I hold two envelopes, and tell you that one holds twice as much money as the other. I then let you choose and keep one. You are about to take the envelope in my left hand, but then decide to do a quick probability calculation. Let the amount of money in the left envelope be 
. Then the amount in the right envelope is either 
or 
. Now, since you picked the left envelope at random, these two amounts are equally likely for the right envelope. Therefore, your expected gain in switching to choosing the right envelope is 
, whereas your gain in stick with the left envelope is . So, logically, you should switch. But then, if you’d chosen the right envelope first, then the same logic would dictate that you should switch to the left one. And that can’t be right. So what’s wrong in your reasoning?
This question was posed as an exercise in probabilistic thinking by Marcus Hutter during his tutorial today at the (very interesting and enjoyable) Australasian Computational Intelligence Summer School. I must admit I missed about half an hour of the remainder of the tutorial trying to solve the problem. It’s actually quite a nasty one to throw up in passing — so nasty, in fact, that even Wikipedia is confused (or at least confusing) about it. (Please feel free to try to unravel the paradox before reading on.)
The first solution to the paradox that Wikipedia offers is that the variable is being misused in the above working. To quote Wikipedia:
You cannot denote an unknown amount chosen by chance by a variable as you would a generally unknown amount.
But this objection is easily removed, without fundamentally resolving the paradox, if I let you open the envelope of your first choice, count the money inside it, and then decide whether to switch. At this point, is no longer an unknown amount at all, and we are still left with the paradoxical conclusion that you should switch envelopes, even though your information about the probabilities in the situation has not fundamentally changed from what it was before you opened the envelope.
So how far did I get (without, I hasten to add, looking at Wikipedia)? Well, Marcus Hutter suggested that we should find a “Bayesian solution” to the paradox. So, let’s recast things. Let the amount in the selected envelope be , and in the unselected envelope, 
; and let us assume that we have opened and found 
. Therefore, the first step for us is to determine 
. Let’s throw this into the Bayes’ rule:
which in our case gives:
Immediately, we notice something interesting. What is 
? That is, what was the prior probability that a randomly-chosen envelope held the discovered amount (say, $100)? Well, working with paradox-maximising economics, we might want to say that we regard any (strictly positive) amount as being equally likely. But then we have a probability distribution over an infinite number of items, which means that each item has zero probability of being realized; that is, that 
.
At that point, I went back to listening to the tutorial. For a fuller discussion, I direct you to someone who (unlike me) appears to know what they’re talking about. The general conclusion is, there’s nothing intuitive about probabilities once infinity is involved. But I’m still left (intuitively) unsatisfied.
If switching envelopes multiplies your expected return by 1.25 (which it does), then switching twice multiplies it by 1.5625. So if this expected return is E, we have E = 1.5625 E, so E=0 or infinity (let’s assume the expected return is not negative . . ). Neither possibility involves a paradox, but since you’re the one holding the envelopes, I’ll bet that E=0 . . .
(I promise not to end my next comment with an ellipsis)
(note that my first answer is a joke, if it is not clear from context . . .)
Yes, I was chuckling to myself about it. There should be more jokes expressed in terms of expectations over ill-defined probability distributions.
By the way, all of this brought back memories of you explaining the (related) St. Petersburg paradox to me while walking down Drummond Street twenty-odd years ago.
As your reference points out, the problem’s assuming the expectation is finite. It’s like assuming 1/0 is finite to prove 2 = 3, by noting that 1/0 * 2 = 1/0 * 3.
When in doubt, draw out the event structure. (This is also how to solve the Monte Haul “paradox”.)
The envelopes contain X and 2X units of currency. The two possible events, which we assume are equally likely, are “choose X, then switch to 2X” and “choose 2X then switch to X”. Hence, after switching, you still have a 50-50 chance of each envelope, just like you had before switching.
The paradox arises from infinite expectations, where switching from any finite amount is a good strategy. Thus Danny Calegari’s point is actually relevant, because “infinity” is the only solution (other than 0) for x and y such that x = 1.25 y and y = 1.25 x.
A simple example of a game with an infinite expected reward is the coin flipping game. You continue flipping a (fair) coin until you get a tails result. If the game pays out 2**N currency units, where N is the number of heads you tossed before the first tails, the expected payout is infinite, being
(2**1 * 1/2) + (2**2 * 1/4) + (2**3 * 1/8) + …
= 1 + 1 + 1 + …