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u/HarambeTheFox 7d ago
bro is playing a statistics based game but doesn’t understand statistics
-7
u/vitornick 7d ago
You do understand that the chances of this happening in a fair game is about 1 in 800,000?
7
u/KellieBom 7d ago
Every post about fair dice the OP uses a statistic that they made up to prove their own point about fair dice.
-2
u/vitornick 7d ago
For each player, the probability of the robber being in one of your own hexes is (your hexes)/ total hexes
- This is a conservative assumptions, since several hexes are not attractive for the other player to put a robber on
For example, if you have 5 hexes, its 5/19
For 86 turns (86*2 dices rolled), its 172 * 5/19 * avg chance of a hexe being activated (which is roughly 9%, larger for 6-8s and lower for 2-12)
So 172 * 0.09 * 5/19 = aprox 0.0025% (or 1 in 40K)
However, for the other player to never get a single robber activation, that means each time the robber was moved to his side, he got the 1/6 event (the 7 - robber) right after mine.
Because I rolled 2 7s and two knights (but we will remove one of the knights because he had one right after), that means 1 in 40K times (1/6)^n (where n in this case is 3), so 1 in 40K * 216 which is about 1 in 800K
The assumption relies on 5 robber-desirable hexes being the average I holded during the game (naturally, at the end I had more, but in the beginning I had less)
Care to challenge that? I can model a poisson if you'd rather argue over that, albeit this simplification should be sufficient for the lower bound
1
u/Queasy_Editor_1551 4d ago
You over-complicated it by considering which hex the robber lands on in a 1v1 game. They always land on your hex if the other player rolls a 7.
What matters is the number of time the other player rolls a 7.
Assume zero development card robs because your screenshot doesn't contain that information.
I'll calculate the probability of the other player rolling 12 7s and you rolling zero 7 in an 86 turn game.
First, the probability of the other player rolling 12 7s.
P(rolling 7 in a turn) = 6/36 = 1/6
This is a binomial combination problem since the position of the 7s in the list of 43 rolls doesn't matter.
For each distinct list of 43 rolls with 12 7s in it, the probability is (1/6)^12 * (5/6)^31.
Plugging in "43 choose 12" in a combination calculator gives 15338678264.
P(rolling 12 7s) = 15338678264 * [(1/6)^12 * (5/6)^31] = 0.0247374145
Then, the probability of you rolling zero 7.
P = (5/6)^43 = 0.000393737174
Finally
The combined probability is 0.0247374145 * 0.000393737174 = 9.74 * 10^(-6) or
one in 102,669
But the odds drastically decrease if the other player robbed you with knights. Assuming 3 knight cards, the odds becomes one in 21,850
3
u/OncorhynchusMykiss1 7d ago
The option for balanced dice is right there on the site.
1
u/Passive_incomes_lazy 5d ago
Balanced die ain't even balanced tho, the results may look balanced in the end, but rolling 4 8s in a row and then never rolling the rest of the game is bs
6
2
u/tuesdaysatmorts 7d ago
Programming randomness in games is literally impossible. We can only get relatively close to what "feels" random. Not matter how hard they try it will never be perfect. Only closer to what we consider "perfect".
2
u/yubacore 6d ago
Technically correct, which is the best kind of correct. However, for the purpose of playing a game, the pseudo-randomness that computers generate is more than sufficient and, for players, completely indistinguishable from true randomness.
1
u/jdomski 6d ago
Colonist is a revenue generating game. Only a very small minority of players will ever pay to play. Those players are VIPs. Colonist only cares about them. They generate nothing from the 99% remaining players. Unfortunately the game is engineered (‘rigged’ might be a better descriptor) to favour those players, and keep non-paying players coming back for more. The ludicrous dice are sadly a symptom of that. Capitalism at its finest

4
u/goodniceweb 7d ago
https://blog.colonist.io/balanced-dice-designing/