Author Topic: Mathematics Challenge by John Baylis  (Read 879 times)

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Geoffw

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Mathematics Challenge by John Baylis
« on: March 29, 2015, 07:54:32 PM »
POSTED ON BEHALF OF JOHN BAYLIS

Each month we propose to offer a Mental Challenge. Kindly, or not so kindly <grin!> devised by John Baylis.

We hope to offer a small prize or at the very least a round of applause at the April monthly meeting so get your thinking caps on! LAST ONE WINS!

"Two players alternately take stones from a pile. The only rule is the number of stones taken at each turn is 1, 2, 3 or 4. The winner is the player who takes the last stone."

Question One: “Does either player have a winning Strategy?”

Question Two: “Under what Circumstances?
« Last Edit: March 29, 2015, 08:02:50 PM by Geoffw »
Geoff

Geoffw

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Re: Mathematics Challenge by John Baylis
« Reply #1 on: March 29, 2015, 08:02:06 PM »
POSTED ON BEHALF OF JOHN BAYLIS

A Clarification

At the monthly meeting on the 5th March, someone raised the point that I hadn't said how many stones were in the pile.  This is true, but no, it wasn't an omission!

Which player, if any, has a winning strategy depends on the size of the initial pile.  So to re-phrase the question; for which size piles can the 1st player force a win, and for which can the 2nd players do so?  And what are the winning strategies?

And just to show I'm a generous sort of guy, here's a hint:

"Start mentally experimenting with small piles, and that should give you a feel for what is going on in general.........................."
Geoff

JanM

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Re: Mathematics Challenge by John Baylis
« Reply #2 on: March 29, 2015, 09:00:25 PM »
Would that be one pile each, or one shared pile.

Geoffw

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Re: Mathematics Challenge by John Baylis
« Reply #3 on: March 29, 2015, 09:02:03 PM »
Would that be one pile each, or one shared pile.

John says "Two players with one pile"
Geoff