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--
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--
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?