# Roulette Mathematics, relationship advantage and bankroll

Roulette Mathematics, relationship advantage and bankroll

Let’s say a roulette player plays a wheel where he has 10% advantage.
He wants to play 100 spins each time covering 7 numbers section with \$10 on each number, \$70 per spin or in total he would place on the table \$70 x 100 = 7k value in casino chips.

Based on 10% advantage in average after 100 spins he should win \$700.
Of course that’s an average and his final winnings can be different.

I made some calculations to show what possibilities of different outcomes.

For example the player has chance;

• < 0.1% chance to lose all bankroll
• 5% to win \$2800-\$3500 or more
• 10% to win \$2100-\$3000
• 15% to win \$1400-\$2100
• 20% to win 700-\$1400 • 20 to win \$0-\$700
• 15% to lose \$0-700 • 10 to lose \$700-\$1500
• 5% to lose \$1400-\$2100

Simply the player has 70% chances to win or 30% chances to lose some of his bankroll.
If the advantage on the wheel was 20% then his chances to win increase to 85%.

That’s if the player has bankroll to play all 100 spins and he plays 1% of his bankroll each spin, but what if he doesn’t?

What if the player walks in the casino with \$1400, enough to play 20 spins without single win? He would play each spin with 5% of his bankroll.

For example he would have a chance of about 3% to play first 20 spins and have no win.
That would take him out of the game. Same as if for 25 spins he has just one win or for 30 spins only 2 wins… etc.

That would take him out of the game. Same as if for 25 spins he has just one win or for 30 spins only 2 wins… etc. Basically at any moment during his 100 spin game if he stays 4 wins shorter he would lose \$1400 bankroll and would have to leave the table. Adding all possibilities that could lead to minus 4 wins during 100 spins game definitely would increase 3% chance. I am not sure if I got it right but it comes to 7.2 % chance that it will happen. Bottom line is if playing even with an advantage the relationship in between bankroll, amount of played numbers and advantage is important.

The mathematical way of betting is very important, because some losses we can accept as very unlucky ( say not hit by one number ) and that do some psyhological pressure on player. When i do similar calculations I usually look too more small distance , because even 10 miss in a row can unsettle player, after what he can begin to do mistakes. So if look to distance of 10 spins then max value after, can be 3,600-700=2900 and minimal -700. When we cover 7 numbers mathematically we have 7/37 chances to hit, if our edge is 10% then we must hit (7/37)*1,127=7.89. Now lets say that our edge is a bit higher and we hit 8 times from 37. Here we can count probability not to hit all 10 times (29/37)^10=8.7%.

So we will meet such case in every 11-12 attempts.This number must remember who plays about 7 numbers per bet. But now lets count what will be if we bet 15 numbers and say 2 times less advantage. So we bet 10 times 15 with chances to hit in every say 16/37. Now not hit all 10 times we will have chances (21/37)^10=0.35% or about 1 time from 300 attempts. Our possible values after 10 spins are 3600-1500=2100 and -1500. So worst possible result is only two times higher and chance to get it is about 30 times less. Best result is only by 25% less. So all these numbers is worth to remember for some players, especially for these which play with not enough bankroll.

I would like to calculate exact percentage of possibilities which would cause loss of bankroll.

By the way, I’ll be away for a week.

Very GOOD topic!!! Txs!

...at any moment during his 100 spin game if he stays 4 wins shorter he would lose \$1400

but 4 spins each time covering 7 numbers section with \$10 on each number, \$70 per spin or in total he would place on the table and lose \$70 x 4 = \$280 value in casino chips but not \$1400 or it’s me that do not understand smth. please?

for this reason I would like to play only 1 number each spin and to put the maximum bet and like this I have 36 spins to make 1 win and go out with any winnings depends of the edge I have on the wheel, but I do not know if that method of play is good or so much discovered or some another factors that I do not know and to prevent?

for this reason I would like to play only 1 number each spin and to put the maximum bet and like this I have 36 spins to make 1 win and go out with any winnings
If you play only one number then your play is more similar to gambling - you can easy wait 100-200 spins with any hits and that will not be nothing strange with your say 10% edge. If you play table max even on one number you can loss quite many. Such is not good way.

