending off a dragon, went to the king for his reward. The king tells him the following:

1) The king holds two pieces of paper in his hands. One with a large number written on it, the other with a number that is half the first number written on it.
2) One of the two numbers is 1000, but the king is not telling whether 1000 is the bigger or the smaller number.

The knight was then asked by the king to pick one of the two pieces of paper. The king then revealed that the piece the knight chose has 1000 written on it, and offers the knight the chance to choose the other piece. Whatever is written on the piece the knight finally choose would be the amount of gold pieces the knight will get. Should the knight change his choice?

The answer is, again, yes.

Knowing that one of the number is 1000, the two numbers in the king's hands are either (500,1000) or (1000,2000). Assume it is equally likely (as implicit in the problem) for either of those cases to occur, the raw expected value for a random choice is

(500+1000+1000+2000)/4 = 1125

However, knowing which piece has the 1000 written on it, means that the expected value of choosing the other piece would be

(500+2000)/2 = 1250

which is better than the 1000 which the knight currently holds.

Lastly, we come to a problem which is slightly different:

The knight and his squire fended off the dragon. The king promised them rewards. Again, the knight and his squire went it to the king, and the king had two pieces of paper in his hand. Again, the king tells the knight that on one piece of the paper he has a large number. On the other he has a number half as large. The king would allow the knight to look at one piece of the paper, and the knight should decide whether he wants the reward on that piece of paper or on the other piece.

At this point, the knight consults with his squire. He says: "If I look at the paper and saw amount X. Half the time the other piece of paper would be X/2, half the time the other piece is 2X, so if I take the other one as a reward, I will, statistically speaking, be better off."

The squire, however, disagrees: "Then why don't you just take the offer on the other piece of paper to start with?"

Who is right?

This question is slightly more difficult. At first sight, you may agree with the knight. However, this readily leads to a paradox. The knight's reasoning is basically this:

If he knows the value on a sheet of paper, the expected value on the other sheet of paper is 1.25 times that.

But that is equivalent to stating that

The expected value on the second sheet of paper is 1.25 times the value on the first one.

But of course, the situation is symmetric. There's nothing inherently different about the two sheets of paper, so that also means

The expected value on the first sheet of paper is 1.25 times the expected value on the second sheet

Doesn't this lead to a contradiction?

Yes and no. The logic is correct there, and the only way a contradiction can be avoided is that on the two pieces of paper are written either both 0 or both "infinity".

What's the problem? The concept of expected value. If, the king is a generous as he makes himself out to be, then he would be given out, with equal probability, any amount of money. Since the amount of money given out is a non-negative real number, which is an unbounded set, a constant probability measure defined on it would either have weight 0 or infinite weight. Which means that you either expect the two pieces of paper to have infinity on them or 0.

The contradiction lies in the assumption that the king can write any large number he like on the piece of paper with equal probability. This is clearly an unphysical assumption.

The probability distribution must taper off for the notion of expected value to make sense. And if the probability distribution does taper off, that change would offset the expected gain from switching by the naive calculations made by the knight.

In the previous question, by the king's own admission that one of the numbers is 1000, we have a discrete probability function on the possible numbers for the other sheet of paper. From that we can calculate an expectation value and arrive at the conclusion that switching is better.

In this question, without that constraint, consider a L^1 (absolutely integrable) probability distribution u on the positive reals that is positive normalized (i.e. int u dx = 1). Let u denote the distribution of the smaller of the two numbers in the king's hand. Randomly picking a piece of paper would yield the expectation value

3/2 * integral[from 0 to infinity] u x dx

What about the knight's look and switch idea?

1/2 * integral[from 0 to infinity] u(x) 2x dx

is the contribution from when he picks the smaller one and

1/2 * integral[from 0 to infinity] u(x/2) x/2 d(x/2)

is the contribution from when he picks the larger one (the reason we have d(x/2) is that technically speaking u is a probability measure, so when we change variable for u we also need to change for dx).

(E-mail me if the above explanation doesn't make sense)

If you add up the two pieces, you will miraculously realize that with his strategy, the knight would do no better than just picking randomly.