So many repeats

[Proclivitas]

I'm seeing way too many repeats. It's definitely possible with the randomization, but it's extremely unlikely. The physical game has 110 cards, with 5 options each.

That makes 550 words possible.

I should not be seeing the same word every 2-3 games, sometimes back to back. That's just extremely unlikely.

Is the card randomization working, so the full 550 words have equal chance of being chosen? Or do you just have the top word of each card or something?

With complete randomness it should be way more rare to see repeats than what I've seen.

Or is this intended to be some kind of "demo" version with what feels like 40 words and doesn't have the full word list from the regular game?

[Jellby]

There are also many repeated threads:

https://boardgamearena.com/forum/viewtopic.php?t=26699
https://boardgamearena.com/forum/viewtopic.php?t=27626
https://boardgamearena.com/forum/viewtopic.php?t=27690

[aghagh]

There are also many repeated threads:

I guess it makes sense that repeated words generate repeated threads. Oh, irony!

[Proclivitas]

Given the games I've played, I find it hard to believe all 550 cards are actually in the game, or that the randomization algorithm is working correctly. Maybe just really bad luck. Today was better, maybe they fixed something or it's just luck.

[Jellby]

Note that our intuition of what is likely or not may be very seriously off (https://en.wikipedia.org/wiki/Birthday_problem).

There are 110 cards, and in each game you choose 13. There are 26515202245096950 possible "games" (where order doesn't matter). (https://www.hackmath.net/en/calculator/ ... 0&repeat=0)
In a second game, if you don't want to repeat cards, you can only choose between 97, so 4660989191504120 2nd games without repetition. This leaves the rest, i.e. 82.42% where there is at least one repeated card between the two games.

Now, assume there's only 1 repeated card. In one game you choose one of the 5 words. The probability of choosing the same word in the other game is 1/5 = 20%.
If there are more than 1 repeated cards, the probability of repeated words must be higher: 1-(4/5)^n for n cards.

So, if in 82.42% of the games there's at least 1 repeated card, I state that the probability of repeated words in two independent games is at least 82.42%/5 = 16.48%. It doesn't look too low. In a real-life game, when you find a repeated card, you'd probably choose a different word, so the word selection is not truly random.

[Jellby]

More complete:

Total card selections: Ct = 110!/(13!*97!)
Card selections with n repetitions: C(n) = 13!/(n!*(13-n)!) * 97!/((13-n)!*(97-(13-n))!) (we choose n out of the 13 previous selected cards, and 13-n out of the remaining 97)
Probability of n card repetitions: C(n)/Ct
Probability of at least one repeated word among n repeated cards: R(n) = 1-(4/5)^n
Probability of at least one repeated card: Pc = sum[C(n)/Ct,1,13] = (C(1)+C(2)+C(3)+C(4)+C(5)+C(6)+C(7)+C(8)+C(9)+C(10)+C(11)+C(12)+C(13))/Ct
Probability of at least one repeated word: Pw = sum[C(n)/Ct*R(n),1,13]

This gives Pc = 82.42% and Pw = 26.96%.

So the probability of at least 1 repeated word in two independent games is actually much higher than 16%, and higher than one in four. Maybe an appropriate suggestion would be to allow players to choose a word from each card, instead of having them randomly selected.

[aghagh]

Jellby, I haven't done the math myself to check if what you put is correct or not, but I think you may have found the key idea here, good thinking.

What most players assume: "13 words out of 550 are chosen randomly at the beginning of the game".
(although some players also think that perhaps "1 word out of 550 is chosen randomly every round" and that's why they don't realize seeing the same word as in previous round is a bug)

What might be actually happening, as you say: "13 cards out of 110 are chosen randomly at the beginning of the game and afterwards 1 word from each card is selected".

This second approach would also mean that some words would be incompatible with each other in the same game. After seeing the mystery word in a round, you could cross out another 4 words from its same card that will not appear during the rest of the game (if you knew the exact word distribution amongst the 110 cards).

We should not forget that the physical game comes with 110 cards with 5 words each, so the second approach is actually the accurate implementation.

[fruktansvärt]

IS the game here on BGA coded with "110 cards with 5 words each"?
If so, since the word is chosen randomly from the 5 supplied, what would be the difference from "550 cards with 1 word each"?

[Jellby]

As aghag says, the difference is that you can't have two words from the same card, because you have to choose 13 different cards first. If you were choosing a random card each time, without discards, then yes, it would be the same... but you could also end up with the same words 13 times in a single game (very unlikely, but possible).

Note that if you forget about cards, and just choose 13 random (different) words out of the pool of 550 words, the probability of repetition in two independent games would be:

Total number of games: N1 = 550!/(13! * (550-13)!)
Games without repetition: N2 = 537!/(13! * (537-13)!)
Probability of at least 1 repetition: 1-N2/N1 = 26.98%

which is not much different.

[aghagh]

I see. Then it's not that important whether the selection of words comes from a one step (13/550) or a two step (13/110->13x1/5) process.