, 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