There's a very easy way to estimate ROR - you get ruined when the deviation from your estimated edge equals your bankroll plus your winnings.
B + n*EV - k*S = 0
B = bankroll, in units
n = number of hands played
EV = expected win per hand, in units
z = how safe you want to be, in standard deviations (more below)
S = one standard deviation for your play over n hands (generally m*sqrt(n), where m is between 2 and 3)
The lifetime ROR can be derived by simply finding the point where your expected winnings equals your original bankroll; this is the point at which the bankroll you will need actually starts going down.
If you like derivations ...
df/dn = EV - (z*m)/(2*sqrt(n)) == 0
n(max) = [(z*m)/(2*EV)]^2
B(max) + [(z*m)/(2*EV)]^2*EV - k*m*(k*m)/(2*EV) = 0
If you don't ...
B(max) = (z*m)^2/(4*EV)
On the flip side, you can calculate ROR for a bankroll by solving for z:
z = 2*sqrt(B*EV)/m
Let's say you want a 2.3% ROR (z = 2) and you play a fairly moderate game (m = 2.5) with moderate EV (EV = 0.01). Your lifetime bankroll is easy to estimate: (2*2.5)^2/(4*0.01) = 625 units. And for a 900 unit bankroll on the same game, z = 2*sqrt(900*0.01)/2.5 = 2.4, which corresponds to a ROR of 0.8%.
z is easy to find because it's a standard parameter for confidence intervals of normal distributions.
Here is a calculator, and
here is a table of z's. Note that the table gives deviation from the middle, so ROR is 0.5 minus the value in the table.
Or, just use this chart (compiled from the statsoft link).
Code:
z ROR
0.1 0.460
0.2 0.421
0.3 0.382
0.4 0.345
0.5 0.309
0.6 0.274
0.7 0.242
0.8 0.212
0.9 0.184
1.0 0.159
1.1 0.136
1.2 0.115
1.3 0.097
1.4 0.081
1.5 0.067
1.6 0.055
1.7 0.045
1.8 0.036
1.9 0.029
2.0 0.023
2.1 0.018
2.2 0.014
2.3 0.011
2.4 0.008
2.5 0.006
2.6 0.005
2.7 0.004
2.8 0.003
2.9 0.002
3.0 0.001
Is this the most precise way to find ROR? Probably not. For one thing, outcome distributions aren't normally distributed, and I bet there's a lot else that can be nitpicked.
But keep in mind that ROR calculations are extraordinarily sensitive to the parameters you put in - incorrectly estimating your edge, for instance (a requirement for every ROR calculation regardless of formula), will cause erroneous ROR.
Furthermore, calculating your exact ROR is pretty futile, since you won't run enough trials to tell the difference. I bet a majority of people only go through one bankroll: either they succeed and never ruin, or they ruin and quit. Few people go through three, and probably only gambling addicts would go through 5 or more. You'll never tell the difference between a 1% ROR and a 2% ROR!
So what about replenishable bankrolls?
This is easy to estimate as well, simply by adding a term to the original equation. Give R as the replenishment rate per hand, or the replenishment rate per time divided by the hands you play per time.
B + n*R + n*EV - z*S = 0
You can rederive or just simply note that everywhere you saw (EV) in the first set of equations, you replace that with (EV + R).
B(max) = (z*m)^2/(4*(EV+R))
z = 2*sqrt(B*(EV+R))/m
If you accept a 16% ROR (z = 1) on an aggressive game (m = 3) with high EV (0.02) and replenish 100 units per month over 5000 hands per month (R = 0.02), your bankroll only needs to be (1*3)^2/(4*(.02+.02) = 56 units. If you have a bankroll of 120 units on a high EV = 0.02, low variance (m = 2) game and can replenish 50 units a year playing 10,000 hands (R = .005), your risk of ruin is z = 2*sqrt(120*(.02+.005))/2 = 1.7, or 4.5%.
Another way to think about ROR is to calculate the "effective" bankroll (B'), which means for a given ROR (constant z), you just define B' = B*(EV+R)/EV such that:
B' = (z*m)^2/(4*EV)
Or, in other words, simply add your replenishment rate (per hand) to your EV and treat your replenishment as a win. If you can replenish as fast as you win, you can play as if you doubled your bankroll.
