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We are playing a game with my friend and trying to determine who is the better player.

The game has an luck element to it.

The current score is 9-7.

The following questions have risen:

1) How many games we should play to get statistically significant result that one is better than the other? We can assume p=9/16 or p=0,5 and something standard for z.

2) What can we infer from the current game score? We tried binomial proportion confidence interval (we used Wilson score, because N<30). Here are the results. Not sure how to interpret them.. does it mean that with 95%, the probability, at which the game score is 9-7, is within 34%-76%?

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    $\begingroup$ The simple answer is that if the players are at all well matched, more than they are willing to play. $\endgroup$ Nov 13, 2010 at 4:52

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I'm not sure I understand your notation, but if we assume that one player wins 9/16 of the time, and we want a confidence of 99.5% (p=0.5%), then we can write the probability that the ratio of 9/16 would arise by chance as a function of the number of games, then solve for the number of games that gives a probability of 0.5%.

Edit: I wrote a function to calculate this for 9/16 and p=5%. The result is that it takes 265 trials (149 wins by one player) to exceed confidence of 95%.

For part (2), you can infer that you can't yet say who is better with 95% confidence. All you can say with 95% confidence is that the "true" outcome is somewhere between 34% and 76%. As long as your confidence interval includes 50%, you can't say who's better.

Here's my sample code in Python (you need the scipy.stats package):

s = 9
n = 16

while binom_test(s, n) > 0.05:
    n += 1
    s += 9.0/16

print n
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  • $\begingroup$ Ah, my bad. I used p to determine probability, which seems to be used for confidence. Sorry about that. We would be very interested to see the function you used to calculate the number of trials. Thanks! $\endgroup$
    – randomguy
    Nov 12, 2010 at 21:00

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