Jeff Warrender
United States Averill Park New York

I have a game that is nearly complete and I'm trying to balance out the different scoring options.
The game is about writing books by accumulating cards. Each card represents an event, and it becomes a page in your book. Thus, your finished book is a row of cards on the table in front of you. Players typically acquire 79 cards over the game.
Each card has a number (136), one to three "motifs" (there are five different motifs), and one to three "witnesses" (there are nine different witnesses). Each motif is present on 9 different cards. Each witness is present on either 8 or 11 cards (5 of the former, 4 of the latter). (It's ok for simplification of the math to just average that number to 9 cards per witness and call it close enough).
You can get points for four different configurations:
1. A cluster of cards in your book is arranged in increasing numerical order. 2. A cluster of at least 3 cards all share a witness. 3. At least four cards all share a single motif. 4. An arrangement of four cards in which the first and fourth share one motif and the second and third share a different motif (an ABBA structure)
You get VP for each condition you meet, and if you meet a given condition more than once you get the appropriate multiple of the points for that condition.
I'm trying to figure out how many points to assign to these different configurations other than just trial and error, and it seems like one way is to figure out how many possible ways there are to arrange the cards so as to produce each arrangement, with the "harder" arrangements being more valuable. (Of course, in reality the cards aren't arranged randomly so this is a slightly coarsegrained approach for assigning the relative "degree of difficulty" for each arrangement). But I'm having trouble with the combinatorics.
For example, for item 3, assuming a tableau of 7 cards, you want at least four that share a motif, so for each motif, (I think) you can get this 9*8*7*6*32*31*30 ways, and there are 5 motifs so multiply that by 5 to get the total number of arrangements in which that condition is met. (And divide by 36!/29! to get the probability, I guess)
But then for item 2, it seems you should be able to do the same thing except there you need the cards to be adjacent, and I'm not sure how to set that up to account for that. And then item 4 is harder still.
So, if there are any combinatorics pros who would find this interesting, your help is certainly welcomed; plenty of GG available for helpful replies!
Thanks very much.

Jeff Warrender
United States Averill Park New York

If it's easier to use a more familiar card set to set up the above problems, it would probably be sufficient to use standard playing cards. So the problems would become, e.g., "how many 7card sets have at least 4 cards of the same suit", or "how many 7card sets have at least 3 adjacent face cards" or "how many 7card sets contain a fourcard arrangement with the first and fourth having one suit and the second and third having another suit".
I can figure out how to analogize to my own deck if I can figure out how those problems are set up.

Nyles Breecher
United States Milwaukee WI

I'm not sure that this will help you all that much. Effectively you want to look at if you had a random layout from your cards, what the probability of having one of your configurations would be, correct? In reality though you won't have random layouts. I'm not sure how players get cards, but I can't imagine every card is equal in value (if you have more witnesses, it's going to be more desirable to more people, right?). If you just look at the raw permutations of cards, you can't really account for the fact that some cards might be more highly contested and so a particular configuration may be harder to achieve.
That said, how do players actually get cards in your game? Also, there are 36 cards in your game total?

Alison Mandible
United States Cambridge Massachusetts

If I'm understanding you correctly...
Your formula for item 3 is missing a factor of 7C4 (7*6*5) in it, I think. For a given motif, you have 9C4 ways to pick the four cards with that motif that the tableau will contain, and 27C3 ways to pick the other three cards. But then you've got four scoring cards and three nonscoring cards; how do we interleave them? So you need to multiply by the number of ways to arrange AAAABBB.
Ah, wait, I see, you've also used 32C3 there instead of 27C3, to take into account the fact that you could have MORE than four cards with the motif. That's not going to work; it'll doublecount many arrangements. I think you might just need to calculate the chance of 4 matching cards, the chance of 5, the chance of 6, and the chance of all 7 having the motif.
BUT there's a bigger problem.
You can't just multiply by 5, because that will doublecount tableaus that have two or more motifs. For instance, if you have three cards with motif A, three with motif B, and one with both A and B... tada! Two motifs.
Item 2 is even murkier, because how much overcounting you're going to do depends on how the witnesses are arranged on the cards. Let me explain with a toy example.
Suppose you have four cards and four witnesses. Each witness is on two cards. You want to know what the odds of a random pair sharing a witness is.
If the cards are
AB AC BD CD
then four of the six possible pairs share a witness. But if the cards are
AB AB CD CD
then only two of the possible pairs share a witness! And both of those pairs share TWO witnesses. See what I'm saying?
Do you have a hunch about how to score them? Can you prototype it and just try out some scores to see if they're fun? I love combinatorics puzzles but you don't actually want mathematical perfection here. You want scoring that's fun. That requires testing.



At least the first problem is rather trivial. There are 34 different configurations. x= no. of total cards  (length of cluster1)
As for the other three problems, if I'm not seriously mistaken, your questions aren't solvable unless you specify how many cards have one witness, two witnesses, three witnesses on them (same for motifs).
That said, I don't think that combinatorics alone will get you very far: When the items that I'm trying to match are distributed unequally and this distribution is known to me, then I'm intelligently searching for combinations to beat the odds.
For example, if I'm trying to match witnesses, I will eagerly grab cards with 3 witnesses on them, if I cannot get one of these, I'll go for the 2s, and use the 1s only at the very end to complete my set.
If I'm going for a numerical order, I'll immediately reject the rows 123, 234, 333435, and 343536 because the cards 1, 2, 35, and 36 have a lesser probability to be used in such a row. After all, I already know that 36 has no successor and that 35 has one successor, but that all other cards have two.
Additionally, I suppose that I already know a subset of the cards that are not available (at least I know my cards, what about those of the other players?). That may drastically reduce the possible number of combinations left in the game.
Last but not least, most of your proposed clusters seem to be allornothing. However, if I'm going for the 4motifcombination, there is higher possibility for me to complete a 2pairscombination of motifs than if I wouldn't be collecting motifs in the first place. (Think Poker: I'm trying to get a full house, so I end up with two pairs; if I go for a straight, an incomplete straight doesn't get me anything.)
Instead, maybe you could simply calculate the probabilities for completing a set, i.e. the chance to draw the last missing card for a complete set out of the cards that are still available?

