Math Is Fun Forum

Problem:

1) Both

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EDIT (21 OCT 07):

Let me clarify what my actual question is. Sorry for not making this clear enough early on.

The question I am asking is this:

1) What restrictions (i.e. the general rule) are needed on

3) How can you prove your answers to questions 1) and 2) ?

Two friends, A and B, are at a beach for two days.
There is a probability p that the water has sharks. If any sharks are present, the swimmer will be attacked for sure.
A peaceful swim gives a payoff of 1.
Being attacked by a shark gives a payoff of of -c<0
Sitting on the beach gives payoff of 0.
If a swimmer is attacked by sharks on the first day, then any swimmer will definitely be attacked on the second day.
However, if at least one person swims on day one and is not attacked, then it is clear that there are no shaks, and no attacks will take place on day two.
If no one swims on day one, then nothing is learnt, and so the probability of shark infestation remains at p.
Each player seeks to maximise the sum of his expected payoffs over the two days.
Assume that 2(1-p)/p > c > (1-p)/p

Question:
In the above scenario, what is the best strategy? In other words, what should the peson do on day one? Please explain your answer.

Hint:
Consider first the case where only one person is present at the beach over two days. What is the best strategy for this person?

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?