Expert: Sherry Wallin - 7/13/2009

Question
QUESTION: Hello, I am trying to discover if there is a math formula that already addresses my problem.  Perhaps you can help or at least shed some light on it.

In most competions, the results are zero sum (i.e. there has to be a winner and a loser - and usually expressed in some winning percentage).  However, I'm trying to devise a way in which ALL competitors would have a winning average (say 50% - 100%).   

Is this a solved problem?

Thanks.

ANSWER: Mark if there are n players and only one winner then there are n-1 losers. Every player has an equal chance for winning so the chance of winning would be 1/n. I'm not sure is if this is what you are attempting to get at?

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Thank you. Actually, I was trying to find a way to show results, expressed in terms of 50% - 100% rather than on a scale of 0% - 100% (so everyone can be perceived to be a winner and average at worst).  Actually, I think this is it: .5X + 50 (where X is the winning percentage).  Sound correct? Thanks.

Answer
Mark~
    That won't work because the probabilities need to add up to 1. If you use .5x + 50 you will get much more than 1 (50 + x>= 0) and if you wrote it as .5x + .5 then the only time it would be accurate is if x + 10, which is 1000%. I don't think you can write it the way you want without violating the sum of the probabilities = 1.

Math Prof