If you have 1:33 hit rate instead of 1 in 37.

Then in 99 spins your average wins on a single number is 99 / 33 = 3
If you play 7 numbers then it is 3 x 7 = 21
But what if somewhere from the start you miss 4 wins?
4 wins is 4 x 35 x \$10 = \$1400

If that’s your bankroll you would have no money to play.
It is expected that you will win 21 times but it can be more or less.
If you play a single number you are more exposed to risk.

While to have no hits playing 7 numbers is almost impossible, if you play just one number for 100 spins there is 2.5% chance that it may happen and there is a 44 % chance to lose some money.

Playing a single number for 100 spins most of the time you wouldn’t see any difference if you have 10% advantage or not.

But what if somewhere from the start you miss 4 wins? 4 wins is 4 x 35 x \$10 = \$1400
Here i not exact understand, it seems that you talk about theese wins with must be in between every 33 spin ? And instead 1 win and 32 loss - you have 33 loss and so 4 times ? Realy i very much researched what is efect of play in diferent amount of numbers and found that here are some optimal amounts of numbers, and better to bet that optimal amount and have less disperssion, than to bet less with slightly even better edge but also with higer disperssion. But about all i can write after week because i am also away.

I haven’t say that my edge is 10%, simply I don’t know what kind of edge will give me in one or another ocasion the FF, but what I do is to search in the graph “predict number-winning number” which is the position with more winnings and play only that number, upper I guess you both speak that if we just do some bets on the table that is better play sector than one number but what about if the graph I say is so strong concentrated in only one position and just play this number because about the history it seems like coming very often?! whatever about some my calculations I discovered that maybe at all is very very good maybe the best and the optimal one decision to play 5 numbers sector - the predicted number and its neighb.

In this case is the optimal win/inversion scale maximum that I have make for myself reached by one cuadratic eq. on some examples of edge of the prediction on that kind of graphs. maybe I wrong smth. but pls what is according you the optimal and the best amounth of number to bet each spin? just for simplify I like to put only 1 chip and still have some fair to put any more, haahhaha:-)) it was shocked when it says “4” and just I put on the table 4 and go out with 1800€, just incredible in one moment thought that they will not pay it, hahahaha:-)) but also depends of the table, the next nigt just betting smallest amounths because there weren’t any so so tilted table and just play to one that seems more or less good but not like the previous night I say:-))

Nice info Forester.
How did you calculate all these scenarios? I´d like to learn more math

bebediktus: when you know how to get the edge the only thing that matters is math.

Normal Distribution Bell Curve for roulette math

Calculate standard deviation, calculate average for 100 spins with 1/33 hits, and 1/36 average (needed to stay equal), check difference, or check difference to any win/lose ratio, then use this chart, or create one with XL. or use Normal Distribution Calculator at;
http://easycalculation.com/statistics/normal-distribution.php to see how many standard deviations it is from 1/33.
I can show you an example when i come back next week.

SD for 100 spins 1/33 hits 7 numbers played is 4.08 , an average is 21 hits so to don’t get a single hit it would be 21/4= , 5 SD , from the graph you can see that even 4 SD is close to impossible.

SD for 100 spins 1/33 hits 1 number played is 1.5 , an average is 3 hits so to don’t get a single hit it would be 3/1.5= , 2 SD , from the graph you can see that even 2 SD is 2.5% chance to lose.

• 68% of values are within 1 standard deviation of the mean
• 95% are within 2 standard deviations
• 99.7% are within 3 standard deviations

That’s why when you have 3SD number hits on biased wheel it is almost certain (99.7%) it is not a coincidence.

