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11 replies to this topic

#1 superprismatic

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Posted 04 September 2011 - 08:14 PM

Two players play a simple game called
Odd Todd. They start with a pile of
an odd number of stones. Players take
turns alternately. On his turn, a
player may remove one or two stones
from the pile. The player who ends up
having taken an odd number of stones
wins. Under what conditions can the
first player force a win? When can the
second player force a win? What are
the winning strategies in both cases?
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#2 James22

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Posted 04 September 2011 - 10:31 PM

Spoiler for I think

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#3 RESHAW

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Posted 05 September 2011 - 08:57 AM

Spoiler for Caffeine free ramblings

Edited by RESHAW, 05 September 2011 - 08:58 AM.

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#4 wolfgang

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Posted 05 September 2011 - 11:03 AM

Spoiler for Let the roles to be like this:

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#5 thoughtfulfellow

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Posted 05 September 2011 - 12:39 PM

If 1 coin: first person to play wins:
Spoiler for More:

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#6 LVanToren

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Posted 05 September 2011 - 09:05 PM

"With three sTones, First player loses! If he takes 1, then second player takes 1, forcing first player to take last stone" and then the first player wins because he "ends up having taken an odd number of stones". If the second player takes 2, then it is a draw.
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#7 LVanToren

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Posted 05 September 2011 - 09:16 PM

Spoiler for Let the roles to be like this:

It is stated that "On his turn, a player may remove one or two stones from the pile."

I think there is no winning strategy.
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#8 plainglazed

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Posted 05 September 2011 - 09:16 PM

"With three sTones, First player loses! If he takes 1, then second player takes 1, forcing first player to take last stone" and then the first player wins because he "ends up having taken an odd number of stones". If the second player takes 2, then it is a draw.

Spoiler for but...

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#9 fabpig

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Posted 06 September 2011 - 09:38 AM

Spoiler for I suspect

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#10 phillip1882

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Posted 06 September 2011 - 10:01 AM

the goal is not to take the last stone but simply to have an even number of stones at the end.
Spoiler for read the original question again

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