## Monday, 22 March 2010

### Mathematical Problem 1: Solution

Here's a solution to Mathematical Problem 1, posted on 13 March 2010.

With 2 players, A wins a match of 29 rounds when A has won 15 rounds. Even if A loses all the remaining rounds B will only have won 14.

The principle is that A would still win even if all the remaining games were to be won by B (A's closest rival).

With 3 players, A wins when a is greater than b added to the number of rounds still to be played (assuming B has the second highest number of wins so far). The number of rounds still to be played is 29 – (a + b + c). So, algebraically:

a > b + [29 – (a + b + c)] which simplifies to 2a > 29 – c.

With 4 players, A wins when

a > b + [29 – (a + b + c + d)] which simplifies to 2a > 29 – (c + d).

The general solution , with any number of players and n rounds, is that A wins when

a > b + [n – (a + b + c + d + e + ...)] which simplifies to 2a > n – (c + d + e + ...)

The surprise in this result is that b does not appear in the simplified rule! Provided B has won more rounds than C, D, E and so on, the actual number that B has won does not make any difference to when A is declared the winner.