LOL!

this graph I haven’t seen from the exam of the buildings materials in the university, it was the distribution of the sand and the rubble of different sizes of sieve, hehehe:-)) really do not understand all would you please explain it for stupid people like me, pls:-)

SD of what?? predicted number-winning number or…

why 1/33 hits?

average 21 hits??

what about 3 and 5 numbers play?

why 7 numbers is 4.08 and 1 number 1.5?

comulative percent graph??

2.5% chance to lose is a lot of?

FINALLY CONCLUSION: WHICH IS THE BEST AMOUNTH TO BET NUMBERS PER SPIN??? or another conclusions?

thanks, regards!

Let´s start from the beginning:

Let’s say a roulette player plays a wheel where he has 10% advantage.

The steps to know the accurate edge aren´t easy, let alone when you guess your edge is lower than 3%.
It takes many probability test to determine it, or taking 15 to 20k trials.

A 10% edge Wheel is not well understood. You can have 1 to 20 number sections. +10% on a single number yields less than a 7-number -section or a 18 number-section. Fluctuations aren´t the same and you win easier.

Gambler´s ruin situations become more comfortable for an AP.

Forester, we must separate the SD measure wether we have already played on the sections or we have just recieved the data.

In the last example: 3 to 4 sd on a 7-numbers-section is not strong to be certain

Things change when you actually play the section and get +3 to 4sd, the next session we could use these numbers you said.

We should add in how many trials it got the sd. The later it get large SD, the worse.

The 68-95-99.7 rule is applied for spins played on a pre-selected section

A nice topic to talk.

BR

Nice post, Forester and good follow up by toby!

Toby, care to expand on this? For example we could include the strength of math to look for to suggest an edge exists (discounting experiance on playing conditions and certain wheels) and for which we will base a priori knowledge as our first step maybe, to try eliminate wasted time to a minimum. So I am curious as to which kind of data you are happy with, eg SD for various sector sizes, times amount of trails before you decide to discount a wheel or proceed to the next step?

Just curios as I have been looking at this abit as I crunch the data and see much room for improvement, over just relying on the SD as we see it at first glance as you have already mentioned.

There´s no study about this subject, just some searchs on equivalence testing and prediction stuff. Frey, Klotz, Ethier and other writhers.
A hard realm to catch.

[quote=“x1x0, post:11, topic:948”]LOL!

this graph I haven’t seen from the exam of the buildings materials in the university, it was the distribution of the sand and the rubble of different sizes of sieve, hehehe:-)) really do not understand all would you please explain it for stupid people like me, pls:-)

SD of what?? predicted number-winning number or…

why 1/33 hits?

average 21 hits??

what about 3 and 5 numbers play?

why 7 numbers is 4.08 and 1 number 1.5?

comulative percent graph??

2.5% chance to lose is a lot of?

FINALLY CONCLUSION: WHICH IS THE BEST AMOUNTH TO BET NUMBERS PER SPIN??? or another conclusions?

thanks, regards![/quote]

1/33 it is only an example where in average you get one hit every 33 spins , it is close to 10% advantage.
If you play 99 spins then based on winnings every 33 spins you would win 3 times if you play single number and 3 x 7 = 21 if you play 7 numbers.
Why 7 numbers is 4.08 is because that is what it is. Standard deviation is calculated as a square root of mean multiplied with odds 7/33 and 1- (1/33)and multiplied with amount of spins. Where 1-(1/33) is acualy a chance to not get a hit.

FINALLY CONCLUSION: WHICH IS THE BEST AMOUNT TO BET NUMBERS PER SPIN???

It depends to what do you want. If playing a single number your advantage will be the highest, but you have high chance to have no hits for some time. I may not agree that playing the widest possible sector is the best. For example playing a single number 99 spins you may have no hits at all and you going to lose few thousands, but next time instead of average of 3 hits you may have 5 , because your bets are high per number it makes few thousands more then what you should win. So the deviation of your outcome based on season you play is greater. If your pocket can handle it on the end after lets say 990 spins you should be better of then someone who played 15 pocket wide sector.

