AIM: To analyse the win/loss ratio of two friends, A and B, when they toss a coin/play a similar game of equal chance.
THEORY: There are numerous games of chance played when trivial decisions must be made, and while these may be fair, one player will often try to ‘game the game’, by suggesting a ‘best of 3’ or ‘best of 5’ upon losing once or twice.
PROCEDURE:The people were asked to toss a fair coin, after a decision on which player would be heads and which would be tails. A was set to be the winner, and A and B had to play until out of an odd number of tosses, A had won a majority.
|Number of trials||Frequency|
The mean of the data above is 8.0625, while the median is 3.
Looking at this logically, if we assume that the first game is lost, at each stage, you must win both games, with probability 0.25. A summation of powers of (0.25) multiplied by the original probability of 0.5 [(0.5) + (0.5)(0.25)1 + (0.5)(0.25)2 + …] results in 0.66….. = 2/3.
[A proof of this can be shown as follows:
x = 1 + 0.25 + 0.0625 + ….
4x = 4 + 1 + 0.25 + 0.0625 + …. = 4 + x
3x = 4
x = 4/3]
Therefore, the data show that if you do not lose all of your games in a best-of-3 (or any successive best-of-2n+1), you will eventually win two-thirds of the time.