I'd love some Math help. This Math question is probably too complicated to be solved. I've created a board game that has one important part that uses cards. I'm having trouble coming up with the number of possible unique games.
Each card effects an important track on the board, so it really does have a big role to make a unique game. I will try to explain as succinctly as I can and post some numbers below, those numbers will probably be wrong though but it could help explain it better.
THE DETAILS
Total cards: 25
Setup Take 12 cards out of 25 = 5,200,300 combinations
Round 1: Split the deck of 12 in half. Pile A has 6 and Pile B has 6 cards each. 924 combinations for pile A and 924 combinations for
Pile B.Show all 6 cards from pile A (Only round One shows all 6 cards, the other only show 3 cards and follow a pattern (show 3, show 3, remove 2 and add 2 new cards).
Round 2: Then remove 2 from a pile A (15 combinations) and add 2 From Pile B (15 combinations) to Pile A 15×15 = 225
Pile B has 4 card remaining.
Shuffle and and show 3 from Pile A = 20 possible combinations.
Round 3: Show the remaining 3 from pile A = 20 possible combinations (by the numbers in previous round).
Round 4: Remove 2 from a pile A (15 combinations) and add 2 From Pile B (15 combinations) to Pile A. 15×15 = 225
Pile B has 2 card remaining.
Shuffle and show 3 from Pile A = 20 possible combinations.
Round 5: Show the remaining 3 from pile A = 20 possible combinations (by the numbers in previous round).
Round 6: Remove 2 from a pile A (15 combinations) and add 2 From Pile B (15 combinations) to Pile A. 15×15 = 225
Pile B has 0 card remaining.
Shuffle and show 3 from Pile A = 20 possible combinations.
Round 7 (END) Show the remaining 3 from pile A = 20 possible combinations (by the numbers in previous round).
Do I add up all those combinations numbers listed above or multiply them or are they all wrong?
P.S. I don't know if this changes things but the order of the cards shown during an individual round don't matter because it all happens instantly before the round starts. But for example if the cards of Round 2 and Round 3 were switched that would matter. For example Round 2 had the cards 1,2,3, and round 3 had the cards 3,4,5. Having the cards 1,2,3 in Round two come out 3,1,2 doesn't matter. But if 2,4,5 came out in Round two and 1,2,3 came out in Round three that would matter.
Best Answer
You got most numbers right, just a few corrections are needed.
Let's follow the progression af a game and see how many distinct ways there are to perform each step:
Setup:
Choose 12 cards from 25: $\binom{25}{12}=5200300$
Round 1:
Choose 6 cards for deck A: $\binom{12}{6}=924$
Other 6 cards go to deck B: $1$ possibility only.
Round 2:
Pick two cards to discard from deck A: $\binom{6}{2}=15$
Pick two cards to move from deck B to deck A: $\binom{6}{2}=15$
Pick three cards from deck A to show: $\binom{6}{3}=20$
Round 3:
Show all cards from deck A: $1$ possibility only.
Round 4:
Pick two cards to discard from deck A: $\binom{6}{2}=15$
Pick two cards out of four to move from deck B to deck A: $\binom{4}{2}=6$
Pick three cards from deck A to show: $\binom{6}{3}=20$
Round 5:
Show all cards from deck A: $1$ possibility only.
Round 6:
Pick two cards to discard from deck A: $\binom{6}{2}=15$
Pick two cards out of two (!) to move from deck B to deck A: $\binom{2}{2}=1$
Pick three cards from deck A to show: $\binom{6}{3}=20$
Round 7:
Show all cards from deck A: $1$ possibility only.
Total number of different games
$T=5200300*924*15*15*20*1*15*6*20*1*15*1*20*1$
$T=11676337596000000000$
$T\approx1.168*10^{19}$
There is little risk of playing twice exactly the same game in a lifetime.