Page 1 of 1

5 or 6 player possibility?

Posted: 14 September 2020, 23:11
by FT3
Would it be possible to include more players as an option for this but only need two columns to win?

I have no idea if this is a good idea or not but I feel like it could work - if I had the real-world game I’d definitely try it.

(I suppose I could make my own version and try it when my game club restarts..)

Re: 5 or 6 player possibility?

Posted: 15 September 2020, 00:31
by Jest Phulin
BGA in general does not make "house rules" like this. Many of the license agreements state they must respect the published rules to the best of a digital capacity. If an authorized source (game developer, publisher, etc.) agrees to a change, then it may be implemented.

Re: 5 or 6 player possibility?

Posted: 15 September 2020, 01:53
by FT3
Oh that’s a shame, but also perfectly fair and understandable.

I just looked at the Wikipedia page for the game which mentions a Sid Sackson devised variant where you can leapfrog other players but I don’t know what I think of that.

It also says the game can last up to 45 minutes...

Re: 5 or 6 player possibility?

Posted: 15 September 2020, 21:52
by CaractacusPots
5-6 players would be unbearable imo. The wait time between your turns would be too long.

Add to that the fact that 2 columns can be made super quick by the first people who get the 678 combo. Game would be over in 2-3 rounds.

Re: 5 or 6 player possibility?

Posted: 30 September 2020, 09:13
by voriki
Someone would need to devize a variant. These are just fictional numbers, for balance it needs lots of tweaking and playtesting.

4 die turn into 6 die, these are split into 2 groups of 3.
3 temporary black stones turn into 4 or 5 stones. Preferably 4. Or it remains at 3 to keep it challenging.
The board goes from 3 to 18(three 1's and three 6's), and obviously the rows become longer. But you will get more combinations to advance faster.

I do think we would have too many combinations with 6 die interchangeable with each other. So maybe every dice has a counterdice they cannot be matched with.
Dice 1A and Dice 1B can never be in the same group of 3
Dice 2A and Dice 2B can never be in the same group of 3
Dice 3A and Dice 3B can never be in the same group of 3
That leaves you with 4 possible outcomes to select from.

Someone into statistics willing to check the odds?