Hi,

I’ve been an avid lurker for a while here and have found this forum very informative. I have a question for you stats wizards - apologies in advance if I’ve misunderstood/misapplied any concepts in the below; look forward to being educated

I’d like to propose the following hypothetical betting strategy for single 0 roulette (I understand that roulette is a –EV game, this is not a “I have a winning system” post!)

1. Starting bankroll of x flat bets (let’s say x=36 for exemplary purposes).

2. Flat bet numbers straight up until bankroll is >=2x bets. Leave if busto.

3. If at any point bankroll is >=2x bets, pocket the change from 2x and repeat, with the betting unit doubled in size.

4. Repeat 1-3 until busto or bankroll is y times its starting size (let’s say y=100).

My questions are as follows:

1. How do I calculate risk of ruin?

2. Is chance of success (leaving with y times starting bankroll) simply 1-RoR, or am I misunderstanding how RoR works? If not, how do I calculate chance of success?

3. What is the relationship between x and RoR (i.e. if I start with more bets is my RoR higher because I’m playing a –EV game for a larger number of trials, or lower because I can have more “goes” before busto)? Is the relationship between x and RoR linear?

4. (Related to point 3 if point 2 is correct) - Is it possible to solve x for a desired chance of success for a given y - e.g. to have a 0.5% chance of leaving with 100 times your bankroll you need to start with 1,000 bets.

5. If this strategy is applied to a 0EV game, do you just get the “fair” odds - i.e. if numbers paid 37:1, would I simply have a 1% chance of leaving with 100x the starting br using this or any other betting strategy, or is it more complex than that?

6. Are these questions easier to answer analytically or through simulation?

Many thanks to anyone who takes the time to reply to this!

First, just realize that a session is a made-up dividing line and that your play is all continuous for your lifetime so what happens in one trip through the casino door doesn't mean anything. Over time you are going to average losing the house edge times the total wagers you have made. No matter what rules you have for a stop-loss or stop-win in a "session".

That said, you can construct rules that allow you to have a majority of winning sessions and an occasional big loss that wipes them out, or vice versa. But the overall result will still converge to the house edge. Guaranteed. But you can possibly increase your enjoyment based on your goals.

If you're still interested in the questions about your "system" I'm sure others will help you with the math. I'm on a mobile right now. But your RoR in roulette is 100% with any system if you play long enough.

I already said it wasn't a "system", and I understand that it's a -EV game, as stated in OP.

Or are you saying that RoR is constant for a given game regardless of the stop-win and betting strategy employed? That seems counterintuitive.

^^ eg. Let's say I start with $100. It seems to me that my RoR is higher betting it on a number straight up than betting it on black, no?

Would it make any difference to the quality of responses if I changed OP to a hypothetical game that pays 38:1 on a 37-number wheel? The questions would remain the same...

You should post the math questions in the Probability thread here:
http://forumserver.twoplustwo.com/25/probability/

ROR is 100% if EV < 0

Eventually yes. But he wants to know the chance of busto before making some multiple of the starting stake, with certain betting rules. So that is less than 100%. But not much less. In his example of 100x it is essentially 100% RoR. Which is why I didn't bother with any calculations.

okay, I am fine with thats
a number
okay too
I like 2 you like 100
same effort to calculate (using a program and not pencil and paper unless you want to be a BruceZ - that is a compliment)
I do not know how you calc RoR
and I'm a thinking you do not know how...
For simple win/lose type bets that pay even money I use the Gambler's Ruin Formula
(Been around since the 1600s)

for many others, with multiple bets too, I can use recursion and Matrix Algebra in Excel most times.
When I feel lazy, I fire up a computer program like R.
Simulations too in WinStats and R (both free programs too)

I agree
with simple win/lose type bets

for even money bets that would be true but when the variance (standard deviation) goes up the curve gets kind of squished up (BruceZ would know the technical name) and I could show photos too
not always. again depends on expected value and standard deviation in my opinion

I would say yes using the the normal distribution, I have not done just that btw,
but not over every type of bet and number of trials too, as most times people do not consider the time involved to hit a win target and it could take a long time for that event to occur.

Fair odds for your example Roulette game = 36 to 1 (36 ways to LOSE and 1 way to WIN)
so with that I agree. Fair odds games are easy to work with in my opinion.
That goes back to the 1600s too. a popular time in History
just take the start bankroll and divide by the target bankroll


depends.
I do both (to me both are easy)
just to see both and catch any errors I may have made, and I do make errors at times

Thank you for asking!

my calculations using 2 programs and a fast simulation:
the chance to double a 36 unit bankroll betting 1 number at 35:1 payoff = 40.527972% = x
average number of spins for a successful double= 44.04992127 (you did not ask this)
RoR = 1-x

you now have $72 and bet $2 per bet you still have that same bankroll size of 36 units (72/2)

But one could do better by just betting all 36 on a color and double with prob = 18/37 (48.648649%)
This is called Bold Play by many.

Using Bold play with a small bankroll and a small winning probability sometimes does not work out too well I must say

Bold Play = bet just what is needed to hit your target or everything, in this case we can easily pass our target and that actually lowers our chances of hitting that target. Our 1st bet has to be $2 as winning just $35 does not exactly double a $36 start bank
that prob = [1] 0.3893195

(and I get that in 2 different programs - R and Excel, BruceZ and me and the chimpanzee agree
That one day soon I'll be a celebrity!
Get it, get it, get it, get it.) Thanks Ray

How about a 72 unit bankroll
the chance to double a 72 unit bankroll betting 1 number at 35:1 payoff = 43.31336% = x
average number of spins for a successful double= 155.5525704 (you did not ask for this)
RoR = 1-x

Hey!!!
easier to double more units

How about a 100 unit bankroll
the chance to double a 100 unit bankroll betting 1 number at 35:1 payoff = 43.227961% = x
average number of spins for a successful double= 295.3630321(still going up)
RoR = 1-x

Hey!!
now less that 72 units and on the way down as the units increase.

