What is Bankroll management (Risk of Ruin — RoR)?
3/3 — Three Key Factors That make a Casino Strategy “Impossible” to beat.
This is the third article of the series:
- Expected Value (EV)
- Bankroll management (Risk of Ruin)
Why would the casino care about variance though?
Someone could assume that since the casino is only accepting bets that are profitable for the house and losing for the player who placed the bet, their work ends there but that’s only part of the truth.
Yes, casino indeed only takes profitable bets but still, there is this commonly used term of RoR% (risk of ruin) which indicates what is the chance of busting.
As you understand casinos need cash to operate so they would be able to accept more +EV bets so they keep making a profit, but for this to happen they have to avoid the chance of letting a player place a bet that jeopardizes the sustainability of their bankroll/cash flow at any cost, even if that bet is “profitable” for the house in the long term.
Casinos protect themselves from the risk of losing all of their money by following strict bankroll-management strategies that make the chance of them busting tend to 0%.
Risk of Ruin (RoR%) formula
So Risk of Ruins aka RoR% is the probability of losing all of the capital that consists of a bankroll (or bankroll decreasing to the maximum acceptable loss → stop loss).
The mathematical formula for RoR consists of the following variables :
- Average win% = (average amount the bet wins) / initial capital
- Average loss% = (average amount the bet loses) / initial capital
- Z = (frequency of winning * average win%) — (frequency of losing * average loss%)
- X = [ frequency of winning * (average win%)² + frequency of losing * (average loss%)²]undefined
- T = ½ (1 + Z/X)
- C = capital/bankroll
And the formula would be:
RoR = [(1-T)/T]^C/X
Would you always take a profitable bet?
The following example will help people to get a grasp of this concept easier and understand a few important points without having to dig deep into the mathematical/statistical theory behind the formula.
So far we have explained what Variance and EV are, hopefully, most people who read up to this point will be able to start calculating EVs for their bets in the future (could be sports betting, trading, poker, stocks, real estate, or whatever else people invest/bet in/at) and will start avoiding taking -EV bets (unless they enjoy gambling)
But does a +EV bet always consists of a great opportunity?
Think of this bet proposal: we flip a coin, if you win I’ll give you 10.1 million dollars, and if you lose you’ll give me 10 million dollars. Using the EV formula from the example above you’ll quickly understand that this bet is pretty profitable for you you’d win on average $50k per flip/toss.
That being said if we just flip the coin once you will either win 10.1M or lose 10M that means that if you take this bet you accept the fact that you have a 50% chance to lose 10 million, likewise if we flip the coin twice you have a 25% chance of losing 20 million dollars, thrice would be 12.5% chance of losing 30M, etc.
This outcome might be “okay” for you if you have “enough” money not to get destroyed by losing all of your capital (thus not be able to repeat the “profitable” bet / “ruined” since even though you have a +EV bet to take you can’t afford to bet with 0$ capital) after you lose once or have an unlucky losing streak but generally speaking if you don’t have an incredibly big bankroll (or an incredibly big urge to gamble) you might be very hesitant at taking this bet.
In a few words, this is a great game for you to play, you are guaranteed to earn a lot of money in the long term (in 100 flips you are expected to win 100*$50k= 5M) all you have to do is to manage to remain in the game for long enough, without losing all of your money (avoid getting “ruined”).
The way to approach the question of if someone would take a bet like this or not is to think of how risk tolerant he is or what his tolerated RoR% is if he takes the bet. RoR% is strictly correlated with the available bankroll (betting units) and the number of repetitions of the bet.
In the prior example, assuming you have exactly 10M (1 betting unit) as a starting bankroll if you lose the very first flip, it’s game over since you don’t have any more money to bet into the next flip. Losing the first flip has a 50% chance of happening, so your RoR%, in this case, would be 50% (if you just flipped the coin once).
This is easy to understand in this specific example but let’s give it a theoretical look so we understand the RoR% calculation in a more generic example.
Assuming someone starts with I dollars, each time he bets 1$ in a fair coin flip he gets back 1$ of profit (total of 2$) or loses his 1$ bet. Let’s say that this person is playing a fair coin flip until he either has 0$ or reaches his target of y dollars. What is the probability p of him ending up with y dollars?
