I'd say the odds of that happening are pretty slim. A good way of looking at this, statistically, would be via the use of the Standard Normal Distribution. Now the mean (average) amount of slays per round is approximately 0.37. Now there is a standard deviation of approximately 0.14. So with this situation specifically, we would want to find the z score that would be used for 2 slays with 8 players on. Then we would want to...
I don't know what the hell I'm talking about
(03-01-2018, 07:43 PM)Prince Nicky La Flama Blanca Wrote: [ -> ]I'd say the odds of that happening are pretty slim. A good way of looking at this, statistically, would be via the use of the Standard Normal Distribution. Now the mean (average) amount of slays per round is approximately 0.37. Now there is a standard deviation of approximately 0.14. So with this situation specifically, we would want to find the z score that would be used for 2 slays with 8 players on. Then we would want to...
I don't know what the hell I'm talking about
when you try to apply what you learn in school :idea:
That bottom guy had a slay from 10 months ago?
Ignoring the fact they both had slays its 12.5%.
Including the slays, I would say... 12.5%.
You cant factor in the chance of a slay because of the immeasurable factors that would include