Quote:
Originally Posted by Setsy
Good post - but how can the above formula be correct given that it doesn't take into account the # of hours played? Any formula testing the difference between two means needs to take into account the sample size, no? In there you had no place where you used OP's 250 hours of play (I understand that you think the SD will converge with more hours, but don't believe that that sufficiently accounts for this).
Yeah, you’re right. Sorry about that; I accidently used the wrong distribution calculations.
I can't edit my original post but here is the updated calculations:
What you need to do is first calculate your critical value for t in excel:
=TINV(2*(1-confidence level),n-1) where n-1 is sample size minus 1 and confidence level is the % you want to be sure in your results, ie 0.95 corresponds to 95% certainty (can never have 100% confidence)
For example, if you want to be 80% confident in the results and your sample size is 265 hours, your calculation would be
=TINV(2*(1-0.8),264)
=0.842985
Then you need to calculate your actual t value:
actual winrate - desired winrate / (standard deviation/sqrt(n))
For example, if your actual winrate was 74.87 and your desired long-term winrate was 50, and you had a standard deviation of 180, and a sample size of 265, your calculation would be
=(74.87-50)/(180/sqrt(265))
=2.24919
If your actual value of t > critical value (2.24919>0.842985), you can conclude with x% of confidence that your actual long term winrate will be >= your desired winrate.
Using these calculations, by modifying your desired confidence level, desired winrate, and sample size, you can determine the likelihood that you are a long term winner at whatever winrate it is you desire.