Possible number of passive tree combinations

I'm doing a youtube analysis of DA2. Why it's such a bad game. Kind of going indepth into game design. But I want to verify my line of thought here.

Path of Exile has how many possible character combinations when only considering class chosen and passive skills picked.

Let's say there are a flat 1200 nodes on the tree.
There are 6 classes.
Let's say we're only counting the first 100 points.

Although the way the current tree is structured removes some degree of choice for your first few passive purchases, players get total freedom, after let's say, 10 points.

Just inside a class's starting zone possible complexity is

choose between 2 nodes (ex: take +10% spell dmg witch starting node)
choose between 3 nodes (ex: take +8% spell dmg, take + 10 int or take +40% mana regen)
choose between 8 nodes (ex take +10 int, +cast speed or +crit, or take +8% mana)

Counting here is messy and confusing, but basically, it's quickly accelerating after your fifth point spent. I'd say, there are 2*3*8 combinations just at your third point - though determining UNIQUE combinations is going to be kind of crazy because you could spend points in a different order and obtain the same build... after a certain number of points, but I guess we can excuse that right? Because even though the end result is the same, if you don't take ANY mana increases until level 50 as a wtich, that's going to be a pretty different playstyle even if you wind up with the same build as me at level 80. A person's passive build has uniqueness at any point in time because it influences play as you level up and grow your passive pool.

So let's say 48 is the number of valid combinations per class and let's say they're all unique. That's almost 300 possible character combinations at level 4. And it's already pretty much impossible to count all these by hand at this point, we're only a little way in. We need a broad way to do this that I can do at one o'clock in the morning (because I'm nuts typing this all out right now).

If there are 1200 nodes and you can pick 100 of them in any order (technically not valid, but the only way I can think of to find out the number of valid combinations is to use a regular expression and brute force the number of combos on a computer app?), then that means the number of combinations is 100! or 9.332622e157 which is more particles in the universe or even possible number of chess boards, and we're not even multiplying this by six yet.

Although this is not a rigorous attempt to determine the possible numbe of passive combinations, is there a smarter way to go about this that anyone knows? Is it fine for me to basically say there are more possible passive skill trees than particles in the universe? Or is there an error in my logic here that's grossly exaggerating the number of possibilities.

Let me show you a more simple way of doing this:

You have a maximum of 111 points to make a character. This isn't particularly likely to ever happen, so let's be conservative and say you will reach full build at 90 points. Another assumption I'm going to make is that any time you gain a skill point, you will be able to choose from 2 different options (this is being VERY conservative). This results in you having to make a choice between two options exactly 90 times. This can be expressed mathematically as

2^90

OR

1 237 940 039 285 380 274 899 124 224

different outcomes.

Oh, and this is only for one class. You should multiply that by six.

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anubite wrote:

...If there are 1200 nodes and you can pick 100 of them in any order (technically not valid, but the only way I can think of to find out the number of valid combinations is to use a regular expression and brute force the number of combos on a computer app?), then that means the number of combinations is 100! or 9.332622e157...

Sorry for double post, just one more thing I had to point out. The number you got for being able to pick any node with out any rules for order in mind is actually A LOT smaller than it actually is. The actual calculation would be

1350!/(1350-111)!

which is 2.80 * 10^345

May seem ridiculous that this number can significantly exceed the number of particles in the universe, but that's the funny thing about statistics and probability, you get ridiculous numbers when you try to calculate things like this.

If you think about it one way, what we're really doing is a sort of model of how 111 of a group of 1350 'particles' can be arranged. The number for all the possible arrangements of all the particles in the universe would make these numbers seem very small.

annnnnnnnd that's enough math for this year.

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MrChristo wrote:

If you think about it one way, what we're really doing is a sort of model of how 111 of a group of 1350 'particles' can be arranged. The number for all the possible arrangements of all the particles in the universe would make these numbers seem very small.

Hmm, good point. Now I feel like I should no longer be amazed by the statement "there are more combinations than there are particles in the universe."

The upper bound would be nC100 for 100 nodes with n possible. n=1200=10^148 combinations maximum

The problem is figuring out how to divide out duplicates and ones that are not possible due to not being connected. This is no trivial task.

From there, counting on graphs gets unbelievably complex... people do research on these topics to this day.

You can attempt to google 'counting on an undirected graph with loops'

wow send your resumes into NASA?

god these forums are so refreshing after getting lost in some of the newer stuff online, basically e-fighting and an endless barrage of queer smiley face things I will never know the meaning of -_-

reminds me of the 36.6k days... when a man defined himself by whether he used a 486 or one of those new pentium things... and weird al did the puff daddy song... and puff daddy was still named puff daddy....

come backs 90s... *sniffle*

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MrChristo wrote:

Let me show you a more simple way of doing this:

You have a maximum of 111 points to make a character. This isn't particularly likely to ever happen, so let's be conservative and say you will reach full build at 90 points. Another assumption I'm going to make is that any time you gain a skill point, you will be able to choose from 2 different options (this is being VERY conservative). This results in you having to make a choice between two options exactly 90 times. This can be expressed mathematically as

2^90

OR

1 237 940 039 285 380 274 899 124 224

different outcomes.

Oh, and this is only for one class. You should multiply that by six.

Thanks for your analysis. That's what I should have done is - assume only two choices every level. That gives us a fairly conservative estimate of the number of outcomes, because we know that once you hit about level 10, you have like 5-10 choices at least, per level.

"

You can attempt to google 'counting on an undirected graph with loops'

Thanks Zeto, I think I will. The number of combinations is probably too enormous to actually count using a computer, and there's probably no hope of making a formula to remove duplicates, so at least I have upper/lower bounds that're meaningful.

Oh and just for comparison, to point out kind of where I'm heading with this:

Baldur's Gate 2:
7 races, 11 classes, up to 9 multi class combinations for each race, 9 alignments, 2 genders, 30 class kits (mage has 9 class kits alone, which I'm treating each as separate classes for human, which should make up for the fact I'm not counting class kits for the other races)

Of course, not all 7 races have access to all 11 classes or their kits, not all 11 classes have access to all 9 alignments

But from my calculations from a race by race basis, where I only make the assumption that each class has a maximum of 3 alignments only (this is actually the minimum, all classes have at least 3 alignments, so I decided to use the minimum here, than to find out every permeation for every race/class combination, which can be up to 9 for some classes)

11*30*2*3+7*4*2*3+9*9*2*3+5*6*2*3+4*3*2*3+4*2*2*3+4*1*3*2 = 2958 possible character creations before attribute dispersal, active skill/spell choice, weapon feats/passive skills, et cetera

Dragon Age 2 offers you how many starting character combinations? Six. Where zero isn't defined, in a positive integer world, BioWare gave us the 5th lowest amount of complexity they possibly ever could.

You do hit a threshold where the number of options stops increasing so much.

It doesn't take too long until you reach options for a build that include choices you could have reached already if you had made different choices.

So it doesn't increase at the same exponential rate throughout your choices.

Interestingly though while raw combinations do face an upper limit, difference in play experience as a character levels has more variety, as choice of when you take what passive does effect things (most notably how your build does against different in game areas.)

Obviously gear and skill selection impact this, but I think this is another aspect that increases the replay ability of path of exile - even if you end up recreating the same build you can get pick your skills in a different order looking for efficiencies or to match the gear you are finding on that run through (or have acquired through other characters.)