The Final Combinatorial Hour
When Arkham Horror: Final Hour begins, you've already lost. The ritual is done, the gates are open and vomiting out monsters, mass hysteria, dogs and cats living together, do you read Sutter Cane? All that.
As players, you are tasked with figuring out how to reverse the ritual. And like all good ritualistic, world-ending spells, this means collecting enough cards that match either of two unrevealed symbols on some tokens.
Here's how it works. There are five symbols. There are two of each. So, carry the one, there are ten tokens. Face 'em down, stir 'em up, and set two aside. These are the two symbols that matter to you. The other 8 (and some extras) are placed on the board for you to reveal.
During the game, you can flip these other tokens over. Revealing a token that was not chosen. This helps you narrow down to what two tokens were chosen. When you feel like it, or on the last turn, everyone plays some cards with symbols on them, you reveal the two chosen tokens, and if you make enough matches you win. If you don't, you lose.
As you play, you have to make decisions about what symbols to keep in your hand. You want to stock up on the things that are more likely to be matches. So when you have no information, it's a crap-shoot. But once you've revealed a token you know a little bit more. Now the probabilities of what's where have shifted. But by how much? Glad you asked.
At the start of the game, you set two aside. The probability that any one particular symbol has been selected is .37777. So 38%. You can get at this a couple ways.
For one, you can just brute force it out. First, the probability that they're both that particular symbol. That can only happen one way, there's only two tokens and you'd have to draw them both out of the 10 available. So when you set the first aside, the chances you grabbed any one specific symbol is 2/10, and then when you set the next aside it's 1/9 (1 because there's only 1 left, 9 because there's 9 tokens left; you've already drawn one of them). The only way both symbols are the same particular symbol is if both of these events happen, so you multiply them together for .02222.
The other possibility for selecting a particular symbol is that only one of them is in there. So you grab one of the symbols first (2/10) and then you don't grab the other one (8/9). Both of those happening is .177777, BUT there's two ways this can happen. If only one token is our particular symbol, it could be EITHER of two available. So it's actually twice that: .35555.
The two tokens being BOTH our specific symbol, or only ONE of them being our symbol, are mutually exclusive (it can't be only one of them AND both of them). So you can add them together to get .37777. So a 38% chance of any one particular symbol being in there.
Like we went over with Star Wars, it's often easier to think in complements. The complement to "at least one of these is this symbol" is "none of these are this symbol." The chances you draw two tokens and completely avoid a symbol is (8/10)*(7/9). Subtract that from one: 38%.
But it's often easier still to think in terms of combinations. In math, a "combination" is when you select a certain number of things from a collection of other things, and the order doesn't matter. A "permutation" is where the order does matter. So, for Final Hour, we're just selecting two symbols from ten, and it doesn't matter which symbol is first or second, they aren't ordered, so it's a combination.
Combinations are typically written with the word "choose" or sometimes just a C. So "ten choose two" or "10C2" means "how many different ways can you select two items out of ten if the order doesn't matter." There's formulas for it you can find, but I'll leave that as an exercise for the reader. The important part is that there's calculators for it...
So how many ways can you choose two items from ten? 10C2 = 45. So there are 45 different possible combinations of the 10 tokens. How many ways can you chose two tokens if we say you can't choose one particular symbol? 8C2 = 28. The probability of an event is "ways event can happen" over "all possible ways." So the probability you choose two tokens, neither of which match a particular symbols is 28/45. Which, remember, is the complement of "at least one of a particular symbol." So 1-(28/45) is, you guessed it, 38%.
Ok, so we've just done the same math three different ways, but have we done anything useful (which makes the false assumption that math for the sake of math isn't useful)?
For the purposes of playing the game, no, not really. At the start of the game, looking at a symbol in your hand, you know there's a 38% chance that that symbol is a useful one. But that's true of every symbol, so it's not actionable.
You also know that there's only a 2ish% chance that both tokens match that symbol (do you see the combination setup for that?). But again, that's true of each symbol. I guess the only real useful thing you could draw from these is to play the field early; don't keep all of one symbol.
But then, one of your teammates does something useful for a change and reveals a token. So now what do we know?
We know that that token was not chosen at the start. So it's as if we shuffled them all up, set that one aside, and THEN chose our two tokens. Then put that one back in and put them on the board. How does that change things?
