I have a feeling it’s not correct, considering that it always creates a value >V for all V>1, which doesn’t make sense that the first player can get more than V marginally.
I’ll do my own check, cuz this isn’t really considered in Combinatorial Game Theory textbooks
[EDIT:] just did the calculations based on S(2^(n-1))-S(2^(n-2))+S(2^(n-3))-…
so if you group them together you get S(2^(n-2))+S(2^(n-4))+S(2^(n-6))+…
so I substituted n for m, where m=n/2, becoming
S(4^(m-1))+S(4^(m-2))+S(4^(m-3))+…
luckily this is still a geometric sequence, so we can just skip right to the that, since you seem to have come up with the same result:
S(4^m-1)/3, expanding that out a little, we’ll make it S(4^m)/3-S/3. I think this is as far as our calculations are similar
Now, how to get that in terms of V, well V=S(2^(n-1)), putting that in terms of m we get V=S(4^m)/2
and if we multiply the result by V/(S(4^m)/2) (which equals one), we get V(S(4^m)/3)/(3*(S(4^m)/2))-VS/(S(4^m)/2), and to simplify that we get 2V/3-2V/(3*4^m)
So let’s test that against cases we know about, particularly when V=1
So if V=1 and m=1, that means there are two coupons 1 and 1/2, so the result should be 1/2
we plug it in and we get 2/3-2/(3*4) = 2/3-1/6 = 1/2
okay, now let’s see V=1 and m=2, corresponding to 1, 1/2, 1/4, 1/8: resulting in 5/8
plug it in for 2/3-2/(3*16) = 2/3-1/24 = 16/24-1/24 = 15/24 = 5/8
If you don’t believe it yet, you can check against other values.
[END EDIT]
the general assumption is linear with exceptions. These exceptions are usually called “tedomari”, the last big points of a certain “temperature” on the board before it makes a drop.
Of course, even that should be considered a somewhat loose approximation, and only really useful for theorycrafting. On the flip side, it’s proven to be one of the most useful approximations for theorycrafting.