Re: Nash games theory
- From: "Peter Webb" <webbfamily@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Thu, 15 Jan 2009 22:19:07 +1100
<pauldepstein@xxxxxxx> wrote in message news:4a40baf4-37c3-42e7-b017-cfc487697d0d@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
A game is being played by more than 1 person. This game has an
associated integer, N > 1.
The game consists as follows: Each player secretly writes down a
number between 1 and N inclusive.
After these numbers have been written and all numbers are known to all
players, each player receives a monetary payout from the banker or
makes a payment to the banker. [In other words, there are no
restrictions on the set of payouts.] This payout depends, by a well-
defined set of rules, on which numbers have been written by each
player. [In other words, I am speaking of a class of games].
All the players are treated in the same way. For example, if two
players have written down the same number, they will receive the same
payout.
Each player has the aim of maximising the expected value of the payout
she receives.
Often the optimal strategy will be probabilistic -- write i with
probability a_i where 1 <= i <= N.
If we can assume that there is a unique optimal strategy, such
problems can sometimes be solved by Lagrange multipliers -- the
constraint is that the sum of a_i = 1, and the function to be maxed is
derived from the set of rules.
It is assumed that each player uses this unique optimal strategy.
However, I have two questions:
When is there an optimal strategy? When is the optimal strategy
unique?
Paul Epstein
I limit myself to zero-sum games in what follows ...
Consider the following three people game.
Each picks a number between 1 and 10. If everybody guesses odd numbers, or everyone guesses even numbers, the stakes are returned. If only one is different, they win the whole pot.
By symmetry, it doesn't matter if you pick an odd number or an even number.
But you have two optimal strategies - pick an odd number, or pick an even number. Both strategies are optimal in the sense they cannot be improved upon.
Now, it seems to me that I can produce two directly contradicting statements.
1. In a finite game, there can never be an optimal strategy.
In general, if every player uses the same strategy (optimal or otherwise), then they will all on average win or lose the same amount (by symmetry). So if everybody is using the same optimal strategy, then the expected return from the optimal strategy is zero.
No move can have an expectation greater than zero, as that move would be better than the "optimal" move which only had expectation zero. If some moves have an expectation less than zero, then there must be moves with an expectation greater than zero, as the total expectation across all moves is zero. So all moves have expectation zero, and it doesn't matter what you do.
2. In a finite game, there is always an optimal strategy
In any finite game, there must always be an optimal strategy. All moves have an expected return, because we can play the game a billion times using that strategy and measure it. Pick the move with the highest expected return; if there is more than one pick one at random.
(Note this is not true for infinite games; consider the game where the winner is the person who picks the largest number.)
These two statements are obviously directly contradictory. I don't know which one (if either) is correct.
.
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