Question : the following is a general question on probability related to Sports TEAM A WINS 55% OF GAMES PLAYED TEAM B WINS 40% OF GAMES PLAYED AGAINST THE SAME OPPONENTS WHAT WINNING PERCENTAGE SHOULD BE EXPECTED WHEN TEAM A PLAYS TEAM B (NEUTRAL SITE)? EXPLAIN WHAT IF TEAM A WIN 65% AND TEAM B WINS 60% Hi, Since 55/40 = 1.375, you might say that team A is 1.375 times as strong as team B. If P% is the percentage of times that B wins when playing A, then A wins (100-P)% of the time when playing B. If A is 1.375 times as strong as B then 1.375 x P = 100 - P thus 2.375 x P = 100 and P = 100/2.375 = 42% Thus B would win 42% of the time and A would win 58% of the time. Of course, the world doesn't work that way. You can't treat "probabilities" in sporting events like you would probabilities with dice. There are just too many variables and too much variation. Which teams did team A beat? Which teams did team B beat? What was the weather? Who had a "good day"? ...??? Even if you could control all the extraneous variables these "probabilities" are suspect. People spend their lives studying the mathematics of tournaments. We all know of situations where A beats B; B beats C; and C beats A. I wouldn't spend 5 cents on a wager based on the "winning percentages" I derived above. Cheers, Harley Go to Math Central