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**tongzilla****Member**- Registered: 2006-06-19
- Posts: 18

There are lions and a piece of prey.

If lion 1 doesn't eat the prey, the game ends.

If lion 1 does eat the prey, it gets slow and fat so lion 2 eats lion 1.

If lion 2 doesn't eat lion 1, the game ends.

If lion 2 does eat lion 1, then lion 2 may be eaten by lion 3.

Etc.

Each lion prefers to eat than be hungry. But it would rather be hungry than be eaten.

What is the logical (i.e. backward induction) outcome for any n number of lions?

*Last edited by tongzilla (2006-11-09 03:17:33)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Interesting puzzle.

Why did the vector cross the road?

It wanted to be normal.

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**tongzilla****Member**- Registered: 2006-06-19
- Posts: 18

mathsyperson wrote:

Interesting puzzle.

Nice answer mathsyperson. If you can be bothered, try to polish your answer in the 3 lion case by using more logic in your backward induction. I.e. if Lion 3 blah blah blah then Lion 2 realises this so blah blah blah, etc.

So when does the game end?

*Last edited by tongzilla (2006-11-09 06:14:28)*

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