Risk of ruin, a different way

OK all you gamblers, it’s time to show you some truths about the probability of going bust that most punters don’t understand. I get particularly annoyed when blackjack card counters come out and say... "The count is positive, that means the odds are in my favour (by 0.1% or something), therefore I should bet as much as I can"

This post, and I will follow up with others, will show that this is madness.

If you have $1, and you bet it on even money shots until you’ve lost it (and all the money that you won with it), then I will show that your probability of losing that $1 is 1. That is, you will lose it, even though it may take time.

Here is a triangle...

Line 1: 1

Line 2: 2 0

Line 3: 3 1

Line 4: 4 2 0

etc.

At the top, you have $1, and you can go diagonally down to either having $2, or $0

In this first model, the probability of going each way will be 50%.

At each point (where the number is), you bet $1, and either add $1 to your bankroll or subtract $1 to your bankroll.

Now you could draw an infinite number of lines and add up the individual paths to 0 to get your probability of ruin.

But there is a better way...

Let PL = your probability of ruin (ending up with 0)

Let Pl = your probability of losing (in our case ½)

Let G = the amount that you win when you win, in our case $2. (Your bankroll goes up by $1, but you bet $1 also, and in this model the amount won must be $2).

It can be seen starting at the top you either go to $2 or ruin. Now if you have $2, then rather than continue with the paths, we can go back to the top with 2 chances, so PL2 (probability of losing $2) must be PL1 squared; because you must lose $1 AND you must lose another $1.

So we can make an expression...

PL = ½ + ½(PL squared)

Solving this...

2PL = 1 + (PL squared)

(PL squared) – 2PL + 1 = 0

(PL-1)(PL-1) = 0

PL = 1

That means... YOU WILL LOSE YOUR MONEY ONE DAY TAKING BREAK EVEN BETS, although I concede that it may take a while.


The general formula is...

PL = Pl + (1-Pl)(PL to the power G) Remember that Pl is the probability of losing 1 bet immediately, PL is probability of losing it, and anything that it won, over as many bets as it takes to lose it.

I have not been able to simplify the general formula, and I would be most grateful to anyone that can do some simplifying for me. The way I solve it is for explicit cases where G is an integer, you can generally divide by (PL-1) until the final root, (PL-x), where x is the probability pops out.

Perhaps I missed the point of your post.


Yes, with a liberal definition of "eventually" you will eventually lose.


But its also true that if you stay outside you will "eventually" get hit by a meteor. Does that mean that if you go outside you will ACTUALLY eventually get hit by a meteor? No, it doesn't. Does this mean that you shouldn't ever go outside? No, it doesn't. You can not live for 128 million years outside.


If you are lucky you will live a long happy life outside and die long before the meteor strikes your grave; likewise with a suitable bankroll you will live a long and happy life wagering for .1% at blackjack.


Assuming that "eventually" always catches up to you is "madness".


- Louie

"likewise with a suitable bankroll you will live a long and happy life wagering for .1% at blackjack"


I don't know about you, but the one time that I was in Vegas, I got told to rack of by 2 casinos when going from a $5 bet to $50. I got to play all of about 10 minutes.


Admittedly on the shoe of 6 packs of cards, I got to play for a few hours, but boy they were watching closely after that few hours and I left.


I need some tips on that side of things because I would like a long and happy life playing blackjack too.


I'll get back to you on the probabilities again later.

It seems like you are concerned with overbetting your bankroll. I think that virtually everyone who posts on these forums understands that overbetting your bankroll will cause you to go broke if you overbet by enough.

The political aspects of professional blackjack (getting along invisibly in the casinos) is the real tough part. Somewhere 2+2 wrote quite a bit about that, but I don't recall where. I think one of the keys is to notice when they are noticing you without them noticing that you notice.


But back to your original post, I think I realized what I wanted to say: you seem to have "proven" that something that CAN happen WILL eventually happen.


- Louie

I believe that amny gambling theory books show that you should only bet the % of your bankroll that corresponds to your edge; ie, you should not bet 100% of your bankroll on a 1% advantage.


Regarding the rest of your post, it looks like mathematical flim flam. I suspect if you change "PL" to "PD" (the probability of doubling your bankroll) the formula would also show a 100% probability of doing this as well (assuming you could go below zero before you got to a double, which seems fair, since you assume you will be able to go well above zero before busting out). Would it not? Is the formula and methodology not exactly the same?

First off, I liked the graph theory thing. Graph theory is cool, but I never got to take a class in it. Do more!


Secondly, like some other posters have said, your audience for a post with "Risk of Ruin" in the title is going pretty much the same people who already have a good handle on it.


Some of the "Ruin" problems I've been playing with lately are:


* How to adapt Kelly betting to the nearly continuous domain of game levels on the internet. Say there is ample evidence of a changing win rate as you go up and down in limits. How long does it take to get a remotely converged function of win rate with game size. I would like to think that your performance would at one level would tell you something about your predicted performance at least for the levels immediately above and below.


