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The better player

  1. Dec 14, 2015 #1
    Below is a question I found in old statistics book of mine, that I really would like to know how to solve:
    Suppose two players, A and B, play a game. If we assume that A has probability pA og winning and B has probability pB=1-pA of winning, the number of wins and losses for player A will be binomially distributed.
    Now let us assume that we a priori don't know pA and pB. Player A and player B play 50 games and it is found that player A wins 30 games. Can he then claim that he is the better player?
     
  2. jcsd
  3. Dec 14, 2015 #2

    BvU

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    He can always claim. Politicians do it all the time.

    Since this comes from a statistics book, there must be more to this exercise. Perhaps you should read the preceding chapter (hypothesis testing is what it may be called). Or you should use some relationships about the binomial distribution to tell you what the likelihood is that you win 30 out of 50 games, even though your pA is only 0.5.

    You've been around long enough to know that PF is about getting assistance in doing your learning/exercises, not about finding someone to do it for you :smile: . So show some attempt and all will be done to help you further !
     
  4. Dec 14, 2015 #3

    micromass

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    You need to do a hypothesis test.
     
  5. Dec 14, 2015 #4
    I have thought about everything you said, I just don't like that as a solution.
    I can do a hypothesis test and use as my null-hypothesis that the players are equally good. Then I can calculate the probability that one player wins 30 games and see if I want to reject that on some significance level or I want to disprove my null hypothesis.
    I just don't see how much is to be learned from that.

    How about instead I look that the general binomial expression:
    P(30,50) = K(30,50) * pA^30 * (1-pA)^(50-30)
    And then calculate something like:

    ∫P(30,50)dpA/∫P(30,50)dpA, where the integral in the numerator extends from 0.5 to 1 and the one in the denominator extends from 0 to 1.

    To me this approach makes more sense... I guess... though it kind of bothers me that pA is not a probability density, so I am not sure about the validity of the approach. The idea of the above, as you might have guessed, is to take sum the probabilities for winning 30 games for all pA>0.5 and then weight them by the total sum of probabilities for winning 30 games for all possible values of pA (hence the integral). What do you think of this approach?
     
  6. Dec 14, 2015 #5

    BvU

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    All far too complicated. You know the standard deviation for such a distribution. A deviation of 5 wins is well within the two sigma for a 95% confidence level .

    And for n = 50 the binomial and the normal distribution are virtually identical.
    My advice: read that old book !


    Binom.jpg
     
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