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

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 2

*a*> 29 –

*c*.

With 4 players, A wins when

*a*>

*b*+ [29 – (

*a*+

*b*+

*c + d*)] which simplifies to 2

*a*> 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 2

*a*>

*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.*
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