Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to understand what is the optimal strategy for a mixed game.

I can illustrate the game as a trading system where you can go Long or Short. Going Long will give 80% win rate and going short will give the remaining 20%. Does the optimal strategy is to go always long or a mixed strategy is more optimal(a mix between going Long/Short with a random percentage)?

How do you approach such games?

share|improve this question
    
Welcome to Math.SE. Thank you for your question. It will help us to better answer it if you give the context of the problem, such as where it comes from, as well as anything you've tried so far. Also, there seems to be some details missing about this game; what are the payoffs? Is it just win/lose or are there degrees of winning? Can the opponent's choices affect the outcome, or is it just whether I choose Long or Short? –  vadim123 May 4 '13 at 15:49
    
This could be a very good question, but as stated by @vadim123 you should add some specifics. Are multiple players involved? Is the objective to win as much as possible or to win over the other players? –  Glen The Udderboat May 4 '13 at 17:24
1  
Thank you very much for the warm welcome. The payoffs are +1 -1, no degree of winning(I want to make it as simple as possible so I could better understand the intuition). Vadim you are right the opponent's can not affect the outcome is all about my decision against the "market". –  Freewind May 4 '13 at 17:35

1 Answer 1

The game as I understand it is:

If I go long, I will win 1 with probability $0.8$. If I go short, I will win 1 with probability $0.2$. Suppose I play the game $n$ times. What is my optimal strategy for those $n$ plays?

This is called a Poisson distribution. For each $i$ with $1\le i\le n$, I can either choose $p_i=0.8$ or $p_i=0.2$. My expected winnings is given by the sum $p_1+p_2+\cdots+p_n$. To maximize my expected winnings, I should always go long, in which case my expected winnings are $0.8n$.

share|improve this answer
    
Thanks Vadim. But if I change the opponent to one that can change his strategy under the "No Free Lunch" principle, how does it change the intuition of the question? –  Freewind May 4 '13 at 19:40
    
"No Free Lunch" applies to two-player games. The game I described is a one-player game, since the opponent's choices do not affect the outcome. –  vadim123 May 4 '13 at 19:47
    
I understood it, my comment was how would you change your answer if this is a two-player game? –  Freewind May 4 '13 at 19:49
    
If the second player affects the outcome, then I will need to know what the outcomes are depending on the choices of the two players. It will then be a completely different game. –  vadim123 May 4 '13 at 19:50
    
This is a zero sum game. My thinking is that the 80/20 is the starting condition and as time pass each player will adjust his decision accordingly, this is why I think it is as a mixed game. I hope I made myself clear enough. –  Freewind May 4 '13 at 19:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.