(09-26-2012 03:38 AM)Harti Wrote:  

(09-26-2012 02:38 AM)Ravernoth Wrote:  The World Cup tournament alone has enough games to be statistically valid. Whether it's 2:1 or 3:1, it's a big issue.

Haha, sorry. While I'm your side it's false to claim that a sample size smaller than 10- or 100,000 shows a statistically valid empirical result.


Good news is that Alex likes the 8 wit idea.
Sounds like we're testing it next week. =)

(09-26-2012 04:42 AM)vivafringe Wrote:  This isn't true! It's a common fallacy that you always need a huge sample size to prove things. It turns out our data is so incredible that there is a huge probability that P1 is advantaged even with our tiny sample size. The standard statistical test for this type of data is, as mentioned in the other thread, to look at the ties. In the other thread, there were 17 tied games where the players won as 1p, and 2 tied games where the players won as 2p. Let's say we want to test whether there is an advantage to playing first. A naive view would be that player 1 has a 50% winrate vs. player 2. Let p be the winrate of player 1. We'll do a 2-sided test, even though a 1-sided could be argued to be more appropriate.

Ho (the null hypothesis): p = .5
Ha (the alternative hypothesis): p does not equal 0.5

The number of tied games where p1 follows a binomial distribution with n = 19, p = 0.5. Our p-value is 0.000729. In other words, if the actual winrate was 0.5, we would only expect a result this unusual 1 out of 1372 times.

(09-26-2012 05:10 AM)TheQwertiest Wrote:  i thought this thread was suppose to be for the praise OML needs xD

(09-26-2012 04:42 AM)vivafringe Wrote:  

(09-26-2012 03:38 AM)Harti Wrote:  Haha, sorry. While I'm your side it's false to claim that a sample size smaller than 10- or 100,000 shows a statistically valid empirical result.

This isn't true! It's a common fallacy that you always need a huge sample size to prove things. It turns out our data is so incredible that there is a huge probability that P1 is advantaged even with our tiny sample size. The standard statistical test for this type of data is, as mentioned in the other thread, to look at the ties. In the other thread, there were 17 tied games where the players won as 1p, and 2 tied games where the players won as 2p. Let's say we want to test whether there is an advantage to playing first. A naive view would be that player 1 has a 50% winrate vs. player 2. Let p be the winrate of player 1. We'll do a 2-sided test, even though a 1-sided could be argued to be more appropriate.

Ho (the null hypothesis): p = .5
Ha (the alternative hypothesis): p does not equal 0.5

The number of tied games where p1 follows a binomial distribution with n = 19, p = 0.5. Our p-value is 0.000729. In other words, if the actual winrate was 0.5, we would only expect a result this unusual 1 out of 1372 times.

(09-26-2012 07:14 AM)awpertunity Wrote:  

(09-26-2012 04:42 AM)vivafringe Wrote:  

(09-26-2012 03:38 AM)Harti Wrote:  Haha, sorry. While I'm your side it's false to claim that a sample size smaller than 10- or 100,000 shows a statistically valid empirical result.

This isn't true! It's a common fallacy that you always need a huge sample size to prove things. It turns out our data is so incredible that there is a huge probability that P1 is advantaged even with our tiny sample size. The standard statistical test for this type of data is, as mentioned in the other thread, to look at the ties. In the other thread, there were 17 tied games where the players won as 1p, and 2 tied games where the players won as 2p. Let's say we want to test whether there is an advantage to playing first. A naive view would be that player 1 has a 50% winrate vs. player 2. Let p be the winrate of player 1. We'll do a 2-sided test, even though a 1-sided could be argued to be more appropriate.

Ho (the null hypothesis): p = .5
Ha (the alternative hypothesis): p does not equal 0.5

The number of tied games where p1 follows a binomial distribution with n = 19, p = 0.5. Our p-value is 0.000729. In other words, if the actual winrate was 0.5, we would only expect a result this unusual 1 out of 1372 times.

How did you get that? I get an even smaller p-value of
0.000364303588866965 for the chances of P1 winning 17/19 or more of the games, or 1 in 2744.

EDIT: Oh nevermind you used a 2-sided test for some reason haha. So mine is exactly half of yours

(09-26-2012 12:20 PM)worldfamous Wrote:  

(09-26-2012 07:14 AM)awpertunity Wrote:  

(09-26-2012 04:42 AM)vivafringe Wrote:  The number of tied games where p1 follows a binomial distribution with n = 19, p = 0.5. Our p-value is 0.000729. In other words, if the actual winrate was 0.5, we would only expect a result this unusual 1 out of 1372 times.

How did you get that? I get an even smaller p-value of
0.000364303588866965 for the chances of P1 winning 17/19 or more of the games, or 1 in 2744.

EDIT: Oh nevermind you used a 2-sided test for some reason haha. So mine is exactly half of yours

So, for us laymen, what you're saying is something like, if you flip a coin 19 times there's a 1/1372 chance that it will land on heads 17/19 times? Henceforth, it's unlikely that its a coincidence that P1 won 17/19 times.