Like Denethor, I'm slowly losing my mind, waiting for the LotR game to finally arrive in stores. However, I don't like to play with fire, as he does. I like to play with MATH! So, if you'll allow me to vent...
I'm always thinking about buying a second core set. I know I don't even have one core set yet, but I have to think about something until the game comes out. But I would like three of each Player Card, so I'll have a full card set to choose from whenever it is time to challenge the shadow.
We know three facts about player cards from the core rulebook:
120 Player Cards (p3), 15 different cards per sphere (p10), 29 total cards per sphere (add Gandalf to get 30, p9 & 10)
So, in each core set, a given player card might appear 1, 2, o-r 3 times. But each sphere has to have 15 different cards total. Representing single copy, double copy, a-nd triple copy cards as a, b, c respectively, we see that a+b+c=15.
Meanwhile, the core set contains a deck of 29 cards for each sphere. using the a, b, c for single, double, a-nd triple copies of cards, that means that a+2b+3c=29!
Taking our two formulas, (a+b+c=15; a+2b+3c=29) I crunched the numbers through my brain and found eight possible card combinations for each sphere deck in the core set, which would satisfy both: 15 different cards per sphere; 29 total cards per sphere deck. And here they are:
1 single copy card; 14 doubles; 0 triples.
2 single copy cards; 12 doubles; 1 triple
3 single copy cards; 10 doubles; 2 triples
4 single copy cards; 8 doubles; 3 triples
5 single copy cards; 6 doubles; 4 triples
6 single copy cards; 4 doubles; 5 triples
7 single copy cards; 2 doubles; 6 triples
8 single copy cards; 0 doubles; 7 triples
It's possible that all the sphere decks will have the same number of single/double/triple cards in each. But maybe not. Regardless of which breakdown we end up with, we will be 64 cards short of a full play set (15 individual cards per sphere times three copies maximum: 15*3=45 needed, 29 per core set, 45-29=16, times four spheres, 16*4=64).
How many of those 64 cards will I be able to get, if I buy that second core set I have been pondering? It depends.
For each sphere that contains 1 single copy card, I can get 15 cards towards my complete play set from a second core set.
For each sphere with 2 single copy cards, I get 14
Each Sphere with 3 single copy cards, 13
4 single copy, 12
5 single copy, 11
6 single copies, 10
7 single copies, 9
8 single copies, 8
So I will get between 60-32 player cards towards a complete play set from a second core copy, seems like a good value. And that will leave me between 4-32 cards short. At that point, it might be time to hit up Ebay.
What do you think? Are you going to try for a complete set of three of each player cards? Did I miss or misinterpret any facts or figures in my post? Should I go back to playing solo BSG o-r catching pokemon to pass the time for the next two weeks? Is there any more math I should be doing? Let me know!