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Oddwins (Posted on 2006-10-03) Difficulty: 3 of 5
A game of 11 marbles wherein each player can either pick one or two marbles from the total. Starting from Player A & then Player B alternatively. This continues till all the marbles are picked. The winner is the one having odd number of marbles.

What is the strategy to be followed for Player A & B to win?. What happens for higher total number of marbles (13, 15 etc )?

No Solution Yet Submitted by Salil    
Rating: 4.0000 (1 votes)

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Solution Maybe this works | Comment 4 of 11 |
My guess is that player A wins if the number of marbles is of the form 4n+1 and player B if it is of the form 4n+3. Generally whether you can win in a particular situation depends on the number of marbles you own modulo 2 and the number of marbles left on the table modulo 4. Here are the winning positions A-E plus winning moves. First number is number of marbles owned mod 2, second is number of marbles left mod 4:

A: 0/1: take 1 : 1/0
B: 0/2: take 1 : 1/1
C: 1/0: take 1: 0/3
D: 1/2: take 2: 1/0
E: 1/3: take 2: 1/1

Whatever the next player does, we will end playing another one of these winning positions in the next move:

A+2 moves: 1/3 or 1/2
B+2 moves: 1/0 or 1/3
C+2 moves: 0/2 or 0/1
D+2 moves: 1/3 or 1/2
E+2 moves: 1/0 or 1/3

Of course in situation C it is possible that no marbles are left, in which case we have won.

I should add that this could be rubbish, cause I am still suffering a little from my cork experiment, so be gentle with your comments...

  Posted by JLo on 2006-10-03 17:27:01
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