Just rework the same math. How many ways can you chose two tokens from 9? 9C2 = 36. How many ways can you chose two tokens if we say you can't choose the one symbol that was just revealed? 8C2 = 28. So 28/36 is the probability that none of the set aside tokens were this symbol, so 1-(28/36) is the probability that this symbol was chosen. And this is 22%.
So, once a symbol is revealed, the probability that it's one of the important tokens drops from 38% to 22%. Once a second one is revealed it drops to 0% (1-(8C2)/(8C2)=0); I wouldn't recommend keeping any of those.
What about the other symbols? Say that you reveal a token and it's a Heart, what happens to the probability that it's a Moon, Star, Clover, or Horseshoe?
Same math. One token's been revealed and you know it wasn't selected, so of the 9 remaining, how many ways can you select two tokens if it can't be your particular token? 7C2. (Remember, we're finding the complement: how many ways can you select none of a particular symbol; since we're interested in a symbol that wasn't revealed, there are 2 of the 9 remaining, leaving 7 to choose from.) And we already know you can overall select two tokens 9C2 ways. So 1 - (7C2)/(9C2) = 41.7%.
So once a token is revealed, the probability that that symbol is still important drops from 38% to 22%, and the probability any one other symbol was selected increases from 38% to 42%.
The problem is, this changes every time you reveal a new token. So there's not just an easy answer because it all depends on differing amounts of different ingredients. But we can make a recipe for it.
The probability of any one particular symbol depends on two things: 1) how many total symbols have been revealed, and 2) how many of that specific symbol have been revealed. We'll call those T (for total) and S (for specific). Given those two pieces of information, we can always figure it out using this:
1 - (((10-T)-(2-S)) choose 2) / ((10-T) choose 2))
This is the generalization of everything we've been doing. The denominator is easy; it's "how many ways can we select 2 tokens?" If T tokens have been revealed, then there are 10-T remaining from which to choose two.
The numerator is a bit tougher. The question here, because we're calculating the complement, is "how many ways can we select two tokens that do not include our specific token"? There are 10-T tokens we're choosing from, but we want to avoid the specific token. If S is how many of those have been selected, then 2-S is how many are left. So (10-T)-(2-S) is how many NOT OUR tokens we have left.
In our examples above we looked at T=1 and S=1 (we've revealed one token, and we're interested in the probability that it's still important) and T=1 and S=0 (we've revealed one token, and we're interested in the probability that some OTHER symbol is important). We've also looked at T=0 and S=0, which is the start of the game. Last, you can see that if S=2, then the numerator and denominator will be the same, so they resolve to 1, and 1-1 = 0; no matter what, once you've revealed two of a symbol, it's impossible that it was chosen.
Ok, so here it all is (in rounded percentages):
T | S
___|____0____1____2
0| 38 NA NA
1| 42 22 NA
2| 46 25 0
3| 52 29 0
4| 60 33 0
5| 70 40 0
6| 83 50 0
7| 100 67 0
8| 100 100 0
So what can we see from this?
1) Once you've revealed two of a symbol, that symbol has a 0 percent chance of being in the ritual. Thank goodness for math.
2) Revealing more symbols always improves your odds. The more tokens you've flipped without revealing the second copy of a particular symbol improves the odds that that symbol is part of the ritual.
3) Once you've revealed 7 tokens, if there is a symbol that has not yet been revealed, you're guaranteed it's a part of the ritual. Once you've revealed 8, you know exactly what the ritual is.
4) Only after 6 symbols have been revealed does the probability of a symbol with one revealed token being in the ritual get to 50%. And by that point, symbols that haven't been revealed at all are up to 83%.
Thing is...all of this is pretty intuitively obvious. Unlike with Black Orchestra, there is not really any actionable results that you wouldn't already likely be doing.
At any point in time after a token is revealed, the symbols that haven't shown up yet are more likely to be critical over the ones that have had one revealed (and those are all equally likely). Because of the way the matching works, you should maybe keep two of those symbols, instead of all of one of them. And I can't think of much more to say about that.
Now, this does not mean the game is simplistic. The decision of the game is, knowing the above, do you hold on to the correct symbols if that means playing a card with too many omens on it? Or playing a card with a value that will likely result in you taking an action you don't want to take? And this on top of the board decisions (where to go, what to fight, etc). Final Hour is not the best game, but it's not because the math.
Did I ever tell you my favorite color was blue?