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Author Topic: The Mathematics of Venture Chaining AND The Sociology of Internet Forums  (Read 28547 times)

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WalrusMcFishSr

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+14

So I was messing around a bit and I came up with this neat formula for the "value" of a Venture:

C(v,N) = (1 + v + N)/(1 + N)

Where C is the "value" obtained *just from venture chaining*
v is the number of Ventures remaining in your deck and discard pile,
and N is the number of non-Venture treasure cards remaining in your deck and discard pile.

This includes the $1 from the starting Venture but *not* the final revealed treasure, so to get the final value you should add d, the treasure density of the non-Venture treasure cards in your deck and discard pile. Note that we don't have to include non-treasures, as we normally would when calculating money density, because everything but treasures is skipped over.

Here's my work:

In Mathematica--

Code: [Select]
In[190]:= Sum[v!/((v - n)! ((g + v)!/((g + v) - n)!)), {n, 0, v}]

Out[190]= (1 + g + v)/(1 + g)

Empirically tested with Python simulations--

Code: [Select]
import random

v=28
N=13
tries=500000

deck=[]
scores=[]

for i in range(tries):
    deck=[1]*v+[0]*N
    random.shuffle(deck)
    going=True
    val=1
    while going:
        card=deck.pop()
        if card:
            val+=1
        else:
            going=False
    scores.append(val)


print float(sum(scores))/len(scores)
           

For example, one run of this code here with v=28 and N=13 gives me 3.000214, which is very close to the predicted value of (1 + 13 + 28)/(1 + 13) = 3. Good luck getting 28 Ventures in your deck.

So the formula seems to work out. Let's look at some selected values:

If you have v=0 and N=whatever, obviously the value will just be 1.

Similarly, if you have v=whatever and N=0, the value will be equal to whatever+1.

What if you have your original 7 Coppers, and 2 Ventures (playing one, one remains in deck). Assuming your only treasure in hand is your Venture:, we'd have v=1, N=7, and C=1.125, which gives 2.125 when you take into account the extra $1 from Copper. Pretty much what you expect--a silver + 1/8 chance of getting gold. And realistically, you might have another one or two of your coppers in hand, so C could be more like 1.14 or 1.16.

What about for more elaborate chaining? Like, for example, let's say we had 5 Ventures. How does that work out for reasonably-thinned decks?

C(5,5) = 1.8333
C(5,4) = 2
C(5,3) = 2.25
C(5,2) = 2.6667
C(5,1) = 3.5

So with 4 coppers left it becomes as good as Gold, and maybe you've got some silvers and stuff too. 5 Ventures (+ the one you played) is kind of a lot though; how about for v=3?

C(3,5) = 1.5
C(3,4) = 1.6
C(3,3) = 1.75
C(3,2) = 2
C(3,1) = 2.5

You can continue to mess around with the function if you want. I was gonna take some derivatives and stuff, but then I was like, nahhhh.

So what's the conclusion of all this? Oh man, I was supposed to come up with a conclusion? I just like math. I know that there are many other considerations with regards to Venture other than its pure dollar value, such as its ability to deal with junkiness and counter certain cards. You guys are better at Dominion than I am--what are the strategic implications of this? Or is it just an amusing diversion?
« Last Edit: August 14, 2013, 07:53:11 pm by WalrusMcFishSr »
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dondon151

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Re: The Mathematics of Venture Chaining
« Reply #1 on: August 14, 2013, 01:35:34 am »
+5

I think it is far more useful to view Venture as a Treasure card that augments the value of another Treasure card in your deck by $1.
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ragingduckd

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Re: The Mathematics of Venture Chaining
« Reply #2 on: August 14, 2013, 01:51:41 am »
+20

I think it is far more useful to view Venture as a Treasure card that augments the value of another Treasure card in your deck by $1.

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WalrusMcFishSr

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Re: The Mathematics of Venture Chaining
« Reply #3 on: August 14, 2013, 02:07:52 am »
+7

I don't think it's scumbaggy to post one's opinions--that's what I was after anyway! He does know more than I on the subject.

I think it is far more useful to view Venture as a Treasure card that augments the value of another Treasure card in your deck by $1.

That is true, except for the chaining ability augments that somewhat. Granted, it may be a fairly insignificant effect, or one that rarely matters, but that's why I went through this exercise, to find out!

