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 06-06-2012, 11:29 AM #1 banned   Join Date: Jun 2012 Posts: 46 Probability I beat this person in air hockey We have played ten games of air hockey, and on average I lose 7-4. Sometimes I lose 7-6, once 7-0, but when we play, I expect to lose 7-4. How much more information is needed to set a line on my beating this person the next time we play?
 06-06-2012, 04:18 PM #2 adept     Join Date: Jun 2005 Location: Playin' It Smart Posts: 745 Re: Probability I beat this person in air hockey You need a standard deviation for the amount by which he beats you. (Along with some basic assumptions about the distribution of that amount.)
06-06-2012, 05:16 PM   #3
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Join Date: Jun 2007
Posts: 1,557
Re: Probability I beat this person in air hockey

Quote:
 Originally Posted by hyachachacha We have played ten games of air hockey, and on average I lose 7-4. Sometimes I lose 7-6, once 7-0, but when we play, I expect to lose 7-4. How much more information is needed to set a line on my beating this person the next time we play?
If we assume that each point is independent, and we can take an average score of 7-4 to mean you win 4/11ths of the points, then the answer comes from the binomial distribution.

The probability of your opponent winning <= 6 out of the 13 points (we may assume every dead point is played without affected the maths) is =BINOMDIST(6,13,7/11,1) = 0.1534

 06-06-2012, 05:16 PM #4 Carpal \'Tunnel     Join Date: Jun 2005 Location: Psychology Department Posts: 7,426 Re: Probability I beat this person in air hockey Another way to approach the problem might be to consider each point an independent event. Thus, what is the probability that you win a point? Based on your 7-4 average score, we can roughly assume that your opponent wins a point with a probability of 7/11 and you win with a probability of 4/11. Then we just need to figure out the probability that you get to 7 points before your opponent gets to 7 points given those probabilities. I am certain there is an exact answer, but for me it is easier to simulate it. R Code Code: ```outcome <- c("win", "lose") sims <- 1000000 wins <- 0 losses <- 0 for(i in 1:sims) { game <- sample(outcome, 13, prob=c(4/11, 7/11), replace=T) winsum <- sum(game=="win") losesum <- sum(game=="lose") if(winsum >= 7) {wins <- wins + 1} if(losesum >= 7) {losses <- losses + 1} } wins / sims losses / sims``` After 1,000,000 trials you won 15.31% of the time. So I'd estimate your probability of winning at about 15%. edit: Ok...pokerfarian beat me to the post...and made me feel like an idiot because it definitely was not easier to simulate it. I just stopped thinking.
 06-07-2012, 06:41 PM #5 adept   Join Date: Aug 2010 Posts: 888 Re: Probability I beat this person in air hockey Just one thing here.. If you play a single game and the score is 7 to 4, then the assumption that the probabilities of scoring the next point are (7/11, 4/11) isn't so great. If the score were 7 to 0, then going about things this way would imply that you have a zero chance of ever beating your opponent. So it's probably best to assert a prior probability distribution of some sort... The easiest option (related to the Beta distribution) is to just add 1 to each score. So the probabilities would be (8/13, 5/13)... Or if you played two games that were 7 to 4 each -- then your opponent scored 14 points, you scored 8 total, so you would use (15/24, 9/24), etc... You basically start with the assumption that you and your opponent are equally likely to score the next point, and adjust this assumption with every observation you make.

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