Originally posted by: Slew Foot
Assume your envelope has $10. This means that the other has either $5 or $20.
If you always keep your envelope, you'll end up with $10 on average.
If you switch, assuming a fair game, youll average (5+20)/2, 12.50 each time on average.
Except, if you think about it - this is clear nonsense. The two envelopes are indistinguishable, so let's say you are playing with a friend, and the friend opens it and finds $20. Should you swap. Your friend performs the same calculation and swaps. The method you use to calculate the amount you are likely to win following a switch is wrong.
This is explained in depth in the links. However, the best explanation was provided by thesurge at
Link. The problem is that this is heavy on mathematical theory and isn't easy to understand if this is the first time you have come across this concept.
Why is your method of calculation wrong? It's because it assumes an impossible set of potential prizes.
Let's say that there is a maximum prize of $100. If you open the envelope and find $60, should you switch in this circumstance? The game is fair - you have a 50:50 chance of having picked the higher or lower envelope. So you should switch right? Of course not. You *know* that you can't have picked the lower prize - and this knowlege is essential to perform a useful calculation. Without this knowledge you don't have enough information to estimate the gain from switching. (This concept of using additional knowledge to refine probabilities is called Bayesian mathematics).
The mistake in your method is subtle - essentially, it assumes that the probability of having picked the lower prize, given that knowledge of what prize you picked, is the same as having picked the higher prize, given the knowledge of what the prize you picked. This is not the same as the probability of picking the higher prize as opposed to the lower prize, in the absence of this knowledge. And, as demonstrated in the example above - knowledge of the prize you picked can substantially affect the odds. In more formal language, the calculation requires the posterior odds, but you only the know the prior odds - and have incorrectly assumed that they are the same
So, is there a situation where you could assume that the odds given the knowledge of the prize are 50:50? The answer is no. The only way in which knowing the value of the prize you picked can have no effect is if the potential prize money is infinite, and any possible dollar value (or fraction) can be chosen. In other words the probability of the host chosing a losing prize of .002 cents would have to be the same as that of chosing a losing prize of $8,819,277,425,299.78. This is unlikely to be a valid assumption.