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

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?

*Last edited by tongzilla (2006-11-09 06:11:56)*

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**Toast****Real Member**- Registered: 2006-10-08
- Posts: 1,321

O_O ...

Someone's created mathematical solutions to not getting eaten by sharks...

*Last edited by Toast (2006-11-10 05:30:10)*

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

Toast wrote:

O_O ...

Someone's created mathematical solutions to not getting eaten by sharks...

???

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,315

Game theory.

I would rather see the game as a mimic of two competing firms at the point to decide whether or not to invest in R&D unprotected. And this makes it less cruel.

*Last edited by George,Y (2006-11-11 23:45:55)*

**X'(y-Xβ)=0**

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