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John and Shane are playing a game, where after each round you get 1 point if you win and lose 1 point if you lose the game. The winner is the one who first reaches 4 points. In how many ways can any one of the players win?

I had this question in the exam yesterday and I was trying to solve it by taking either of the two possibilities game after game and add them all, but I thought it wouldn't work if in general if we have $n$ points for winning.

So can anyone help me show how to solve it?

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I think there are infinite ways of winning. Assume John and Shane win alternately. After 2*n rounds their score would be each 0. Then John wins 4 rounds in a row. And he wins the game. For every natural number n exists 1 solution. (actually a lot more than one) –  user38034 Aug 24 '12 at 10:15
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Do scores drop below zero? Is it always the case one wins and one loses? –  rschwieb Aug 24 '12 at 12:00
    
@Gerry Myerson I wanted to say at least $4$ times. Of yourse as you said it might be more actually $7$ at most. However the idea is that there is a certain number of differences and among them infinitely many oscillation is allowed. –  Seyhmus Güngören Aug 24 '12 at 12:17
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As already said by others, there are infinitely countably ways of winning. However, if the game is fair, it takes 16 rounds in the mean before one of the players wins. –  Did Aug 24 '12 at 12:22
    
Could it be that the loser gets $0$ points instead of $-1$ points? –  Christian Blatter Aug 24 '12 at 12:42

1 Answer 1

If player a is awarded a point then player b loses a point and also the other way around. Therefore the final score is always going to be 4,-4 or -4,4. However there are infinite number of ways this can happen. The game can last 4 turns, 6 turns, 8 turns, any even number of turns. Therefore there are infinite ways it can happen but only two possible outcomes.

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