I also forgot to mention that this formula is very practical for extending Ventures to the real and complex domains. For example, what if you had -4 Ventures and i Coppers in your deck? You would be confident in guessing that you would receive about $-1+2i on average, plus your money density (which I'm not assuming to be positive or real).
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sudgy

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Re: The Mathematics of Venture Chaining
« Reply #4 on: August 14, 2013, 02:18:47 am »
+8

I play my Ventures, giving me 6 + i coins.  I buy a Possession.
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

dondon151

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Re: The Mathematics of Venture Chaining
« Reply #5 on: August 14, 2013, 03:45:59 am »
+1

(image)

I don't know if you're being facetious or actually being a douchebag. It really feels like the latter to me.
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pst

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Re: The Mathematics of Venture Chaining
« Reply #6 on: August 14, 2013, 04:22:07 am »
+8

So I was messing around a bit and I came up with this neat formula for the "value" of a Venture:

C(v,N) = (1 + v + N)/(1 + N)

Where C is the "value" obtained *just from venture chaining*
v is the number of Ventures remaining in your deck and discard pile,
and N is the number of non-Venture treasure cards remaining in your deck and discard pile.

Since you only give empiral test, here is a proof by induction as well.

C(v,N) = (1+v+N)/(1+N) is true when v=0, because then it's just (1+N)/(1+N) = 1, the value of that single Venture.

Now assume C(v,N) = (1+v+N)/(1+N). Is it then also true for v+1? That is, is C(v+1, N) =(2+v+N)/(1+N) ?

Well, C(v+1,N) is always at least 1 (the value of the Venture itself), and then with probability (v+1)/(v+1+N) (the probability the next treasure drawn is a new Venture) we add the value with one less Venture, that we already know about. So,

C(v+1,N) = 1 + ((v+1)/(v+1+N)) C(v,N)   =
1 + ((v+1)/(v+1+N)) ((1+v+N)/(1+N)) = 1 + (v+1)/(1+N) = (2+v+N)/(1+N)

So it's true when v = 0, and if it's true for some v, it's also true for v+1; ergo it's true for all natural numbers.
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GwinnR

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Re: The Mathematics of Venture Chaining
« Reply #7 on: August 14, 2013, 05:58:37 am »
0

I don't know if I understood this all right. But there has to be a difference between the draw and discard pile. For example I play the Venture and I know, the only treasure-card in my draw pile is a Copper. So I know this Venture will draw this Copper; and if there are 9 Ventures in my discard pile or none of them, that makes no difference.
Is this included in your formula?
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popsofctown

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Re: The Mathematics of Venture Chaining
« Reply #8 on: August 14, 2013, 06:25:00 am »
+12

(image)

I don't know if you're being facetious or actually being a douchebag. It really feels like the latter to me.
His five upvoters are douchebags too, by extension.


I really think your one line post was unconstructive and dismissive. 
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pst

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Re: The Mathematics of Venture Chaining
« Reply #9 on: August 14, 2013, 06:34:14 am »
0

I don't know if I understood this all right. But there has to be a difference between the draw and discard pile. For example I play the Venture and I know, the only treasure-card in my draw pile is a Copper. So I know this Venture will draw this Copper; and if there are 9 Ventures in my discard pile or none of them, that makes no difference.
Is this included in your formula?

It is not. It is meant to give an indication of how much value your Ventures will give in your deck. In a specific situation where you play a single Venture, set N and v according to what's just in your draw deck instead (as long as there is some "terminal treasure" there. (Playing several Ventures will be a bit more complicated.)
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RTT

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Re: The Mathematics of Venture Chaining
« Reply #10 on: August 14, 2013, 07:21:07 am »
+2

And what if you start your venture Chain with an Adventurer  :o

 ;)
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GwinnR

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Re: The Mathematics of Venture Chaining
« Reply #11 on: August 14, 2013, 07:38:22 am »
+2

And what if you start your venture Chain with an Adventurer  :o

 ;)
They don't get played directly. You only put them into your hand and then play them in your buy phase.

I don't know if I understood this all right. But there has to be a difference between the draw and discard pile. For example I play the Venture and I know, the only treasure-card in my draw pile is a Copper. So I know this Venture will draw this Copper; and if there are 9 Ventures in my discard pile or none of them, that makes no difference.
Is this included in your formula?

