A candidate is about to win a prize in a quiz.
He is presented with three curtains; one of them hides the prize, the other two are empty.
The quizmaster asks the candidate to select a curtain and make his choice known.
Before opening the curtain of choice, the quizmaster opens up one of the other curtains, one of which he knows that it is empty. He then asks the candidate if he wants to change to the other curtain.
The dilemma is: does changing from curtain improve his chances of winning the prize or not?
The way the problem is stated, there is no answer.
For example, suppose the quizmaster only does this - opening one of the other curtains - if the contestant has chosen the right curtain with the prize. Then, obviously, the contestant shouldn't change. Or if he only does this when the contestant had chosen the wrong curtain - then he should change.
If, on the other hand, the quizmaster always does this - opening one of the other curtains - every time without fail, then one can indeed make a probability calculation without considerations of intent being involved.
Since the quizmaster can, and does, always do this, regardless of which curtain the contestant chooses, this act gives no information about the curtain the contestant chose. The chance of that curtain having the prize was 1/3, and it stays 1/3.
But for the other two curtains - before, the chance for each of them was 1/3. Now, opening one of them up has given information about those two curtains - the 1/3 probability for the one that has been opened has now moved to the one that wasn't.
It was a 1/3 chance of the one you picked, and a 2/3 chance for the two others - and, after the curtain is opened, it still is. But now, if it's the two others, it has to be the one that wasn't opened.
By coincidence, I've been looking at this page. I know there's the correct way of working out the probabilities, but the incorrect way doesn't seem to be flawed (though it obviously is.)
I blame Bayes.
Blaming Bayes is right for this one.
With the two envelope paradox, the problem isn't "the probability is 50% that this envelope will have the check for the larger sum of money in it".
Once the envelope is opened, and you see the amount of money you chose, you have new information now. The check in that envelope is for 40,000 pounds.
So, while the
a priori probability of the larger check being in the envelope you picked was 50%, with your new information, what you need to decide whether to change envelopes is...
the probability that they filled the envelopes with checks for 40,000 pounds and 20,000 pounds respectively, versus
the probability that they filled the envelopes with checks for 80,000 pounds and 40,000 pounds respectively.
The prize distribution is the distribution that counts now, not the envelope distribution. And that distribution is, given the conditions of the problem, a complete unknown. One data point - a 40,000 pound check - isn't enough to give meaningful insights into that distribution in a Bayesian sense.
Yes, one stands to gain twice as much as one could lose if the amounts of those checks were totally random. But if they were, then a check for 40,000,000,000,000,000,000 pounds would have been just as likely as a 40,000 pound check. So somewhere along the line, the probability of a check being for X pounds has to be less than that of a check for X/2 pounds.
A single check for 40,000 pounds does indicate that the midpoint of the distribution of check sizes is somewhere vaguely around there: that's all that Bayesian statistics will buy you.
