This is a well known problem known as the "two envelope paradox." Sometimes it is asked where the player may open the first envelope before switching. The argument in favor of switching is that the other envelope has either half or twice as much, with 50% probability each. If there is $x is the first envelops then the expected value of the other envelope is ($2x+$0.5x)/2 = $1.25x. However if this argument could be used to go back and forth infinitely if the first envelope is not opened. If you know the option will be granted in advance you could use it to save the trouble of switching and pick the other envelope in the first place, but then you could use it again to switch back again, infinitely.
Whether or not the first envelope is opened you can not blindly say the other envelope might have twice as much. To make an extreme example if the total amount of money in the world is t and the first envelope has more than t/2 then it is impossible for the other envelope to have twice as much. More practially if the person offering the money must have a finite amount to give. So we can't blindly say there is a 50% chance of doubling if the first envelope. For example if the person offering the envelopes has exactly $1,000,000 dollars the most he can offer is $500,000 and $1,000,000. So if the first envelope had $1,000,000 it would be impossible for the other envelope to have $2,000,000.
However lets take the case where a billionaire is hosting the game and the first envelope has less than $100. Any amount under $10,000 would mean nothing to the host. So bumping up against the maximum is apparently not an issue. A million players could use the expected value to switch, but would they be better off? Indeed on average half would end with 200% of the first envelope and half 50%, for an average amount in the second envelope of 125% the first. However what is important is not the rate of increase but the amount of the increase. As a group switching will not have helped. This is the hard part to explain but when the switch was the right decision the player is doubling off of the small amount, and when it is wrong the player is halving from a larger amount. Either way the decision to switch will result a win or loss of the amount in the smaller envelope. So the expected rate of increase is 25% but the expected amount of increase is $0. So although the expected rate of increase is 25% that does not mean the other envelope will have 25% more. The second envelope has a 100% chance of having 100% more if the first envelope is the smaller amount, and a 100% chance of having 50% less if the first envelope is the larger amount. So, once the first envelope is chosen the second envelope is no longer a random variable.
Adam Atkinson adds another interesting observation he credits to a graduate student at Cambridge University, whose name he forgot. Using an increasing function, mapping values of x from 0 to infinity to 0 to 1, the player can improve his probability of ultimately choosing the correct envelope to above 50%. For example if the amount in the initial envelope is x and the player keeps his original envelope with probability x/(c+x), where c>0, then he will be more likely to not switch with the larger envelope. Note that the function must be decided upon before the first envelope is chosen and a method must be available to choose a precise random number. To add my own comment you could also improve the probability of a successful switch above 50/50 by deciding in advance at what point you will switch. For example I'll switch with $1000.005 (I add the 0.005 to avoid ties) or less in the first envelope. This will produce a 100% success rate if $1000.005 falls between the two envelope amounts, although only 50% otherwise.
Links
Two envelope problem at Wikipedia.
Exhaustive analysis of this paradox: www.u.arizona.edu/~chalmers/papers/envelope.html.
Michael Shackleford, A.S.A.MathProblems.info home