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[Guide]How many bp's are needed to form a set? -the monte carlo approach

DeletedUser109250

Hello.

First let me start by stating the assumption that when given a bp you are equally likely to hit in all the 9 different spots, and when you trade two for one you are equally likely to hit in any of the 8 other spots.

Also, lets define set as one of each of the 9 different bp's on any given gb.

Intro; why monte carlo?

The probability to finish a set in only 9 bp's is 8!/9^8, this equates to a litle less than one tent of a percent. I left out the first bp as that is guaranted to hit an empty spot, and the probability will decrese for each one as the spot gets taken.
Now the analytical prosess get quite tricky from here on out, as there are 9 ways to finish a set with 10 bp's and a lot more ways with 11 and then you have to factor in the 2:1 trade as well.
My analytical skillset is not equipped to handle that problem, but my programming is.
Thus, I wrote a program to run a number of simulations and the law of large number tells us that the probability goes towards the relative frequence as the number of tries increases.

My program will deal a bp at random to the nine slots, it will trade two for one if any slot has three and will return the number of bp's dealt to complete the set.

I ran 100 000 000 simulations and this is my findings:

complete in 9 : 93984 (this is then 0.094% the analytical value gives us 0.093665% so spot on so far!)
complete in 10: 450383 this would be 0.45%
complete in 11: 1174678 this is 1.2%
and the list goes on.

Now I want to know, how many bp's are needed to be done by 50% probability, then we look at the cumulated (added) finishes and we pass 50 million when going to 21 bp's

That means it's equally likely to use 20 bp's or fewer as it is to use 21 bp's or more!

some more interessing results:
with 95% probability you should be done in 39 bp's or less
(still on average 1 out of 20 will use 39 or more!)

with 99% probability you should be done in 50 bp's or less.

1 845 out of 100 000 000 runs used 100 or more tries. So it's more likely to win the lottery than it is to use more than 100 bp's but given enough attemps it will happen :)


If anyone wants my program it's 40 lines of java, and I'll be happy to clean up some variable names and add a ew comments.

Best regards, SlakterT
 

DeletedUser106685

Nice work! Another question for you.

Say you're looking for one last Alcatraz blueprint. You have 8 filled, no doubles.
You need it quickly, so you polish/motivate only progressive era buildings manually.
How many BPs will you need to find on average to complete your set?
(Don't forget that you can also get BP from the 2nd GB that you don't want :P)
 

DeletedUser15432

The guaranteed method is to invest in another Alcatraz - the actual average is 22 blue prints - with a statistical variance of +/- 25% per individual great building
 

DeletedUser109250

I prefer the median as opposed to the average in choise of measure of Central tendency in this dataset.
The outliers that are the long bad runs skew the average. So when more than half are done within 21 bp's I don't think saying the average is 22 is as good as saying that the median is between 20 and 21.

Eksample:
Say the salaries from a factory are the following:
1k, 1.3k, 1.4k, 1.4k, 2k, 2.2k and 10k.
There is one manager making a lot more than the others, now the 2 different Central tendency measurements will give you:
average: 2.76k
median: 1.4k

now which measure does best describe the usual salary?
 
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