So many repeats

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Jellby
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Re: So many repeats

Post by Jellby »

Tip for the designer: if you want the probability to drop below 10%, you need 1617 words, and 16828 words to drop below 1% (always selecting 13 words).
pjt33
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Re: So many repeats

Post by pjt33 »

Jellby wrote: 06 November 2022, 10:12 This gives Pc = 82.42%
But that's misleading because in most games the players don't see all 13 words. According to the statistics for "all players", the average is 11 rounds played per game (which is higher than I expected, to be honest).
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Jellby
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Re: So many repeats

Post by Jellby »

Well, yes, if you don't play/see all 13 words, then the probability of repetition is lower. I just played the game a couple of times and don't remember the exact rules. But the point was that the probability is much higher than what one would "intuitively" believe. With 12 words it's around 23%, with 11 words 20%, with 10 words 17%, with 9 words 14%...
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KuWizard
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Re: So many repeats

Post by KuWizard »

fruktansvärt wrote: 06 November 2022, 12:49 IS the game here on BGA coded with "110 cards with 5 words each"?
It is in fact coded as 110 cards with 5 words each. And yes, game randomly selects a card on the game start. After that it randomly chooses a number from 0 to 4 as $wordID. Then it picks this shuffled cards list and uses this one $wordId 110 or 13 times with random $cardId. If 110 words is not enough (Infinite words option), it does $wordID+1 (or $wordID = 0 if previous $wordID was 4) and takes 110 words 4 more times. Then it puts everything in a database. So what you will see in a database after game setup is 13 (or 550) combinations of $cardId-$wordId with the first 110 $wordId being the same. But cards are shuffled which means it's still random distribution after all.
Hope this helps
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KuWizard
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Re: So many repeats

Post by KuWizard »

Jellby wrote: 06 November 2022, 10:12 ...

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.
Wow, thanks a lot for the maths, that makes this mostly discussed problem of words randomness to another level!
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Jellby
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Re: So many repeats

Post by Jellby »

KuWizard wrote: 15 November 2022, 14:48It is in fact coded as 110 cards with 5 words each. And yes, game randomly selects a card on the game start. After that it randomly chooses a number from 0 to 4 as $wordID. Then it picks this shuffled cards list and uses this one $wordId 110 or 13 times with random $cardId.
Wait... what? Are you saying that it chooses 13 random cards, and then, for instance, the 2nd word of every one of these cards (not necessarily in this order, but effectively)? If that's the case that's totally wrong, that would mean that once you know it's the 2nd word of a card, you can rule out all words which are not 2nd in their card. I think it would also increase the probability of multiple repetitions.
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KuWizard
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Re: So many repeats

Post by KuWizard »

Jellby wrote: 15 November 2022, 15:42
KuWizard wrote: 15 November 2022, 14:48It is in fact coded as 110 cards with 5 words each. And yes, game randomly selects a card on the game start. After that it randomly chooses a number from 0 to 4 as $wordID. Then it picks this shuffled cards list and uses this one $wordId 110 or 13 times with random $cardId.
Wait... what? Are you saying that it chooses 13 random cards, and then, for instance, the 2nd word of every one of these cards (not necessarily in this order, but effectively)? If that's the case that's totally wrong, that would mean that once you know it's the 2nd word of a card, you can rule out all words which are not 2nd in their card. I think it would also increase the probability of multiple repetitions.
I can't see how exactly it would increase the probability. However during game was in alpha and in beta the algorithm was different (before Infinite games were introduced). Even then the amount of complaints about repetitions was not lower than it is now. So I can fix that, no problem, but this won't solve the problem.
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Jellby
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Re: So many repeats

Post by Jellby »

Well, let's see... I won't take the trouble of computing the probabilities with random card and word selections, I will instead compare the probability of your method (random card selection + a common word index selection) with those of total random word selection (i.e., from the pool of 550 words).

