From the EQLive boards.
Now, my statistics and probability are a bit rusty, but here goes.
The chance to succeed, "S", is a binomial random variable, with (in theory) P=.95.
Let E(X) be the expected value of a random event, V(X) be the variance, SD(X) the standard deviation, and C(X) be the 95% confidence radius.
(by definition, SD(X) = sqrt(V(X)), and C(X) = 1.96*SD(X))
More notation: S^1000 means "1000 trials of the random event S".
The following are "stat 101" formulas, given that successive trials are independant:
E(S^1000) = 1000 * E(S) = 950
V(S^1000) = 1000 * V(S) = 1000 * .95 * .05 = 47.5
SD(S^1000) = sqrt( V(S^1000) ) =~ 6.892
C(S^1000) = 1.96 * SD(S^1000) =~ 13.51
So, 19 times out of 20, we would expect after 1000 trials to get between 936.49 and 963.51 successes. (ie, 950 - 13.51 and 950 + 13.51).
Note that
966 is outside of this interval. In other words, the above post demonstrates evidence that with a JC skill of 300 and no AAs and no MOD, on a trivial 335 item, the chance of success is greater than 95%.
Now, is it my math that is off, or did Ngreth incorrectly claim there was no statistical evidence for an anomoly?
Going deeper into the rabbit hole, let us compare the differences between trials.
2 and 3 are the furthest apart:
The success percent on 1000 combines is a random number.
The null hypothesis is that both are 1000 trials at an independant 95% success rate.
If the null hypothesis is true, the expected value is 95% and the SD is 0.689%.
The difference between two independant random variables is a random variable. The variances (not SDs) add, and the expected values subtract.
So, the expected value of the difference between trial 2 and 3 is 0%, with a standard deviation of 0.975%. This gives a confidence interval of 1.91%.
Ie, for any of the above two trials, the difference should be less than 1.91%, or we have an anomoly.
The observed difference between trials 2 and 3 is 1.7% -- within the confidence interval.
So maybe that is what Ngreth did -- compared the differences between the samples, and assumed the 95% success chance was accurate?
On one hand, the samples are not statistically different from each other. On the other hand, one of the samples is statistically different from a 1000 repeated 95% chance trial.
Combining all 4 trials together, we get 3836 successes out of 4000 attempts.
Assuming the null hypotheses (95% chance of success), the expected number of successes would be 3800. The variance would be 190. The SD would be 13.78. The Confidence Radius would be 27.
3836 has a z-score of (3836-3800)/13.78 = 2.6. The chance of this happening by random luck, assuming the null hypothesis, is 1 in 215.
Ngreth's post is strong evidence that the chance of success at that recipie is not 95%. It is higher. This matters to the discussion at hand because a higher underlying success chance gives a lower expected variance, so the difference between trials 2 and 3 becomes more statistically significant.
If you assume a 96% chance of success, the difference between trial 2 and 3 becomes equal to the confidence radius -- ie, there is only a 2.5% chance that a given two trials would be that far apart by random chance, assuming the 96% null hypothesis.
My stats are rusty. Anyone want to tear my work apart?
Originally posted by Ngreth
The chance to succeed, "S", is a binomial random variable, with (in theory) P=.95.
Let E(X) be the expected value of a random event, V(X) be the variance, SD(X) the standard deviation, and C(X) be the 95% confidence radius.
(by definition, SD(X) = sqrt(V(X)), and C(X) = 1.96*SD(X))
More notation: S^1000 means "1000 trials of the random event S".
The following are "stat 101" formulas, given that successive trials are independant:
E(S^1000) = 1000 * E(S) = 950
V(S^1000) = 1000 * V(S) = 1000 * .95 * .05 = 47.5
SD(S^1000) = sqrt( V(S^1000) ) =~ 6.892
C(S^1000) = 1.96 * SD(S^1000) =~ 13.51
So, 19 times out of 20, we would expect after 1000 trials to get between 936.49 and 963.51 successes. (ie, 950 - 13.51 and 950 + 13.51).
Note that
Jewelcraft Skill of 300, No AA's, No mod, trivial 335
Expected Success Rate: 95%
1000 combines - 966 = 96.6% Success Rate
Expected Success Rate: 95%
1000 combines - 966 = 96.6% Success Rate
Now, is it my math that is off, or did Ngreth incorrectly claim there was no statistical evidence for an anomoly?
Going deeper into the rabbit hole, let us compare the differences between trials.
2 and 3 are the furthest apart:
Jewelcraft Skill of 300, No AA's, No mod, trivial 335
Expected Success Rate: 95%
1000 combines - 966 = 96.6% Success Rate
Jewelcraft Skill of 299, No AA's, 15% mod (maxed trophy), trivial 335
Expected Success Rate: 95%
1000 combines - 949 = 94.9% Success Rate
Expected Success Rate: 95%
1000 combines - 966 = 96.6% Success Rate
Jewelcraft Skill of 299, No AA's, 15% mod (maxed trophy), trivial 335
Expected Success Rate: 95%
1000 combines - 949 = 94.9% Success Rate
The null hypothesis is that both are 1000 trials at an independant 95% success rate.
If the null hypothesis is true, the expected value is 95% and the SD is 0.689%.
The difference between two independant random variables is a random variable. The variances (not SDs) add, and the expected values subtract.
So, the expected value of the difference between trial 2 and 3 is 0%, with a standard deviation of 0.975%. This gives a confidence interval of 1.91%.
Ie, for any of the above two trials, the difference should be less than 1.91%, or we have an anomoly.
The observed difference between trials 2 and 3 is 1.7% -- within the confidence interval.
So maybe that is what Ngreth did -- compared the differences between the samples, and assumed the 95% success chance was accurate?
On one hand, the samples are not statistically different from each other. On the other hand, one of the samples is statistically different from a 1000 repeated 95% chance trial.
Combining all 4 trials together, we get 3836 successes out of 4000 attempts.
Assuming the null hypotheses (95% chance of success), the expected number of successes would be 3800. The variance would be 190. The SD would be 13.78. The Confidence Radius would be 27.
3836 has a z-score of (3836-3800)/13.78 = 2.6. The chance of this happening by random luck, assuming the null hypothesis, is 1 in 215.
Ngreth's post is strong evidence that the chance of success at that recipie is not 95%. It is higher. This matters to the discussion at hand because a higher underlying success chance gives a lower expected variance, so the difference between trials 2 and 3 becomes more statistically significant.
If you assume a 96% chance of success, the difference between trial 2 and 3 becomes equal to the confidence radius -- ie, there is only a 2.5% chance that a given two trials would be that far apart by random chance, assuming the 96% null hypothesis.
My stats are rusty. Anyone want to tear my work apart?


Ngreth Thergn
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