[pdurrant]

Originally Posted by clarknova

This seriously depends...

Spoiler:

...on whether your goal is to get more wins, or to lose less money. You always lose money (games of chance are always to the houses advantage).

If you change your bet everytime, you will win the game more often, however you will lose more money than if you stay with the original bet.

I suck with statistics, so I just ran some monte carlos of the game with both strategies (each time the player starts with 1 coin). Here are four runs (of a million games) when sticking with the original bet:
Games: 1000000 Wins: 333402 Losses: 666598 Player Money: -333195
Games: 1000000 Wins: 333292 Losses: 666708 Player Money: -333415
Games: 1000000 Wins: 333540 Losses: 666460 Player Money: -332919
Games: 1000000 Wins: 333916 Losses: 666084 Player Money: -332167

And here are four runs when changing the bet (doubling your wager, but not doubling the return).
Games: 1000000 Wins: 499503 Losses: 500497 Player Money: -501490
Games: 1000000 Wins: 499822 Losses: 500178 Player Money: -500533
Games: 1000000 Wins: 500125 Losses: 499875 Player Money: -499624
Games: 1000000 Wins: 499347 Losses: 500653 Player Money: -501958

While changing your bet allows you to win around 50 percent of the time, your losses increase. If the game was a real double or nothing (where the dealer also doubled his bet when you doubled yours) then the strategy of always changing your hand would be break even.

In other words: The only winning move is not to play. How about a nice game of chess?

Sorry, you're not right. The chances of winning by sticking to your original choice and winning by switching to the other remaining cylinder must sum to 1. You have it adding up to 5/6ths.