exactly 1 card repetition: C(13,1)*C(97,13-1)/C(110,13) = 34.95%
exactly 1 word repetition (common word index): 34.95% / 5 = 6.99%
exactly 1 word repetition (random word): C(13,1)*C(537,13-1)/C(550,13) = 23.51%

exactly 2 card repetitions: C(13,2)*C(97,13-2)/C(110,13) = 29.26%
exactly 2 word repetitions (common word index): 29.26% / 5 = 5.85%
exactly 2 word repetitions (random word): C(13,2)*C(537,13-2)/C(550,13) = 3.22%

exactly 3 card repetitions: C(13,3)*C(97,13-3)/C(110,13) = 13.57%
exactly 3 word repetitions (common word index): 13.57% / 5 = 2.71%
exactly 3 word repetitions (random word): C(13,3)*C(537,13-3)/C(550,13) = 0.25%

Do you see a pattern? Higher repetitions are much more unlikely when you choose everything randomly. With a common word index, once N cards are repeated, if one word is repeated, there's no escape, N words must be repeated.

If we sum up all probability from 1 to 13 repetitions we get:

at least 1 card repetition: 82.42%
at least 1 word repetition (common word index): 82.42% / 5 = 16.48%
at least 1 word repetition (random word): 26.98%

Or, summing only from 2 or 3 to 13:

at least 2 card repetitions: 47.47%
at least 2 word repetitions (common word index): 47.47% / 5 = 9.49%
at least 2 word repetitions (random word): 3.48%

at least 3 card repetitions: 18.21%
at least 3 word repetitions (common word index): 18.21% / 5 = 3.64%
at least 3 word repetitions (random word): 0.26%

So yes, a common word index reduces the probability of 1 repetition, but greatly increases the probability of multiple repetitions, which seems to be what people are complaining about.

That is... if I didn't make any mistake.
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aghagh
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Re: So many repeats

Post by aghagh »

Interesting. After your last posts I went to my played games to do the following check: if the game uses 5-word cards and then selects the same word position (1-5) for every of the 13 cards for a particular game, it would mean each word would always appear in a game along with a subset of the other 109 words in the same position on the other cards. However, what I found with my test word was that there were around 120 words that had appeared along with it throughout my different games.

Am I missing something ? Or perhaps the random method has been modified at some point and i mixed data from different periods ? (I've used data from all these months we've been playing)
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Mathew5000
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Re: So many repeats

Post by Mathew5000 »

Jellby wrote: 06 November 2022, 09:25In 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.
This point is key. If the implementation were altered so that the "judge" each turn could select the word (from among five on a card), then complaints about repeated words would be reduced, presuming that groups usually play a few games per session with only small changes in the composition of the group of players. Because the judge, presumably, would avoid any word recently seen in a game with the same players.
Proclivitas wrote: 05 November 2022, 09:49 I should not be seeing the same word every 2-3 games, sometimes back to back. That's just extremely unlikely.
I ran two simulations of a million trials each, each trial consisting of three games. Algorithm 1: 13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly. Result for algorithm 1: in my simulation of a million three-game sets, 613,151 three-game sets (61.3%) contained at least one repeat. Algorithm 2: 13 distinct words out of 550 are chosen randomly. Result for algorithm 2: in my simulation of a million three-game sets, 614,219 three-game sets (61.4%) contained at least one repeat.

Either way it's clear: if you play three games, you probably will see at least one repeated word.

Detailed results:
Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly):
38.7% no repeats
38.2% exactly one word repeated once (with no word repeated twice)
17.0% exactly two words each repeated once (with no word repeated twice)
4.5% exactly three words each repeated once (with no word repeated twice)
0.8% exactly four words each repeated once (with no word repeated twice)
0.1% five or more words each repeated once (with no word repeated twice)
0.7% at least one word repeated twice (i.e. there's a word that's in all three games of the set)

Algorithm 2 (13 distinct words out of 550 are chosen randomly):
38.6% no repeats
38.3% exactly one word repeated once (with no word repeated twice)
17.0% exactly two words repeated once (with no word repeated twice)
4.5% exactly three words each repeated once (with no word repeated twice)
0.8% exactly four words each repeated once (with no word repeated twice)
0.1% five or more words each repeated once (with no word repeated twice)
0.7% at least one word repeated twice (i.e. there's a word that's in all three games of the set)

So it seems that for a set of three games, there's no major difference between those two algorithms in the probability of repeats.
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