The Math Of Farming

[GoneBlank]

I was wondering if somebody with stronger math skills than me could check this logic.....

 

“…the following form of geometric distribution is used for modelling number of failures until the first success:

 

                Pr(Y = k) = (1 – p)^k p

 

for k = 0, 1, 2, 3,….”.

 

https://en.wikipedia.org/wiki/Geometric_distribution

 

where k denotes the number of independent trials each with success probability p. Translation,

 

k = number of failed missions run (before success)

p = item drop rate of interest

 

So the above probability starts by giving a probability that we get the item straight away, Pr(Y = 0) = p. Next, we have the probability of getting the item after one failed mission Pr(Y = 1) = (1 – p)p. The probability of getting the item after two failed missions Pr(Y = 2) = (1-p)^2 p etc.

 

What we are really interested is the aggregation of all these scenarios so we can answer questions of the following form:

 

How many missions do we need to run for a: (i) 50%, (ii) 90% and (iii) 99% chance of getting the item?  Here we apply the Cumulative Distribution Function and solve for k

 

                k = ln(1 – CDF)/ln(1 – p) - 1

 

Let’s consider the case of Armored Agility mod which – according to the wiki – has a drop rate of p = 0.67%. Hence,

 

k              CDF        p

102         0.5          0.0067

342         0.9          0.0067

684         0.99        0.0067

 

So 102 runs sees an even (50%) chance of the item dropping. 342 runs sees a high (90%) chance of the item dropping. 684 runs sees an almost certain (99%) chance of the item dropping.

 

Let’s contrast this to a more common mod, such as Shred which has a p = 8.431% drop rate (according to the wiki).

 

k              CDF        p

7              0.5          0.08431

25           0.9          0.08431

51           0.99        0.08431

 

In this case 7 runs sees an even (50%) chance of the item dropping, 25 runs sees a high (90%) chance of the item dropping and 51 runs sees an almost certain (99%) chance of the item dropping.

 

If the above numbers are correct - my desire to farm for Armored Agility has been destroyed.

 

[(PSN)Juzam666]

Seems easier to go into the trade chat and type, WTB r0 armored agility 30p

[321agemo]

Seems easier to go into the trade chat and type, WTB r0 armored agility 30p

Either that or wait for it to appear on alert again

[Agentawesome]

From what I studied in mathematics the logic is correct.

But in a human perspective, regardless of how high or low probability, its all still luck.

[(PSN)T_ravenis]

My head hurts now

[H3RRD0KT0R]

I feel like a grineer reading that. 

[ThePredicament]

Just have to farm in an intelligent manner. Farmed thirty times? Chances are, there are a bunch of items that can be quicksold (some plats under average) that can help pay for that part you need. Calculating probability just works out when there aren't any other alternatives to actual farming.

 

Makes me miss calculus classes though.

[Sitchrea]

But doesn't the entire equation get reset upon each mission? RNG doesn't compensate for your previous attempts, only the current attempt.

[AoiiToori]

Math of Farming?

 

 

Farm=RNG

 

done.

[GoneBlank]

@Sitchrea: Hello. The above math assumes the result of each mission is independent of any previous mission and that the drop rate for the item being farmed is constant. I don't think there is any type of "reset upon each mission" (but I'm not 100% sure - it's been years since I thought about this stuff seriously). All I'm looking at is the range of losing-streaks you need to consider for a given probability of success.

 

I know that RNG is all about luck and that low drop rates = unlikely outcome. Saying that drop rate for X = 0.67% is one thing - but saying that you may have to run up to 342 missions to have a 90% chance of successfully getting the item casts the issue in whole different light (well at least it did for me).

[elele]

Pretty much correct although I didn't check the details. Note you actually need to run (k+1) times, because you'll only stop running UNTIL you get the desired item.   Nvm. I didn't see that you already mentioned that k is for failed missions. 

Edited July 22, 2015 by elele

[Lightsmith]

Don't you just want to know "Drop Rate"

Although RNG things are weighted. i.e. so on a 1-100 roll, 1-98 is an Orokin Cell, 99-100 is what you want.

The Codex does break down on rare drops are (not rewards) Gold is less common.

I'm sure there is a site that breaks down drop rates, but maybe not since DE can adjust the drop tables without notification.

 

To get the actual odds you'd have to run tons and tons of missions. Depending on the size of the matrix it would take longer as well. if its 1-10 its easier to get a breakdown for odds than a 1-1000 matrix. Your deviation will be very high until you do a large number of runs.

 

DE could further throw this off by having dynamic loot tables, where they can change odds throughout the day. (I doubt they do). Since Loot Tables are server side DE can do them without anyone having to even log out.

 

The amount of data you would need to for a reliable table would be immense, you'd be better off just asking DE, what are the odds of each thing.

 

I won't even bring up that there is nothing that is actually random.

Edited July 22, 2015 by Lightsmith

[GoneBlank]

Hello Lightsmith.

 

I am explicitly assuming that the drop rate is known and constant. Is that the truth? I don't know. The drop rates I list in the original post come from the Warframe wiki (lookup Nightmare Mode).

 

Do you want to know "just" the drop rate? Well, if you are only going to play the mission once then the drop rate pretty much tells you the whole story. If the drop rates are very low (and in some cases extremely low) I think it is interesting to look the issue from a slightly different angle - how many times will have to run this mission to hit a given probability of the item dropping? I find this easier to contextualize. However, I think this ultimately boils down to personal preference.

