New Variant : Dice Deck

[The-True-KingofTokyo]

I can't get this to launch despite enabling it in the table options.. any ideas?

[twiddly2]

Does this format reduce the number of complainers who play this game? Not being facetious, but joy from playing is being genuinely sucked dry by the players who constantly do the following in the game chat, none of which are actual reportable offenses.

1) accuse you of robbing ONLY them
2) tell you how to play your turn
3) accuse you of conspiring with the other player against them

Hello players,

A new option have been added to Catan.
This optionhas been directly designed by the Catan Studios team for the BGA adaptation.

This is the new "Dice Deck" option, which replaces the dice with a stack of cards. Visually, nothing changes — you will still see the dice rolling on screen — but information about the card stack (such as the remaining cards and when the deck is shuffled) is now available in the logs.

This is how the dice deck works:

- A stack with 36 cards is build. On these cards the results of 2 dices are written based on the expected statistical probabilities: 2=1; 3=2x; 4=3x; 5=4x; 6=5x; 7=6x; 8=5x; 9=4x; 10=3x; 11=2x; 12=1
- 18 cards are removed randomly.
- Now a second stack is build in the same way, so that you have two stacks with 18 cards in the end.
- Both stacks are shuffled together to form a new stack of 36 cards.

Each turn a card is drawn from the virtual stack forming the dice number for that turn. If the stack runs out it will be created again in the same way.

Note: This procedure is different than the one explained in the Traders & Barbarians expansion, because in online play it may be possible that a player can has a higher chance to predict the next number, if the dice results are written down secretly. This would not be so easy in a game at a table. Therefore the procedure was changed especially for online games.

[Samba12345678]

Seafarers is coming, it's being alpha tested at the moment.

Great news!!! ETA?

[Ingoschi]

There is still a probability of 0.02% that the deck contains not a single 6, so expect complains every 5000 games

How many 6s does the deck contain? (Same for 8)
# Probability
0 0.02%
1 0.36%
2 2.43%
3 7.40%
4 13.56%
5 16.68%
6 13.56%
7 7.40%
8 2.43%
9 0.36%
10 0.02%

[Jellby]

As you say, same for 8, so expect complaints at least every 2500 games

[The-True-KingofTokyo]

Still can't get this to launch. Has anyone else been successful yet?

[euklid314]

There is still a probability of 0.02% that the deck contains not a single 6, so expect complains every 5000 games

How many 6s does the deck contain? (Same for 8)
# Probability
0 0.02%
1 0.36%
2 2.43%
3 7.40%
4 13.56%
5 16.68%
6 13.56%
7 7.40%
8 2.43%
9 0.36%
10 0.02%

How did you calculate the percentages? I arrive at different numbers:
# Probability
0 0.05%
1 0.66%
2 3.64%
3 11.18%
4 21.31%
5 26.31%
6 21.31%
7 11.18%
8 3.64%
9 0.66%
10 0.05%

Short explanation for P(X=0)=0.05%. Of the 36 cards, 5 are "6s" and we randomly draw 18 of the 36 cards. The probability distribution is hypergeometric and the probability to get 0 "6s" out of the 18 chosen cards is 2.27% = nCr(5,5)*nCr(31,13)/nCr(36,18).
Then this is repeated and we again need to draw 0 "6s". Thus 0 "6s" out of 36 in the Dice Deck has a probability of 2.27% * 2.27% = 0.05%.

Just for comparison. With random dice (=standard Catan) we have the following number of 6s within 36 rolls:
# Probability
0 0.46%
1 2.67%
2 7.53%
3 13.76%
4 18.31%
5 18.90%
6 15.75%
7 10.89%
8 6.37%
9 3.19%
10 1.39%
11 0.53%
12 0.18%
13 0.05%
14 0.01%

The most striking difference are the cases that "6" will appear more often than expected. With random rolls the "bell curve" is not symmetrical of course (we can theoretically have 20 "6s" out of 36 independent rolls) and the cases of, say, 9+ "6s" within 36 rolls currently has a probability of 5.37%. With the new variant Dice Deck this goes down to 0.71%.

[Ingoschi]

How did you calculate the percentages?

Thank you for checking the calculations, I've to admit that I was mislead by Meta AI and posted wrong results.

Meta AI also suggested the hypergeometric distribution but claimed that the result of
nCr(5,0)*nCr(31,18)/nCr(36,18) or
nCr(5,5)*nCr(31,13)/nCr(36,18) would be 1.48%.
Wolfram Alpha calculates 1/44 = 2.27%
When I confronted Meta AI that its result was wrong, it correct it.

0.02 % for the combination of two stacks was then 0.0148 * 0.0148 = 0.0002.

So the correct values for the first stack of 18 cards (now calculated with Wolfram Alpha, the sum of Meta's "corrected" results is still not even near 100%) are
k=0 nCr(5,0)*nCr(31,18)/nCr(36,18) 2,27%
k=1 nCr(5,1)*nCr(31,17)/nCr(36,18) 14,61%
k=2 nCr(5,2)*nCr(31,16)/nCr(36,18) 33,11%
k=3 nCr(5,3)*nCr(31,15)/nCr(36,18) 33,11%
k=4 nCr(5,4)*nCr(31,14)/nCr(36,18) 14,61%
k=5 nCr(5,5)*nCr(31,13)/nCr(36,18) 2,27%

The results for the combination of two stacks are then the same as yours
0 (k=0) * (k=0) = 0.05%
1 (k=0) * (k=1) + (k=1) * (k=0) = 0.66%
2 (k=0) * (k=2) + (k=1) * (k=1) + (k=2) * (k=0) = 3.64%
3 (k=0) * (k=3) + (k=1) * (k=2) + (k=2) * (k=1) + (k=3) * (k=0) = 11.18%
4 (k=0) * (k=4) + (k=1) * (k=3) + (k=2) * (k=2) + (k=3) * (k=1) + (k=4) * (k=0) = 21.31%
5 (k=0) * (k=5) + (k=1) * (k=4) + (k=2) * (k=3) + (k=3) * (k=2) + (k=4) * (k=1) + (k=5) * (k=0) = 26.31%

Sorry I should have known better not to trust a LLM with calculating.

[Shivaware]

Still can't get this to launch. Has anyone else been successful yet?

It looks like you got it going a few times, but quit the games after a few turns.

Here's a sample game. Did the rolling dice animation make you think it wasn't using the dice deck?

[euklid314]

5 (k=0) * (k=5) + (k=1) * (k=4) + (k=2) * (k=3) + (k=3) * (k=2) + (k=4) * (k=1) + (k=5) * (k=0) = 33,11%

The last line must be a typo, it should be 26.31%.

You are a new generation, compared to me. I would not even consider using an AI for such a question - but in fact I am not using AI for anything at all.