Drop rates, loss runs and transparency
None of what follows is new, but I hope it encourages EA to have a think about how they handle artificially stochastic functions in the game, and gives some of you a bit more info about how to gain d...
Forum Discussion
"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.
