A gambler starts with a bankroll of b dollars and will repeatedly bet on a coinflip, wagering L dollars to win W with probability p of winning and q=1-p of losing. Given W≠L and p≠.5, what is the probability of busting if the gambler will play until bust?
For now, let's ignore short-stack situations. Maybe we say the minimum bet is L and the game is over if b<L. Or maybe we allow the gambler to go all-in when b<L. But let's deal with that later because the problem is hard enough without that wrinkle. We'll pretend the gambler can still win W even if b<L.
After taking a few unsuccessful stabs at this, I skimmed lots of books, journals and preprints for the answer. The closest thing I could find is
this paper by Cong & Sato, but their solution only works when giving/receiving integer odds (n:1). What I seek is a formula that works for any parameters. Given how many sources I've checked (including the few that cite the linked paper), I'm led to believe this is an open problem in mathematics.
So far I've attempted a basic combinatorial approach and I chose the parameters W=2, L=3 and b=2. The gambler is embarking on a one-dimensional random walk (aka a drunkard's walk), but it's more convenient to think of it as a restricted lattice path which I'll describe. (I also encourage you to try thinking of a better way to visualize the problem, since the lattice approach hasn't much helped me in my attempts at this.)
On a piece of graph paper, draw the line y = 1.5x.
The lines of the graph paper are the lattice. Hero starts 2 units left of the line. Hero moves one unit up on a winning flip or one unit right on a losing flip. To touch or cross the line is to go broke.
Another way to depict it is with disjoint line segments:
(0,0) to (2,2)
(2,3) to (4,5)
(4,6) to (6,8)
and so on.
That 2nd representation shows why this problem isn't like the 1:1 case (whose diagram is just y=x). There is an easy pattern for a 45° line, but in this problem the line keeps shifting and changing the pattern.
Suppose Hero starts at (-2,0).
Hero can reach (0,0) with probability q².
Hero can reach (3,1) via 4 different paths, but 2 of the paths cross through (0,0) and thus were already counted. So there are only 2 new paths to count.
There are then 5 new paths to (2,2) and then 14 new paths to (2,3). So far the coefficients have been Catalan numbers, but next the pattern breaks: there are 28 new paths to (3,4).
Here's a list of the coefs: 1, 2, 5, 14, 28, 76, 227, 488, 1391, 4308, 9647, 28301, 89617, 205893, 615444, 1978069, 4624876, 14007082, ...
(After the first several coefs I reverse-engineered the rest by coding a transition matrix.)
No hits on
OEIS.
For demonstration, the probability of going broke within the first 17 or 18 flips is as follows (letting v=pq):
q² + 2q²v + 5q²v² + 14qv³ + 28qv⁴ + 76qv⁵ + 227v
6 + 488v⁷ + 1391v
8 + 4308v
8p
If you can find a good way to get those coefficients, or find another combinatorial solution altogether, then all power to you.
A better route might be to express the problem recursively: R(b) = p*R(b+W) + q*R(b-L)
That's a variable-degree difference equation, which I know nothing about and IDK if it's solvable. Specifying W and L might make it easier, though I'm not even sure, because when the odds are irrational it's still a fractional-order equation.
Even with rational odds, we probably need a bit of sophisticated machinery to solve the recurrence because there doesn't seem to be an algebraic shortcut (unlike the 1:1 case). For W=2 and L=3 there is a way to form a closed system of 6 equations, but I think it would only give solutions to specific b's, not a formula for arbitrary b.
Specifying W and L sacrifices generality, but maybe after separately solving for multiple values of W and L we can combine the formulas into one or at least gain clues for how to think about the problem.
Closing remarks:
- I get the impression that problems like these are where
analytic combinatorics (generating functions etc) really comes in handy and that without understanding that stuff, one can only have scratched the surface of combinatorics.
- I wouldn't be surprised if the pro move here is to find tight bounds instead of an exact solution.
I'm interested in people's thoughts on the problem.