Subject: An infinite flippage of coin
Date: Wed, 12 May 2000 14:11:15 +0000
From: Clifton Baron
Dear friends -
Several of us are trying to solve a mathematical puzzle, and there seems to be convincing arguments in two conflicting directions:
Let us say we are playing a game called "Heads". The game consists of the flipping of a coin. If you flip the coin and get "heads" on the first "try", you win.
If you get "tails" on the first try, you don't lose, you get another "try". But since this is the second "try", then, in order to win you must get, not one, but two "heads" in a row.
If you fail to get two heads in a row on the second "try", once again, you don't lose, you get another "try". However, since this is the third "try", then, in order to win you must get, neither one nor two "heads" in a row , but rather three "heads" in a row. Etcetera
Our question is: Given an infinite number of "tries", are you guaranteed to win? We have developed our answer to this question to the following dilemma:
On the one hand, it would in no way be untrue to say that each time you were faced with another "try" there would be a finite possibility that you would "win", and given that you have an infinite number of "tries" it seems one might be able to conclude that therefore, someday, you would have to "win".
Conversely, the odds of winning on the first "try" are 1/2. The odds of winning on the second "try", which would mean a line-up of total flips from the beginning of the game are "tails-heads-heads" is 1/8 - which brings the total odds of winning in two "tries" to 5/8.
The third "try" adds 3/32 to this series, and the fourth "try" adds 21/1024. We have not specifically formulated the total of the series, however, it is clear to us that were to do so, the odds of winning, given an infinite number of tries are less than one.
Therefore we might conclude that even if you are given an infinite number of tries, you won't win every time. What sayest thou?