Quote:
my conjecture at this point is that it doesn't matter whether you bet or not when the game is not a lock, so any sensible strategy which bets everything when the deck only has cards of one color should average the same $9.08. I should be able to prove this shortly.
Proof:
By induction, we will show that your equity when the deck is n:k is 2^(2n+k) / (2n+k choose n).
At the base of the induction, this is true for small counts.
Inductively, this equity is how much you get on if you bet everything, or if the deck is n:n. What must be shown is that you can't do better (in fact, you average exactly the same amount) if you bet nothing when the deck is (n+k):n.
n/(2n+k) of the time you end up with a deck that is (n+k)
n-1), in which case your equity is
2^(2n+k-1) / (2n+k-1 choose n-1)
by induction.
(n+k)/(2n+k) of the time, you end up with a deck that is (n+k-1):n, in which case your equity is
2^(2n+k-1) / (2n+k-1 choose n)
by induction.
So, your equity if the deck is not a lock, and you bet nothing, is
2^(2n+k-1) (n/(2n+k) / 2n+k-1 choose n-1) + 2^(2n+k-1) ((n+k)/(2n+k) / 2n+k-1 choose n)
= (2^(2n+k-1) / (2n+k-1)!) * ( (n /(2n+k)) (n+k)!(n-1)! + ((n+k)/(2n+k)) (n+k-1)! n!)
= (2^(2n+k-1) / ((2n+k)(2n+k-1)!)) * (n (n-1)!(n+k!) + (n+k)n! (n+k-1)!)
= (2^(2n+k-1) / (2n+k)!) * (2 n! (n+k)!)
= 2^(2n+k) / 2n+k choose n.
QED
Interestingly, both of the cases contribute half of your equity.
I've been trying to post this for the past 30 minutes. Is the server being unresponsive to anyone else?