Clarifying a couple of things that are sometimes misunderstood
The modifiers for multicraft chances work differently from the ones used for quality chances. Whilst quality modifiers are multiplicative (well, actually weird form additive but that’s a topic for another time), the multicraft ones are sensible additive, which is:
A) Cool and
B) Useful as hell.
To give an example, let’s take a look at Druid Laurels:
By default an item has a multicraft chance of 0%. As you can see from the picture we’ve ascended the item, giving us a 10% chance to trigger multicraft each time we make it.
At 21 stars in the ascension tree for an item you unlock a further 5% multicraft chance. We now have a 15% chance to trigger it each craft.
To go even further, the multicraft event adds another 25% so during that we would have a 40% chance to trigger multicraft (or: lots)
You might think “that’s insane” or “you’re talking complete rubbish Reiga” but you don’t have to take my word for it, you can take the word of the developer 😛
Small Talk success rate in events
Whenever we get an event like Better Deals some players feel that the success rate is lowered for Small Talk. During the past weekend I set out to test this to be able to confirm or deny if that is a thing:
After recording just over 2.5k small talks I can safely say: no, it doesn’t change. 74% is actually higher than what the original small talk sample test found before (over 6k small talks were taken, the pass rate was estimated to be 70%.)
If you wish to contest the numbers above, here is the footage I recorded to go along with it:
Two things to note
1) I didn’t record the first 100 sales cause it only occurred to me then to do so. You can subtract 100 off the pass column if you want, still puts us at 73.52% pass
2) At one point I managed a string of 16 successes in a row. If you could find the timestamp and send that to me I’d be very grateful.
2 replies on “Misinformation Clarification”
You can estimate a confidence interval for your small talk success rate. This is a binomial distribution; the observed success rate is p = 0.7445 over n = 2517 trials. For example, if we want an error rate of 0.05, then we can use the equation p ± 1.96*SQRT(p(1-p)/n). In your case, this tells us that our 95% confidence interval is approximately [72.25%, 76.15%].
Hi, thanks for that calculation, learning new things every day 😛