Truth is the answer to this question is not very easy in theory and requires some sort of statistical/probability background to be explained in depth. But if you think about it, at any point in time the only thing that matters is the player’s available money. If the player’s available money is neither 0$ nor y then the player will take another bet which will increase/decrease his available bankroll by exactly 1$ depending on if he lost or won that next bet. He is going to keep doing so until he reaches 0$ or Y$.
So how to calculate what is his chance of actually reaching his Y$ target given he starts with a bankroll of i$? Well obviously if his starting bankroll i= 0 then he has a 0% chance of reaching Y$ cause he can’t even bet, on the other hand, if he starts with a bankroll = Y$ he has a 100% chance of reaching his Y$ target without having to make a bet (risk capital).
So this question gets interesting if his i bankroll is > 0$ but < Y$ ( 0 < i < Y ). This type of problem is usually represented in what is known as Markov’s chain.
Assuming we are at the starting state of I where 0 < i < Y, the player has a 50% chance (0.5) to reach the i+1$ state if he wins the flip and also another 50% chance (0.5) of reaching the i-1 state if he loses the flip. So with this diagram (it’s named the Markov chain), it gets easier to understand that the chance of reaching the Y$ target is higher the closer the distance between i and y is and vice versa. In this example since the chance of winning/losing is 50% the probability p of the player reaching the state of Y$ equals p= i / y. This calculation is not easy at all to understand but it indicates the fact that: the number of available betting units and the distance from 0$ or the target (bankroll), and the probability of win/lose, are the only factors that will dictate what’s the possibility of reaching the start/end point of that chain.
So since casinos are pretty much deciding all of those 3 factors (they even set a minimum/maximum bet limit) they are choosing their RoR%. If you have control over those 3 factors you can decide what RoR% you are willing to accept and give it whatever value you want, from 100% down to tending to 0%.
Bibliography / useful sources
There are plenty of sources online explaining the mathematics behind this p = i / y in a bigger depth, you will find my favorite ones by searching:
- MIT probability lectures, Brandon Foltz Finite Math: Markov chain,
- Kevin Delaplante: The Gambler’s Fallacy
My favorite one is a paper coming from Columbia university that is very clean and explanatory giving specific examples and practical applications in a short but detailed way that you can find here → http://www.columbia.edu/~ks20/FE-Notes/4700-07-Notes-GR.pdf.
BetSwirl’s Coin Toss example
Let’s end this article by showcasing a few examples of BetSwirl’s Coin Toss so we can see in practice how values change when we tweak the variables.
Set of variables: EV/edge, bankroll, number of iterations.
So House edge in BetSwirl’s coin toss is set at 3% for AVAX/BNB/MATIC, this means the EV of each player betting against the platform would be -3% * his betting amount.
Bankroll: the capital/bankroll of the house.
The number of iterations: how many bets will players place?
Assuming the house edge is the same in every example let’s see how results differ by changing the available bankroll and how the RoR% of BetSwirl is influenced in every case.
Let us see 3 different examples where the house edge is 3% and the player’s maximum bet is 1 AVAX. In those 3 examples, we’ll assign 3 different values of BetSwirl’s bankroll with 10, 25, and 100 AVAX and see the distribution of the results given that 1000 different Players will place 1000 individual bets of 1 $AVAX.
So starting with a bankroll of C AVAX and allowing 1 AVAX bet on coin toss results for BetSwirl would look like this.
On these following graphs, we can see 20 random samples of BetSwirl’s Bankroll including the best and the worst runs in every scenario.
C= 10 AVAX
C = 25 AVAX
C = 100
We can easily see that when C = 10 there are quite a few occasions where BetSwirl’s bankroll would drop below 0 (-10 AVAX ruin), with C = 25 we can see only 2 out of the 20 random samples falling below 0 AVAX and with C = 100 we can see that even in the worst run scenario BetSwirl’s bankroll won’t bust.
The RoR calculator estimates (pretty accurately) that:
- a starting bankroll of 10 AVAX would give a RoR% of 50%,
- a bankroll of 25 AVAX would give a RoR% of 15%,
- a bankroll of 38 AVAX would give a RoR% of 5%,
- and a bankroll of 50 AVAX would give a RoR% of 1%
At about 70–75 AVAX bankroll BetSwirl is already “impossible” to get ruined since RoR% tends to 0%.