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.

OBSERVATIONS:

Number of trials |
Frequency |

1 | 16 |

3 | 7 |

5 | 2 |

7 | 3 |

9 | 1 |

21 | 1 |

69 | 1 |

91 | 1 |

The mean of the data above is 8.0625, while the median is 3.

CONCLUSION:

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.