It is not. It is meant to give an indication of how much value your Ventures will give in your deck. In a specific situation where you play a single Venture, set N and v according to what's just in your draw deck instead (as long as there is some "terminal treasure" there. (Playing several Ventures will be a bit more complicated.)
Ok. I think as long as there is one other treasure card in the draw pile, you just take "N" and "v" for the cards in your draw pile.
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DG

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Re: The Mathematics of Venture Chaining
« Reply #12 on: August 14, 2013, 09:06:32 am »
+1

Unfortunately there are a number of ways of looking at ventures and the chain value isn't necessarily the most useful. You've essentially looked at the potential sequences of treasures and ventures in the deck and counted how many ventures you expect to draw before the first other treasure (TVXXX, VTXXX, VVTXXX, etc). This doesn't take into account the value of those treasures and the number of non-treasure cards in the deck, both of which are important to actually using ventures. It can be more useful to take the value of a venture as 1 + average treasure value of the draw deck (as Dondon said).

There are still plenty of complications since a working Dominion deck has cards both in hand and the discard pile that invalidate any generalized mathematics about the draw deck. I hereby forbid anyone to discuss the farming village effect in this thread since people always misunderstand it and the thread quickly gets full of nonsensical claims and counter claims.
« Last Edit: August 14, 2013, 09:08:07 am by DG »
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BadAssMutha

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Re: The Mathematics of Venture Chaining
« Reply #13 on: August 14, 2013, 09:17:35 am »
+1

Neat stuff! However, we all know that Dominion is a game of big turns and small turns, not one of averages. I'd be interested to see a probability distribution over the Venture values instead of just the average value of a Venture. If I play a Venture in a deck that contains of 4 Ventures and no other treasures, I will always get $5 total. If I play Venture in a deck of Venture, Copper, Silver, Gold, I get 3/12 chance of $2, 4/12 of $3, 4/12 of $4, and 1/12 of $5. The distribution of these values says a little more than just the average of $3.25. I'm imagining a situation where you don't want to trigger a reshuffle with Venture, but the $ from it could be enough to end the game in a win. The probability distribution would tell you a lot more about whether it's a good idea or not. In the second example, needing only $3 (just a quarter under the average) gives you a good 75% chance of making it. Needing $4 (a full $0.75 more than the average) still gives you a 41% chance. I did the example by hand, care to figure out the formula for the distribution? I suppose it would require a list of Treasures in the deck, rather than just the number of them, though.
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Witherweaver

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Re: The Mathematics of Venture Chaining
« Reply #14 on: August 14, 2013, 09:43:37 am »
0

...
It can be more useful to take the value of a venture as 1 + average treasure value of the draw deck (as Dondon said).

But how do you compute that when you have more ventures in the deck?  Don't you also have to consider the possibilities of where the other ventures lie? 
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RTT

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Re: The Mathematics of Venture Chaining
« Reply #15 on: August 14, 2013, 09:47:18 am »
0

Isnīt a Venture just a Pedler wich is guaranted to draw a treasure card (as long as you have one remaining in your deck) and forcing you to play that treasure.

oh ant it canīt be drawn dead like peddlers can.
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Witherweaver

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Re: The Mathematics of Venture Chaining
« Reply #16 on: August 14, 2013, 09:53:30 am »
+10

Isnīt a Venture just a Pedler wich is guaranted to draw a treasure card (as long as you have one remaining in your deck) and forcing you to play that treasure.

oh ant it canīt be drawn dead like peddlers can.

I prefer to think of it as a Ruined Village that also draws a treasure card, forces you to play that treasure card, and gives +1 coin.  And can't be drawn dead.  And counts as a treasure.
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GwinnR

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Re: The Mathematics of Venture Chaining
« Reply #17 on: August 14, 2013, 10:14:13 am »
0

Isnīt a Venture just a Pedler wich is guaranted to draw a treasure card (as long as you have one remaining in your deck) and forcing you to play that treasure.

oh ant it canīt be drawn dead like peddlers can.
But it also can't be throned and kinged and it doesn't help for Peddlers and Conspirators. But it can be counterfeited.