[quote=“toby, post:14, topic:948”]There´s no study about this subject, just some searchs on equivalence testing and prediction stuff. Frey, Klotz, Ethier and other writhers.
A hard realm to catch.[/quote]

Actually I was talking in terms of building a solid hypothesis from the data collected in the quickest way possible to confirm a potential edge. I am sure an experianced AP will be more insighful here adding in to the equation a physical wheel, then a phd published study on random numbers, but never mind.

FINALLY CONCLUSION: WHICH IS THE BEST AMOUNTH TO BET NUMBERS PER SPIN??? or another conclusions?

You have to make statistics how far the fallen number in pockets is away from your
predicted number.
That sector of numbers where you have the most hits , that sector you should bet, there you
Mosstime 5, 7, 9 numbers.
And what Scott and Basieux say : the better your predictions , the more numbers you could bet.

E.H.

yep, txs evrb.

maybe there is smth. like a fear when thinking that if I bet more amount of numbers have to take a win often. if we can take a look the differences are so drastic: if we bet only 1 number then we have a chanse to try it 36 times to fail, that advantage with FF it has to be so bad setted up or smb. who even really doesn’t know what is doing with it or really so bad wheel(but in this case we have to recognise it and go away from that wheel) to not to reach such a result. then let’s see what is happened when 2 numbers play: we have to reach a win 1 time in 18 spins - THE DIFERENCE IN THE TRIALS IS SO DRASTIC COMPARED WITH THAT OF WHAT 2 NUMBERS BET COULD HELPS US MORE TO WIN THAN 1!! I think that really our chanse to win it not so elevated if instead of 1 number we bet on 2 numbers, really the bet zone is not grow so much, and if take a look on the next table:

BET ON NUMBER/S and WIN THAT HAVE TO REACH IN SPINS to not to have a lose

1 number - 1 in 36
2 numbers - 1 in 18
3 numbers - 1 in 12
4 numbers - 1 in 9
5 numbers - 1 in 7
6 numbers - 1 in 6
7 numbers - 1 in 5
8 numbers - 1 in 4/5
9 numbers - 1 in 4

and so on! The interesting part here is that from 1number bet to 2numbers bet - there is a big gap between 1in36 to 1in18 - DOUBLE and lose in total 18spins-so big amount of trials only putting on just 1 chip more!!!
when from 5numbers bet to 9numbers bet from 1in7 goes to 1in4, putting on more DOUBLE of chips we lose ONLY 3 trials!

all I mean with this we have to find the optimal play we have to do from this table but also it depends from the % of advantage we have with FF which is in relation of the tilt of the wheel and another outside factors and most of the “predicted number - winning number” graph, I think then there will be several cases that are most often seemed than another and for example some result like instructions simple to follow to reach all this difficult explinations like: if the tilt of the wheel and the DD are … then most likely play XX numbers bet per spin. - or smth. in this sense(I mean not exactly but expect you could follow my idea)

smb. could give some instructions of this or this is not so easy, or smth. in my theory is wrong:-))

bebediktus: when you know how to get the edge the only thing that matters is math.
Matter is not in math, or say not main in math. Main thing is to know which numbers have more chances to cach ball. Answer to that can give you physick if you play some kind of VB, or mathematick or statistick if you play some kind of bias. But even if you play VB and know prediction as zone which is more favorable than other zone you must still decide which part of favorable zone to cover. Usually you not have enough time to cover ideally and here choosing diferent amount of numbers which you cover can give you diferent results.

As I say before - if you choose to cover very small amount of numbers - you can get quite long serie non hits simply because you are unlucky. Of course all can be oposite and you will have hits with theese small amount of numbers, but this part is not worth to disscus because it not do damage for player, but part with long serrie with no hits is more interesting at least for me. And in this part is very important how changes amount of averidge amount non hits if we increase amount of numbers which we cover.

Top quality post. I’ve always worked off the 1% rule so it’s good to see it represented mathematically.