Jeff Warrender
United States Averill Park New York

nbreecher wrote: I'm not sure that this will help you all that much. Effectively you want to look at if you had a random layout from your cards, what the probability of having one of your configurations would be, correct? In reality though you won't have random layouts. I'm not sure how players get cards, but I can't imagine every card is equal in value (if you have more witnesses, it's going to be more desirable to more people, right?). If you just look at the raw permutations of cards, you can't really account for the fact that some cards might be more highly contested and so a particular configuration may be harder to achieve.
That said, how do players actually get cards in your game? Also, there are 36 cards in your game total?
It's true; even a mathematically rigorous answer to the question won't give a perfectly balanced scoring system! The intent is really to just get a coarsegrained sense of "this is a bit easier to achieve than that which is easier than this other thing". There are other aspects to the scoring beyond these, there are placement rules that somewhat limit your ability to arrange an arbitrary configuration of cards, and as you note, you don't acquire the cards randomly anyway. (You travel to various cities, each of which contains several cards; you get to pick one and add it to your book.)
There are 72 cards total, two instances of each card.

Jeff Warrender
United States Averill Park New York

grasa_total wrote:
then only two of the possible pairs share a witness! And both of those pairs share TWO witnesses. See what I'm saying?
Do you have a hunch about how to score them? Can you prototype it and just try out some scores to see if they're fun? I love combinatorics puzzles but you don't actually want mathematical perfection here. You want scoring that's fun. That requires testing.
Ah, that's a good point, thanks.
Indeed, I've been testing a fair bit, but have focused more on the turn mechanisms; those seem to be settling into place, so I'm looking to balance the scoring. I've assigned toy numbers to these scoring categories that seem to work ok, but just don't want a situation whereby "you always go for keeping your cards in chronological order, it's always better than making clusters of witnesses" or whatever.
Thanks for the helpful reply!

Jeremy Lennert
United States California

I think you've omitted several details that would be necessary to provide a mathematically rigorous answer.
The obvious ones are the rules for how you acquire cards and how you determine their order. You've suggested those are pretty complicated and so we ought to pretend they're random for simplicity. I think there's a pretty good chance that pretending they are random is going to skew the results enough to make them nearly worthless, but I can't evaluate that without knowing the full rules, and I suppose that even if you wanted to share them I probably wouldn't have the patience to read them all purely for a mathematical exercise. So let's ignore that for the moment.
jwarrend wrote: You can get points for four different configurations:
1. A cluster of cards in your book is arranged in increasing numerical order. 2. A cluster of at least 3 cards all share a witness. 3. At least four cards all share a single motif. 4. An arrangement of four cards in which the first and fourth share one motif and the second and third share a different motif (an ABBA structure)
You get VP for each condition you meet, and if you meet a given condition more than once you get the appropriate multiple of the points for that condition.
You haven't defined what a "cluster" is. You probably have in mind that the cards must be consecutive, though that's not 100% clear. But if I assume it's any sequence of consecutive cards, then these rules are almost certainly different from what you meant.
Consider a 7page book with these card numbers:
1, 2, 3, 4, 9, 5, 6
Does this fulfill condition #1? If so, how many times does it get to score #1? Arguably, it could score it as many as fourteen times! Because it contains all of the following subsequences:
1, 2, 3, 4 5, 6 1, 2, 3 2, 3, 4 1, 2 2, 3 3, 4 1 2 3 4 9 5 6
That's 14 distinct subsequences that are each arranged in increasing order. Is that what you meant? Even if I throw out the "degenerate" sequences of length 1, that's still 7 subsequences with length 2 or greater, which is still probably not what you want.
Consider witnesses (rule #2). From the above example, you already know you should ask whether 4 consecutive cards with the same witness scores this once, twice, or 3 times. But the fact that one card can have multiple witnesses actually makes things even more complicated
Consider these arrangements:
A, A, AB, B, B  does that score twice, with the AB card counted in both clusters?
AB, AB, AB  does that score twice, for the same exact 3 cards, but for a different witness each time?
Rule #3 requires 4 cards of the same motif, but doesn't seem to specify anything about their relative arrangement. So if I have 7 cards with motifs:
A, B, A, C, A, D, A,
Does that score for having four As?

Jeff Warrender
United States Averill Park New York

That's right  it's not so much that the rest of the game is especially complicated as that it's sufficiently /incremental/ that this exercise is somehwat artificial in the context of what actually happens in the game.
I may have misstated what I meant by #1; the idea is simply that you have a set of cards where each card has a higher number than the ones before it. So 1, 7, 15, 32, 26, 35 has a cluster of length 4 and a cluster of length 2.
To your example about witnesses, yes, both of those score twice. The idea with this computation was simply to get a rough sense of how relatively easy it is to get a 3 card cluster for a single witness compared to having any four cards with a given motif. The practical details of the cards of course make a rigorous answer incalculable so the intent is just to get a ballpark to see whether it agrees with my guesstimate of what the values ought to be or whether I'm way off.