But in general, it's just what you said, which of course doesn't tell you anything about the "value" of the Venture.
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DG

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Re: The Mathematics of Venture Chaining
« Reply #18 on: August 14, 2013, 11:01:53 am »
+1

Quote
But how do you compute that when you have more ventures in the deck?  Don't you also have to consider the possibilities of where the other ventures lie?

No. If you consider a sequence of treasures in draw deck of copper, silver, venture, venture, you eventually get the same total income from s.v.v.c, c.v.v.s, v.v.s.c, v.c.v.s, etc. You will get 5 coins by playing 2 treasures from hand. The complications come from shuffles, cards in hand, and the farming village effect.
« Last Edit: August 14, 2013, 11:03:52 am by DG »
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Witherweaver

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Re: The Mathematics of Venture Chaining
« Reply #19 on: August 14, 2013, 11:14:28 am »
0

Quote
But how do you compute that when you have more ventures in the deck?  Don't you also have to consider the possibilities of where the other ventures lie?

No. If you consider a sequence of treasures in draw deck of copper, silver, venture, venture, you eventually get the same total income from s.v.v.c, c.v.v.s, v.v.s.c, v.c.v.s, etc. You will get 5 coins by playing 2 treasures from hand. The complications come from shuffles, cards in hand, and the farming village effect.

This doesn't seem quite right.  If you have, for instance, V C in hand and S V in deck, then playing Venture and Copper gets you four coins.  Or any ordering where the second venture is last (or three coins when both ventures are last).  Or am I missing what you're saying?

Anyway, my point more was that if you consider the value of venture to be 1+<average treasure value of draw deck>, how do you estimate <average treasure value of draw deck>?  Do you count Ventures in the draw deck as 1?  Shouldn't they instead be 1+<average treasure value of (some perhaps modified) draw deck>? 
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dondon151

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Re: The Mathematics of Venture Chaining
« Reply #20 on: August 14, 2013, 11:58:04 am »
+1

I really think your one line post was unconstructive and dismissive.

I'm sure that posts about Dominion strategy have a reputation for being constructive. No surprise that when I post something that I feel is mildly insightful, I get called out for being a Goko detractor. How is that even relevant, anyway?
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DG

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Re: The Mathematics of Venture Chaining
« Reply #21 on: August 14, 2013, 12:12:18 pm »
+1

Quote
This doesn't seem quite right.  If you have, for instance, V C in hand and S V in deck, then playing Venture and Copper gets you four coins.  Or any ordering where the second venture is last (or three coins when both ventures are last).  Or am I missing what you're saying?

It is right and you are missing what I'm saying. I did indeed state that cards in hand change the maths and reshuffling changes the maths. We have to look at simple sequences so in a simple example the order of playing treasures doesn't matter as they will all get played before the shuffle and the number of treasures played from hand is always the same.
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Witherweaver

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Re: The Mathematics of Venture Chaining
« Reply #22 on: August 14, 2013, 12:17:36 pm »
0

Quote
This doesn't seem quite right.  If you have, for instance, V C in hand and S V in deck, then playing Venture and Copper gets you four coins.  Or any ordering where the second venture is last (or three coins when both ventures are last).  Or am I missing what you're saying?

It is right and you are missing what I'm saying. I did indeed state that cards in hand change the maths and reshuffling changes the maths. We have to look at simple sequences so in a simple example the order of playing treasures doesn't matter as they will all get played before the shuffle and the number of treasures played from hand is always the same.

Ah, I thought you were talking about for a given (arbitrary) hand and not throughout the entire shuffle.
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Re: The Mathematics of Venture Chaining
« Reply #23 on: August 14, 2013, 12:23:30 pm »
+20

I really think your one line post was unconstructive and dismissive.

I'm sure that posts about Dominion strategy have a reputation for being constructive. No surprise that when I post something that I feel is mildly insightful, I get called out for being a Goko detractor. How is that even relevant, anyway?

To be honest, alot of your posts have this sort of aloof and dismissive tone that vaguely annoys me, regardless of whether I agree with their actual content.
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Re: The Mathematics of Venture Chaining
« Reply #24 on: August 14, 2013, 01:47:57 pm »
+7

(image)

I don't know if you're being facetious or actually being a douchebag. It really feels like the latter to me.
His five upvoters are douchebags too, by extension.
Since it's ragingduckd we're talking about, it's by goko extension.
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