Time for a fun weekend, I can post some photos of what I have talked about if you would like or not, I am a big girl, and fat too.
when is this baby going to pop!

Sally

He asked for the chance to get to 100x the starting bankroll before going broke. Not double.

You can run the ROR calculations with opposite signs.

The chances of reaching 100 units with EV < 0 is the same as the chances of reaching -100 units with |EV| > 0.

I appreciate your responses itt, but it seems you're basically saying "I won't do the maths because in the example you gave, you set y to 100". Neither you nor anyone else on this forum owe me a response, but that reasoning seems somewhat disingenuous; fine, set y to 0.8 or 1.2 or whatever else seems reasonable. The question was really for the formula/methodology that relates the variables set out in the OP.

I appreciate that I may have posted the question in the wrong forum. Mods - if this belongs in the probability forum as suggested by NewOldGuy, please move it as you see fit. Thank you!

Sally, thank you for your response. There are a number of things I don't understand in this post and would like to query further; however, I can't make a detailed post right now, but will try to do so tomorrow. I appreciate the in-depth reply.

Ok, I don't really care about the signs. How do I perform the RoR calculations for the strategy described in OP? Can we take game EV as another variable (which can be positive or negative)?

^ Or 0!

ROR is 100% and EV is not a variable.

To be honest, I think you're fishing around for someone to tell you this isn't a horrible idea and then you can latch on to that 1 opinion and ignore the 99 who say it's a horrible idea.

Your patronising attitude is totally unjustified. I'm pretty sure "RoR" doesn't mean what you think it means; e.g. http://wizardofodds.com/games/blackjack/appendix/12/ (just a quick Google search).

Yeah you're probably right

I think we can use the unfair coin toss by setting it up so that you have a 1/37 chance to win, you start with 35 units and the house starts with 1 unit. If you bankrupt the house then you have doubled your bet.

Using this formula:
http://en.wikipedia.org/wiki/Gambler%27s_ruin

Gives a 2.8% chance to bust the house and double your bet for the first round. I'm not sure how to treat doubling the bet or if this is a correct modification to the formula. I could probably simulate it faster than figuring it out!

the OP wants to double the start bank and pocket the rest and keep doubling to X

i showed him how to start and was waiting for his return,
but really do not want to answer his question asked
as what he wants is a
classic cheating method by some current team members in casino team play.

yep is

I catch team cheats with math, it is fun for me,
team catches them with video red-handed-
for a few teams and think the OP may be or want to be one, just my opinion

============================
the GR formula does not work for these type of problems, it considers even money bets
a formula could be done but i see it could be very unstable too

a Markov chain is easy
i use excel, but some may not have that
so here is a homework code, do not laugh, that works in R

can be done online too here: try it out
Execute R Script Online
for your example of doubling a 35 unit bank
the results for all starting bankrolls

for you
35: 0.40830975

see how close your sim gets
and have fun!
Sally

The Markov chain was an interesting read. I decided just to simulate it. Starting with a 35 unit bank I got a 41.1% chance to pocket some amount, with 12637 units pocketed once out of 100,000 times. Lucky guy.

could be a lucky girl

so you played until ruin.

I only went until at least double and started over

OP wants to double the start bank, pocket the overage and keep going

i get for your 35 to 70 example
this end bank dist for the success rate of 0.408309753
(avg end bank = 82.48979331)
and looks like this


I simmed it too, but actually enjoy the matrix calculation experience more
I did this one in Excel this time

all in good fun
Sally

EDIT -- Pretty much all of below is wrong because I thought OP was betting a color. My formula doesn't apply when you're betting a number.

Given #4, what does #3 mean? If you're just gonna re-wager your winnings, then you haven't really "pocketed" them. So I'm interpreting this the same way as NewOldGuy, that you're just playing until 100x BR or bust.

Let b = bankroll in units of your flat wager
Let z = quitting point in units of flat wager
Let p = chance of winning
Let q = 1-p

Then success rate = [(q/p)^b - 1] / [(q/p)^z - 1]
Or ROR = 1 - that, or [(p/q)^(z-b) - 1] / [(p/q)^z - 1]
(This answers #2.)
Since your target (what I called z) is a constant factor of x (what I called b), your target increases proportionally to x. You are correct that ROR increases as x does, provided that z increases proportionally to x. But no, the increase in ROR is not linear (as you can see by my formula).
Yes, by rearrangement of my formula and taking logs. WolframAlpha can probably do it easily.
It's as simple as that, but only for the case of 0EV. For 0EV, my formula becomes an indeterminate form 0/0, so a different formula is required: P(success) = b/z.
Depends. For me, they used to be easier by simulation because it only takes noob coding skills to write a sim, but not-so-noob math skills to solve this analytically. Now that I already solved it and know the formula, it's much easier analytically
You seem to have a much different interpretation of OP's goal. If I plug 35 and 70 into my formula, and set z = 2b (instead of 100b), I get a success rate of about 0.131. (If the goal is 100b, then as NOG said, it will be incredibly close to 0.)