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 Win some, lose some II (Posted on 2015-03-25)
Three friends organized a ping-pong contest, whereas after each game the winner stayed to play the next game.
The 1st game between A and B was won by A, the last game between A and C was won by C.
Altogether A participated in 24 games, B in 28 and C in 38.

How many games were won by C?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 solution Comment 1 of 1
There were (24+28+38)/2 = 45 games.

Each game except for the first consisted of one player who won the previous game and one player who had sat out the preceding game.

The only player who sits out is the loser of the previous game, except in the first game.

C sat out 45 - 38 = 7 games, one of which was the first game, so he had lost 6 games, meaning that he had won 32.

BTW, B sat out 45 - 28 = 17, making 17 losses and 11 wins; A sat out 21 and so had 21 losses and 3 wins.

But this adds up to 46 winners in 45 games; that's impossible.

The fault in the above reasoning is that the losses for a given player amount to his number of sittings out, except for the first game. But also in the last game, the loser doesn't sit out one of the 45 games. But C wasn't the loser of that last game; rather, A was the loser. So A has one more loss that we hadn't accounted for by the sit-out count. It was A who had only 2 wins rather than 3; C's record still stands.

A won 2.
B won 11.
C won 32.

 Posted by Charlie on 2015-03-25 08:41:08
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