25 April 2009 UTC Here is a little something that I came up with myself Let the boxes equal the letters A V Let the prize that you want to win be behind box A It doesn t matter which box the prize you want to win is in also assume biggest prize Now if you pick box A then the probability of getting through to the end of the game with the prize you want still in the game is 1 certain This occurs is 1 22 cases 1 1 22 1 22 If you don t pick box A then the probability of you getting to the end of the game without taking out the prize you want is 20 21 19 20 18 19 2 3 This can be re written as 2 20 1 21 There is one very important thing to note here In the sum you don t multiply by 1 2 because you don t need to eliminate all of the boxes just all but one plus yours This means that instead of dividing by 21 you must divide by 21 2 which is why there is a 2 on the top The sum works out with a value of 2 21 0 9523809524 This occurs is 21 22 cases so 21 22 2 21 1 11 Therefore you are twice as likely to be on the path where you didn t pick the prize you wanted i e grand first which means that you should swap Now there may be a flaw in my logic and if there is then please point it out 82 26 43 91 15 46