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May 2005

Bottom dollar

roulette wheel

You're in a glitzy casino in Las Vegas. Having tried your hand at everything from Roulette to Black Jack, you've managed to lose most of your money and have only one dollar left.

What's worse, with all the champagne and everything, you've misbehaved and the management has made it very clear that you're not allowed any more games. But you need two dollars to get the bus back to the hotel.

Two shady-looking characters at the bar offer you a game: they have a pile of 15 stones. Each of you in turn is to take your choice of 1, 2, 3, 4 or 5 stones from the pile. The person who takes the last stone gets one dollar from the person who drew previously, and the third person neither wins nor loses.

You're to draw first. You're sure that both of the other players will play to their best personal advantage and won't make any mistakes. Should you agree to play the game?

If you are stumped by last issue's puzzle, here is the solution.

For some challenging mathematical puzzles, see the NRICH puzzles from this month or last month.


Only two winning "leaves" are 7 and 14
So since he can go first and can leave 14 he can't lose

Leaves :
1 to 5 obviously lose
6 draws
7 wins
8 to 12 all lose - next player wins with 7
13 draws - next player loses - last player wins

Take 1 and leave 14 wins - next must take 1 (2 to 5 lose) and leave 13 to last player for a draw - last player also loses with 2 through 5 so must also take 1 to leave 11 for a draw - first player takes 4 for the win by leaving 7 - next player must take 1 for a draw leaving 6 for last player who loses to the first player who left the 14

JW Three Rivers CA

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