Roulette Math help please

Can someone please help with the math on this scenario;

Single zero roulette wheel, bank roll 510 units

Bet selection, start with any number and bet that number
plus 2 neighbours either side

Add another adjacent number each time a bet loses

Play this series of bets until a win occurs as follows;

1st bet, 5 numbers, 2 units on each
2nd bet, 6 numbers, 3 units on each
3rd bet, 7 numbers, 4 units on each
4th bet, 8 numbers, 5 units on each
5th bet, 9 numbers, 6 units on each
6th bet, 10 numbers, 7 units on each
7th bet, 11 numbers, 10 units on each
8th bet, 12 numbers, 15 units on each

Can someone please show me the math to calculate

1) Risk of ruin (ie: loose all 8 bets)
2) How many times I would expect to bust if I played
this whole scenario 50 times

I ask because I did my own calculations but they are not even
close to some actual spins I played at DB on a live wheel (5 sessions)

Many thanks for your help

Why would you want to play a rule based system
against random outcomes? It never works in the
long run, it can't work.

Quote: Pando

1st bet, 5 numbers, 2 units on each
2nd bet, 6 numbers, 3 units on each
3rd bet, 7 numbers, 4 units on each
4th bet, 8 numbers, 5 units on each
5th bet, 9 numbers, 6 units on each
6th bet, 10 numbers, 7 units on each
7th bet, 11 numbers, 10 units on each
8th bet, 12 numbers, 15 units on each

Probabilities of loss for that spin:
p1 = 32/37
p2 = 31/37
p3 = 30/37
p4 = 29/37
p5 = 28/37
p6 = 27/37
p7 = 26/37
p8 = 25/37

Quote:

Can someone please show me the math to calculate

1) Risk of ruin (ie: loose all 8 bets)

p1*p2*p3*p4*p5*p6*p7*p8 = 12.07%.

Quote:

2) How many times I would expect to bust if I played
this whole scenario 50 times

12.07% * 50 = 6 times.

Quote: Pando

Can someone please help with the math on this scenario;

Single zero roulette wheel, bank roll 510 units

Bet selection, start with any number and bet that number
plus 2 neighbours either side

Add another adjacent number each time a bet loses

Play this series of bets until a win occurs as follows;

1st bet, 5 numbers, 2 units on each
2nd bet, 6 numbers, 3 units on each
3rd bet, 7 numbers, 4 units on each
4th bet, 8 numbers, 5 units on each
5th bet, 9 numbers, 6 units on each
6th bet, 10 numbers, 7 units on each
7th bet, 11 numbers, 10 units on each
8th bet, 12 numbers, 15 units on each

Can someone please show me the math to calculate

1) Risk of ruin (ie: loose all 8 bets)
2) How many times I would expect to bust if I played
this whole scenario 50 times

I ask because I did my own calculations but they are not even
close to some actual spins I played at DB on a live wheel (5 sessions)

Many thanks for your help

Isn't the first bet : 3 numbers, one unit on each

Thankyou MangoJ
That's exactly what I wanted to know

Thank you Mango... the math is nice and I shall explore it more fully this evening.

I think the original poster with his rules about expanding to the neighbors should realize that epanding to those numbers that are most distant, rather than nearest, would work out to the same math.

This is the part of the discussion I am really interested in

I played 50 games on a live wheel (over 5 sessions) at DB

In those 50 games I busted only one time, much less than I should have
done according to the math calculations from Mango (who suggested 6 times)

I played neighbours every time

So was I just lucky or is playing neighbours better than random number selections

I do believe that some dealers are prone, from time to time, to throw long runs of
repeats or near repeats. I have seen it in every casino I have played in

Quote: Pando

In those 50 games I busted only one time, much less than I should have
done

The standard deviation of a single 12% event is 0.325. For 50 games you expect 6 events with a standard deviation of 2.3

If you only get 1 event, that is 2.6 standard deviations below expectation. Lucky yes, but nothing that would raise an eyebrow.

Quote: Pando

So was I just lucky or is playing neighbours better than random number selections

I have seen this term on a E-Roulette machine I thought it meant bet on your neighbors bets.. lol..... what dose it mean?

Roulette is the only game where you can win on a constant basis. You do that by switching your bets AFTER the ball began to spin. Is it ethical? That's another question.

In this context neighbours are numbers that sit beside each other on the wheel layout.

For example, on a single zero wheel, number "0" has 26 on one side and 32 on the other side.
They are the neighbours of "0"

Quote: Pando

..."0" has 26 on one side and 32 on the other side. They are the neighbours of "0"

I haven't been to the US for ages and never go to the roulette tables; presumably they also have racetracks for "Neighbour" bets like we do in the UK. The reason is that the "Neighbour" bet covers five numbers, i.e. two each side, and the chip(s) is placed on the racetrack and any winners, the 1/5th value chip is put straight up on the number by the dealer. Interestingly when I started playing in casinos there were no racetracks, although some casinos used plaques to accept various bets.

Quote: Dicenor33

Roulette is the only game where you can win on a constant basis. You do that by switching your bets AFTER the ball began to spin. Is it ethical? That's another question.

Hi, I was wondering if I could ask a few questions regarding your post?

1) Are you talking live roulette or an E-game?
2) Why would changing your bets after the ball has been spun be either unethicala or illegal? As long as you have placed or changed all your bets before the dealer calls "no more bets" you are free to do whatever you please.
3) Did you mean place bets AFTER the ball has dropped?

Roulette is a game of independent events and if you know and understand probability you know that each new independent event has a 1/37 (single zero wheel) probability of occuring so unless you are using wheel bias or flat out cheating, I don't 100% understand your post.

Don't take it personally, for the most part I don't understand my own posts either :)