* How to combine numbers fro different games. If you are playing at two tables, one at $4.99 - $9.98 and one at $5.02 - $10.02, your risk would be nearly identical to two tables of 5-10. However, if you are playing 1 @ .01 - .02, and one at $9.99 - $18.98, your risk is almost completely determined by the higher game. The Kelly BR formula is Units = Var/expectation. Given Var1, Var2, Exp1, Exp2, and alpha, a fractional time spent at table 1, (Table 2 time would be (1-alpha)), what would be the proper expression for the Kelly BR?


best,


zooey

***The general formula is...

PL = Pl + (1-Pl)(PL to the power G) ***


I'm not sure I understand your logic here. So if G is 2, PL is {1 +- sqrt[1 - 4*(1-P1)*P1]}/(2*(1-P1)). That would mean that if P1 was 1/4, then PL = (1+-sqrt[1/4])/(3/2) = (2+-1)/3, which means that the probability is either 1 or 1/3 in this case. So what does this mean? This is not an even money bet, so does this mean that it only works for even money bets?


What if you are betting $1 for a 1/4 chance of winning $3? Then P1 is clearly 3/4, and G must be 4. So PL=3/4 + (1/4)PL^4. Now, 1 is clearly a solution to this equation, but it is not the only solution, so what is the actual value of PL here?


Also, I don't see a variable in your equation which signifies the bankroll size. Does this mean that the probability of losing $10 is the same as the probability of losing $1?

"But its also true that if you stay outside you will "eventually" get hit by a meteor. Does that mean that if you go outside you will ACTUALLY eventually get hit by a meteor?"


Yes it is true that if you live whatever million years you will be hit by a meteor, but going inside will not help, you will get hit there too, only your house will get destroyed with you. Better to go outside to be hit LOL.


On the more serious note, I guess you can have a lot of fun with a reasonable bankroll taking break even bets; at the end of the day it turns into a matching of bankrolls I guess as to who wins.

Yes, I guess that was my concern when I originally came up with this formala.

At that time I was going for sizeable keno jackpots with a bankroll that wasn't quite big enough, and boy did I turn over this risk of ruin formalae at the time.


I did learn one big thing though...

Your bankroll isn't your bankroll if you get scared after losing 1/2 of it.

I got scared, and now I stick to poker.

Not sure.


PL is the probability of losing your bet at any time in the future, and having to dig into your pocket for another bet when you run out of money.

On the other side (1-PL), which I call PW, is the probability of never having to dig into your pocket again.


PD doesn't quite fit in with these formula I don't think.

What if you are betting $1 for a 1/4 chance of winning $3? Then P1 is clearly 3/4, and G must be 4. So PL=3/4 + (1/4)PL^4. Now, 1 is clearly a solution to this equation, but it is not the only solution, so what is the actual value of PL here?


Good quesion.


I've factorized it myself to

(PL-1)(PL-1)(PL + 1 + sqrt(2)i)(PL + 1 - sqrt(2)i) = 0


I can't right now explain the imaginary roots.


I'll think about it some more and hopefully get back to you.

That's a good point. If you can't mentally take the swings, you need to play smaller or bet less.

But back to your original post, I think I realized what I wanted to say: you seem to have "proven" that something that CAN happen WILL eventually happen.


The general understanding that I hear from a lot of gamblers is that if you get the odds in your favour by a small amount, then it is ok to bet a lot.


This formula if you play with it, shows that...

at break even you still don't bet

slightly above break even, PL is still close to 1, and you would bet just a small amount...


It tells me to take no risk at break even, and from there, as Pl increases, gradually take more risk until you are on a sure thing, where you can then bet your whole bankroll as one bet.


I know that length of time to lose your money is not shown in the solution here, but I am sure that if I was a professional gambler, that the discipline I would strive for is to win surely, not just "my money is lasting a long time, and I'm gonna lose it one day after which I don't know what I'll do".

This is a good post. Yes I agree that you should increase your wager in proportion to your advantage for the reasons you gave: if you always bet your whole bankroll whenever you had a slight advantage (figuring to retire if you make $10mil) then you will go bust the vast majority of the time.


But even making small bets you will "eventually" go bust which you proved in your original post.


- Louie

My guess is that the imaginary roots are some sort of "damping factor", and I'm fairly sure that we can ignore them.

Lets face it, if you drew the graph of PL^4 - 4PL + 3, it only equals 0 at one place (PL=1). The imaginary roots may as well be non-existant.


Now the solution to this is identical to that of the example in my original post; and that is because they are both break even situations.

"Also, I don't see a variable in your equation which signifies the bankroll size. Does this mean that the probability of losing $10 is the same as the probability of losing $1? "


The equation is for a unit bet, and gives the probability of not digging into your pocket for another unit bet if you were to bet with the odds given for ever more.


If you bet $10 instead of $1 wherever you play, then the PL is your probability of losing that $10.


If your bankroll is $10, and you wish to bet in $1 units, then your probability of ruin is PL^10, because you have to lose your bet (and go to a subsequent $1) 10 times.


Note that a PL = 1 raised to the power 10, is still 1. With bets above break even, then your risk of ruin will obviously go right down as you increase your bankroll.

". So if G is 2, PL is {1 +- sqrt[1 - 4*(1-P1)*P1]}/(2*(1-P1)). That would mean that if P1 was 1/4, then PL = (1+-sqrt[1/4])/(3/2) = (2+-1)/3, which means that the probability is either 1 or 1/3 in this case. So what does this mean? "


I believe that this means that your probality of losing your unit bet is 1/3. Why the other solution of PL=1 still turns up I don't know for sure, it always does turn up though. When I first saw this, I took a saying from the Irish... it's "to be sure to be sure". Or "the answer is 1/3 if I exist at all"


However, yes, you have a 66% chance of never going below that unit bet for eternity (provided you keep taking that same risk).