Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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%?

share|improve this question
1  
The simple answer is that if the players are at all well matched, more than they are willing to play. –  Ross Millikan Nov 13 '10 at 4:52

1 Answer 1

up vote 1 down vote accepted

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
share|improve this answer
    
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! –  randomguy Nov 12 '10 at 21:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.