Drop rates, loss runs and transparency

"kobe_juan_canobie;864913" wrote:

"Acrofales;864863" wrote:
Bollocks. Those 9 runs without salvage I was talking about? They just become 15.
E: actually, I'm not quite sure they are equivalent. My statistics is a bit shakey here, but given a long chain of actions you'd probably also have to take into account how many sequences may occur. And there are far far far more 2-sequences than 15-sequences. So starting from 0, your chance of hitting 2 salvages in a row is only slightly higher than higher than hitting 15 shots of nada, but in a long chain of actions, it should be easier to find a sequence (any sequence) of 2 salvages in a row than a 15-sequence of nothing?

The sequence of win/loss trials shouldn't 'remember' previous trials, so the probability of an arbitrary number of losses before a wins should be a geometrically distributed random variable (and so far as I can tell, both loss and win run distributions are distributed this way).

"Acrofales;864863" wrote:

If I got this right, then given a chain of X actions, the number of 2-sequences are X - 1, whereas the number of 15-sequences is X-15. So your chance of finding a 2-sequence is: 1 - 0.96^(X - 1), whereas your chance of finding a 15-sequence of no salvage is 1 - 0.965^(X-15). So it clearly depends on the size of X. If X is large enough, then the influence of X is negligible (the chance trends to 1, of course, in the infinite), but if we look for something like X=100, then we have a 98% chance of getting a 2-sequence with 2 stun gun salvages, and a 95% of having any 15-sequence with 0 salvages, which is a noticeable difference (albeit not really to the human running those 100 sims, because human perception of chance sucks).

If the drop rate is denoted p and the probability of missing a drop is 1 - p then the probability of 2 losses before a win is just:

(1 - p) * (1 - p) * p 

If the drop rate is 0.2, then that gives a 0.128 probability of two losses before a win (three trials in all).

The probability of 15 losses before a win is

(1 - p)^15 * p = 0.007036874

- not very likely at all.

Give it some thought (or check the wiki page and you can work backwards from probability to loss runs. I think the most interesting thing to monitor is the expected maximum loss sequence for an arbitrary number of runs -- if you think about it, the expected maximum loss sequence is more useful to know ahead of time.

I don't think we're disagreeing, just calculating 2 different things. I was calculating (1) the probability of me going on a 15-run dry spell, and then putting it into perspective by calculating there being a dry spell of at least 15 sims in a sequence of 100 sims (pretty sure my 95% on that is correct).

I also agree that being able to compute the expected maximum loss sequence for a number of runs is far more useful (and will let you put it into perspective). I saw you did that for stun guns, and my 15 run sequence is still within the bounds.

Now I accumulated enough energy for another 3 sims. So lets go see what I get!

---

And. Zip. Nada. Squat.

So we're up to an 18-run. Which is out of bounds for maximum expected dry spell (was that a 95% confidence interval?) in 100 sims. Of course, you computed it for a sequence of 100 sims, which is an arbitrary number of sims you chose because it's a pretty round number. And I don't really know how many I have done. I stopped somewhere around 30 stun guns to do light side battles for the daily. I then got up to 46 stun guns before my dry spell started. So how do we count that sequence? In total it's clearly more than 100 sims, but since the light side battles ended, I spent 4 refreshes, so we're only up to 48 sims + natural regen (probably another 30 or so?)