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Author Topic: Quadratic cards  (Read 12478 times)

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yaron

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Re: Quadratic cards
« Reply #25 on: December 27, 2011, 10:26:06 am »
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Laboratories certainly aren't quadratic (each extra Lab played after the first only increases your handsize by 1).

It depends on whether your independent variable is "cards of this type played this turn", or "cards of this type in your deck".

If you're looking at cards played, then Labs are linear - each one is +1 card.

But if you're looking at cards in deck, then each Lab gives you +1 size when you draw, but it also makes your other Labs better by being there to be drawn by them.

Another way to look at it: while "value from Labs" is linear in "Labs played per turn", "Labs played per turn" is quadratic in "Labs in deck".  If you have 6 Labs to my 3, you'll play more than 2 Labs for each Lab I play.

This is true for all engine cards - the more you have in your deck, the more each one is worth.
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yaron

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Re: Quadratic cards
« Reply #26 on: December 27, 2011, 10:40:49 am »
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v - average value of venture
k - number of ventures in deck
n - number of non-ventures in deck
T - total value of non-ventures in deck

v = 1 + (T+kv) / (n + k)

(in words: a venture is worth 1 coin + the average value of all treasures in deck)

This reduces to:

v = 1 + T/n + k/n

So the value of each particular venture (v) grows (in a linear fashion) with the number of ventures in deck (k).

The total value of all ventures grows quadratically.



Venture is quadratic since the average treasure value increases linearly with the number of ventures in the deck.

Maybe I'm misunderstanding, but I don't see how that's quadratic. The average treasure value is represented by (T+v)/n where T is the total value of non-Venture cards and n is the deck size. As you say, this is a linear relationship and not quadratic.

The effect of the card is also linear. Assuming a single Venture results in revealing all your Ventures in succession, the value of your hand would be T + v + t, where T is the total value of non-Venture cards in your play area and t is the value of the treasure drawn as a result of your last Venture. That is not quadratic either.

Although, I suppose that one could argue that a linear equation could be quadratic, as ax^2 + bx + c could see a value of a=0, but I believe that an equation must have a!=0 to quality as quadratic.
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DG

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Re: Quadratic cards
« Reply #27 on: December 27, 2011, 02:13:48 pm »
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The calculation of each venture's value is dependent upon the other ventures so you can't just multiply them up. In other words, the ventures brought into play from the deck do not have the same value as the ventures played from hand.
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jimjam

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Re: Quadratic cards
« Reply #28 on: December 27, 2011, 06:35:11 pm »
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@K: Yeah, a!=0 for a quadratic equation

@DG: Yes, they are dependent, but the leading term is still x^2, per the reasoning that the venture value increases linearly with ventures and the chance of a certain card in hand being a venture is linear. Moreover in the limit, the value difference between v(k,n) and v(k-1,n) is negligible.
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DG

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Re: Quadratic cards
« Reply #29 on: December 27, 2011, 08:34:01 pm »
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Not convinced. If we consider the simplest deck model this would probably be where the treasures are arranged in a series in the deck, and a treasure is considered played "from hand" if the preceding treasure is not a venture (i.e. we flatten out the 5 cards hands). In this case the chance of a venture being played "from hand" would be n/n+k-1. So the number of ventures in the deck that are played from hand may not increase linearly when you add more ventures.

If you want to go further I think you need to define what the "total value of non-ventures" and "total value of ventures" actually mean since they're inconsistent at the moment.
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jimjam

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Re: Quadratic cards
« Reply #30 on: December 28, 2011, 01:08:50 am »
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*
This is my reasoning: If k is the number of ventures and n is the number of coppers (lets keep it simple with just two types of treasure), then in the case where k<<n (if the deck is mostly ventures then the treasure per turn is bounded by the size of the deck, which is linearly related to ventures. If this is what you're talking about, then I agree. If you're in the case where you're playing almost your whole deck, then only Bank or Venture/HoP/Mandarin will yield quadratic treasure).
Then the chance of any given card in hand being a venture is k/(n+k), which is approximately k/n.
Now in the large case, we can sample with replacement. Therefore when a venture is played the ventures hit before copper is geometric, so we expect the value of a venture played from your hand to be 1+ 1/(n/(n+k-1)), which comes out to be 1+(k-1+n)/n
so the value of your hand = expected $ of ventures *expected value of a venture=5*chance of a card being a venture*expected value of a venture. The  highest term is k^2 and it is clearly not canceled out anywhere. Therefore it is quadratic.

Now you might ask, well, if I have a venture as one card in my hand, then doesn't that mean less chance of another card being a venture? Yes, but it balances out because if your first card is a venture, you can just not look at your other cards in hand, and play it. Then the venture has a certain expected value, which was calculated. If you have any other ventures in hand, you can forget about the first one, because the first card (be it copper or venture) isn't likely to take out more ventures or coppers from the deck than the pre-existing ratio, and it will thus have the same expectation.


*Ok, I understand what you mean now. The main point is basically that (k being coppers and n being ventures in your formula) n/n+k-1 is approximately n/k since we're taking k to be significantly larger than n. It's up to debate whether we can have an arbitrarily large constant k that is always larger but doesn't grow with n, but what you said before about the value dependence of ventures is not a reason against it being quadratic.

Spoilered for obsolescence

Edit: Ok, I think you're right. The calculation I'm using for this is that given a large deck of v ventures and c coppers, the total treasure of the deck is v+c and it takes c/5 turns to cycle through it, so ventures indeed increase the treasure per turn linearly.
« Last Edit: December 28, 2011, 01:37:14 am by jimjam »
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Geronimoo

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Re: Quadratic cards
« Reply #31 on: December 28, 2011, 06:11:03 pm »
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If you want to know how much a Venture will be worth in a certain deck you could always use my simulator and more specifically the new feature where you define the start state (edit/create, then choose the "Start state" tab next to the description)

For instance: you want to know how much $ a Venture would give on average if you have 7 Copper, 4 Venture, 3 Silver and 2 Banks in my deck: just put the Venture in the Starting Hand and the  "7 Copper, 4 Venture, 3 Silver and 2 Bank" in the Starting draw deck. Save and simulate and according to the average $/turn graph it will be worth about $2.8 ...

No complex formulas needed (or if you found a formula, you could check its validity with the simulator)
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jimjam

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Re: Quadratic cards
« Reply #32 on: December 28, 2011, 08:20:55 pm »
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Well what we're looking at here is the limiting relationship, not the exact values of the ventures. I was considering writing a quick python code to calculate the values but I decided it would be better to just figure out the solution algebraically. I was wrong at first since I used a ridiculous approximation, since all variables except n should be fixed/bounded while n goes to infinity. And the number of ventures is hand is trivially bounded by 5, so I'm now of the opinion ventures are linear, which I think nobody now disagrees. The final formula for ventures and coppers I came up with, 5(v+c)/c seems to be correct and not require much approximation at all.

As for minion, it is roughly quartic, not quintic as I had said earlier, for the same reason (the value of a hand is bounded by $8), though Minion Sudden Death (reshuffling in the middle of the turn) is an issue. This might be worth running in a simulator.
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