[Soju.]

What... the...

... I think I need to lay down for a moment.... my head hurts..

[(XBOX)Grihaly]

There is no math in farming, only pain and suffering.

[HyokaChan]

The math of RNG is that the chance to obtain anything you want is effectively 0%

 

0% * 1,000,000 tries = 0%

[ShortCat]

Very nice and informativ post. Now I only need to know he droprates of the items.

[SmokeyJesus]

But doesn't the entire equation get reset upon each mission? RNG doesn't compensate for your previous attempts, only the current attempt.

I believe the equation accounts for that. Geometric Distribution is the idea that the more times you accomplish a task, the higher a chance for the desired reward. But it never reaches 100% (there is never a 100% chance for a drop table reward) because the distribution is exponential. If OPs math is right you could have run the mission nearly 1000 times for armored agility, and would still be in the 99.X% range.

Edited July 22, 2015 by SmokeyJesus

[Odadda]

The values are correct.

 

Don't take this statements as gold, because I've done my Calculus and Statistic course like 2 years ago.

 

Geometric Distribution is cool for a distribution of natural values. To be totally sure howerver, I'll use the Normal distribution. It should be the same thing in this case.

 

OP said the chance of getting Armored Agility is 0.67% a.k.a. 0.0067

 

p = 0.0067

 

Which mean the chance of not getting it is

 

pn = 1 - p = 0.9933

 

Now what if I do x runs? Well, we are looking for a single mod a.k.a. a single success. So to get it we need to "break up" a serie of "failure" streak.

 

How much chance I have to not get it in x amount of runs?

 

Pn = pn^x = 0.9933^x

 

And so 

 

P = 1 - Pn = 1-0.09933^x

 

How to find x?

 

If we use the Normal distribution, chances are our probability of success will follow a curve similar to this ones

 

 

As you can see, the highest value of chance is in the middle of the curve. Be careful thought, the probability you see every point of the curve has is for the specific event that is in the x-axis (I'll explain it later).

 

What we are looking for here is the area under those curves: they are Normalized curves, so the area of every of them is 1 ( to represent the 100% chance of getting a positive event after infite runs).

 

The only thing Important from those curves now is to understand that we have the highest chance of success halfway in our runs. So let's see 

 

P(max) = 0.5 = 1 -0.9933^x

 

0.5 = 0.9933^x

 

x= log 0.9933 (0.5) = 103.107

 

almost as you stated.

 

However we have to consider another factor, the standard deviation.

 

You see, thse curves tells us another thing: the more runs we do, the less chance we have to succeed.

 

How's that? Reason is , the more runs we do , the more likely is us to succeed and less to fail. Thus is less likely we'll have to do 684 runs to get the mod.

 

I don't want to make this too long or complicated: basically the rule is that taking your expected value (OP's k, my x) has a 50% chance of success.

 

Standard deviation allows us to see the chance of having such mod after having delimited a range of runs we have to do.

 

To make it quick, I have calculated this deviation supposing the Normal distribution to be rather a Poissonian distribution (a.k.a. a Normal distribution with a very low chance of success for the single run).

 

Under this conditions

 

Standard deviation = ro = srqt(103.107) = 10.14

 

we have 50% chance to succeed after x runs, 68% after x+ro runs, 95% after x+2ro and 99.5% after x+3ro.

 

So our "best bet" is to obtain it after 103.107+3*10.14= 133.57 runs.

 

EDIT: Probably there are some mistakes on the last part, but the essence should be correct. Will take back my book this evening.

Edited July 22, 2015 by Odadda

[helix.hex]

math? in warframe? what kind of magic is this? have you seen or heard what happens when math is involved?

[-CM-Valen]

Then I must be really unlucky to grind 103 runs during the previous TA2 alert but not get Buzz kill. 50% would have gotten it from the above logical derivation

[Lichcontract]

Wait so does this mean the Despair would take like 25k stalker kills to have gotten one at the 99% mark? God what has happened

[Kevlareater]

Great. This may explain why I'll never again see another Dual Decurion/Fluctus reward unless I buy plat.

[-KoreanPotato-]

Your chances keep getting closer to 100% but it will never reach it! Therefore, you will never get it! Muhuhahahahahaha

[GoneBlank]

Thanks for taking a detailed look Odadda (it's been over 20 years since I last looked at stuff like continuous v discrete distributions etc at uni). At least it seems my original logic is right (the grey matter in my head hasn't failed just yet)- which I'm happy about.

 

What do we get out of all this?

 

Firstly, a straightforward explanation to give to someone who posts "...I've done over 200 runs....how come this hasn't dropped yet....". I know the term "straightforward" is quite subjective - however, at least I find this structure useful to get more context.

 

Second, I hope DE thinks seriously about setting minimum drop rates and that the structure presented in this thread helps them think about the implications. For the Armored Agility example in the original post, for every person who manages to farm Armored Agility within 102 runs there will be someone who requires more than 102 runs to get the item. Think about that - half the community farming this item will not get it after 102 runs.

 

PS This is not intended as a rant about Armored Agility (it's just an example). The focus here is really on translating drop rate percentages in to a language of how many times am I likely to have to run said mission to gain the item (to a given level of certainty).