Say, 'A' has so far won 60% of the games (chess, shuttlecock) he participated. But 'B' has won only 40% of the games. If A and B clash in a game, clearly A has 20% winning edge over 'B'. But , if luck favors B, A may lose. So one game is not fair enough to decide the winner.
Suppose A and B plays a set of three games, whoever wins any two games out of three is a clear winner. Let us see the maths behind it.
What is the chance for A to win any 2 games out of 3. There are four possibilities or pattern to win.
Pattern1. Win, win, win
Pattern2: win, win, lose
Pattern3. Lose,win,win
Pattern4. win, lose, win
The winning probability is 0.60 and losing probability is 0.40 for A. Hence substitute the probabilities and use multiplication rule of statistics.
Pattern1. .60*.60*.60 = 0.216
Pattern2. .60*.60*.40 =0.144
Pattern3 .40*.60*.60 = 0.144
Pattern4. .60*.40*.60= 0.144
Total winning probability = 0.648
In a set of three games, the winning chance of A is .648 or 64.8% and the remaining 100-64.8 = 35.2% is the winning chance of B. Now A has nearly 30% more edge over B.
So a set of three or five games will bring out the genuine winner.
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