Recurring numbers rolled - Game is rigged

[Gulchen]

​
https://boardgamearena.com/table?table=451655181 had 56 rolls. ​ ​ ​ For 56, the probability of there being at least one

number occurred few times ​ event at least as surprising as ​ "10 dropped only 1 time" ​ ​ ​ is greater than ​ 24% :

Code: Select all

Python 3.8.0 (default, 
[GCC 8.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> from math import comb,factorial
>>> from fractions import Fraction as frac
>>> threshold = 56*frac(3,36)*(frac(33,36)**55)
>>> comb(56,4)*(frac(6,36)**4)*(frac(30,36)**52) < threshold < comb(56,5)*(frac(6,36)**5)*(frac(30,36)**51)
True
>>> [(n,k) for n in (1,2,3,4,5,6) for k in (0,1,2,3,4) if comb(56,k)*(frac(n,36)**k)*(frac(36-n,36)**(56-k)) == threshold]
[(3, 1)]
>>> [(n,[k for k in (0,1,2,3,4) if comb(56,k)*(frac(n,36)**k)*(frac(36-n,36)**(56-k)) <= threshold]) for n in (1,2,3,4,5,6)]
[(1, []), (2, []), (3, [0, 1]), (4, [0, 1, 2]), (5, [0, 1, 2, 3]), (6, [0, 1, 2, 3, 4])]
>>> lowfrequenciesfor = ((),(),(),(),(0,1),(0,1,2),(0,1,2,3),(0,1,2,3,4),(0,1,2,3),(0,1,2),(0,1),(),())
>>> unionbound = sum(comb(56,k)*(frac(6-abs(roll-7),36)**k)*(frac(36-(6-abs(roll-7)),36)**(56-k)) for roll in (4,5,6,7,8,9,10) for k in lowfrequenciesfor[roll])
>>> pairwisecorrection = sum(frac(factorial(56),factorial(j)*factorial(k)*factorial(56-(j+k)))*(frac(6-abs(m-7),36)**j)*(frac(6-abs(n-7),36)**k)*(frac(36-((6-abs(m-7))+(6-abs(n-7))),36)**(56-(j+k))) for m in (4,5,6,7,8,9,10) for n in (4,5,6,7,8,9,10) for j in lowfrequenciesfor[m] for k in lowfrequenciesfor[n] if m != n)
>>> frac(17,(17*4)+1) < frac(246550,10**6) < unionbound-pairwisecorrection
True
>>> 

(By one of the Bonferroni inequalities, ​ https://en.wikipedia.org/wiki/Boole's_i ... equalities

unionbound-pairwisecorrection is a lower bound on the probability that _at least one_ of the rolls 4,5,6,7,8,9,10 has a surprisingly low frequency.

I don't know a proof of that in general, but I can explain the pairwise special case:

Code: Select all

Let S_0,S_1,...,S_(n-1) be the sets.  For a first attempt at estimating    Probability ( S_0 union S_1 union ... S_(n-1) )  ,
one can use  Probability(S_0) + Probability(S_1) + ... + Probability(S_(n-1))  .
The only issue with that sum is, it counts each event a number of times equal to the
number of the sets S the event is in, rather than counting each event at most once.

Accordingly, one can get a lower bound by subtracting the result of counting
each event that is in _more than_ one of the sets S
a number of times equal to at least
the number of the sets S the event is in  minus 1    .


Adding the probabilities of the pairwise intersections does count those events at least that often:

Events in exactly s of the sets S get counted  s choose 2  times.
When s is at least 2,
s choose 2  =  s*(s-1)/2)  =  (s-1)*s/2  >=  (s-1)*2/2 = s-1    .

)

[alfredddd]

How do we get this fixed? Just played a game where 9s and 5s were easily 60% of the numbers that were rolled. Admins are you listening?

[Stroom]

How do we get this fixed? Just played a game where 9s and 5s were easily 60% of the numbers that were rolled. Admins are you listening?

This is not broken. Catan IS a random game. People who complain, do not understand how randomness works. It is very often unfair. If you don't like randomness, there are many other games for you to play.

[Gulchen]

How do we get this fixed? Just played a game where 9s and 5s were easily 60% of the numbers that were rolled. Admins are you listening?

Which table was that?


(Also, my guess is no: ​ I imagine they only occasionally check in on these threads, to see if an actual analysis has concluded that
there's likely a problem or the reports are still just people saying they got something strange-seeming in one or a few of their games.)

[pdy333]

I have been studying for some weeks now, the proliferation of success of the highest ELO player in each match I play in Catan. So startling have the results been, that I deliberately try to select numbers on the board at the start of the game that are the same as the high ELO player. My results since starting this tactic have improved, but the fact remains, that having focused on the high ELO player, the bias toward that players numbers dropping is astounding, and in short, demotivating.
An example https://boardgamearena.com/table?table=453295782
Even the winning player apologized at the end of that match.
Again, I’ve been recording all number drops of only myself and the highest ELO player for weeks. It’s clearly a pattern, and therefore, I believe not random inasmuch as only one player benefits from the ‘random’ rolls.
Try it yourself…see for yourself.
Please don’t feed me the ‘randomness’ lines, I know what randomness looked like in Catan.

[Panthergirl]

There's a well hidden button below the screen that determines the outcome of all dice rolls. If you want random rolls, you have to disable this feature.

This was mentioned in another thread but I checked the game preferences and cannot find anything. (not sure if you were joking here or not)

[Jellby]

Yes, he was. The best option to disable bad rolls is actually Alt+F4.

[Panthergirl]

Yes, he was. The best option to disable bad rolls is actually Alt+F4.

Ah, you never know around here

[Gulchen]

@ pdy333 ​ :
​
​
Is that ​ ​ ​ the player just has the number or doesn't ​ ​ ​ or ​ ​ ​ corresponding to how much the player gets from the number ​ ?

For whichever of those, is it ​ ​ ​ ​ ​ ​ ​ considering just that player ​ ​ ​ or
also considering ​ whether / how much ​ the other players get from the number ​ ​ ​ ?

[wxmidwest]

It would be interesting to see if the outcomes tend toward the normal dice distribution over time. If one took all the past history of numbers rolled in say all the games played on this platform until now, what would show up. If the process is truly (Pseudo-random), the distribution table will show it, it should head toward the nearly exact Gaussian distribution as the number of dice outcomes heads toward infinity (forever increasing outcomes).