% question

[Toby]

Supose you play on an European single-zero wheel.

You decide to play on 27 numbers each spin, 1 unit each flat bet  with -2.7% edge.

You have a 500 bankroll to invest.

Your goal is to double your BR or leave the casino with 0 dollars.

1)What is the chance to achieve the goal(reach 1000 units) or ruin(lose the 500) after 100, 200, 300 or 500(or more) spins?

2)Another scenario: you decide to split the BR and play on 2 wheels at the same time, 27 numbers with the same conditions.
What are the chance to win 500 between 2 wheels or to lose your entire BR? Same trials.

3)You play on 3 wheels at the same time, you divide your 500 BR in 3(166.6) and try to get 500 profit playing on the 3 wheels.

4)How do you calculate the odds against?

Thanks in advance
toby

[Bayes]

Interesting questions Toby. I can't give you an exact probability at the moment, but intuitively it seems that there should be no advantage in splitting your bankroll and playing multiple wheels. For example, in the case of 2 wheels (using 250 at each wheel), there are many possible ways in which you can double your 500 units between the 2 wheels. The following relationship holds:

Profit from wheel 1 + Profit from wheel 2 = 500 units

And there are obviously a lot of ways in which this equation can be satisfied (e.g. 100 units on wheel 1, 400 on wheel 2, 213 units on wheel 1, 287 units on wheel 2, etc), but all the possibilities are subject to the general "law" that the bigger the profit, the less likely you are to achieve it. So the net effect would be that although on some wheels you might do very well, (ie; make your 500 units) on some other wheel you may LOSE 500 units. Look at it from the casino's point of view; they have many players win or lose during an evening, but the net balance will be  in their favour, regardless of the gain of any particular punter.

[Toby]

LetÃ,´s reverse the question.

Supose you have a device or anything that allow you to play on 10 numbers having 4% edge.

Your BR is huge.

You decide to play 200 spins a day(or whatever) with a starting 500 BR, 1 unit each, total 10 units per spin. Total wagered 2000.

What is the chance to double our BR to reach 1000?

What is the chance to ruin before getting extra 500 units?

What happens if I have the opportunity to play on another wheel with the same conditions having another 500 BR?

We now try to get the +500 but with the effort of 2 wheels(250 each or 300 one and 200 the other).

What are the chances with 2 wheels at the same time? Better, no doubt.

The same scenario opening a 3rd wheel, with another 500 BR trying to reach +500 with the effort of 3 wheels(166 each).


I guess youÃ,´ve got what I want to know.

[Bayes]

There's no simple way to calculate the answers to these 'risk of ruin' type questions. The easiest way is to write computer simulations. I might get around to doing it... or not!   

[Toby]

IÃ,´ll make a try to calculate the game.

LetÃ,´s take a 300-spin-session playing 10 numbers with a 10/37 hit ratio.(regular player)

Total waggered is 300x10=3000/36=83.333, 84 hits is what you need to break even.

The average is 3000/37=81 hits

1SD for 10 numbers on a 300-spin session is 7.692 hits.

68% of the times you play 300 spins this way you hit from 73 to 89 times, so the best days you win 200 units and the worst you lose 370 units.

95% of the times you hit from 65 to 97 times, so you could win 500 or lose 640.

99,7% of the times you hit from 58 to 105 times, you could win 790 or lose 900

So, our BR can go from +800 to -1000 playing 10 units each time on a 300-spin session.

[Toby]

LetÃ,´s see another situation, you hit 10/34.5(4% edge)(an AP) playing 10 numbers in a 300-spin-session.

The break even point is the same 83.3 hits, the average now is 3000/34.5=87 hits instead of 81(regular player).

1SD for this situation is 8.25.

68% will hit from 79 to 95 times, winning 430 or losing 150.

95% will hit from 70 to 103 times, winning 700 or losing 470.

99.7% will hit from 62 to 112, winning 1040 or losing 750.

It changed a lot, what a difference.

[Toby]

Now, we have the information to try to know what would happen playing on several wheels at the same time.

I donÃ,´t know how to combine the info yet.

I supose we could calculate what would happen in a 600-spin-session(2 wheels) or 900 to 1200-spin-session(3 to 4 wheels at the same time).

600(2 wheels), average 174, break even 166. 1Sd for 10 numbers=11.66

68%= from 162 to 186

95%= from 151 to 198

99.7=from 140  to 210 hits


900(3 wheels), average 261, break even 250, 1sd for 10 numbers= 14.29

68%= 247 to 275

95%= 232 to 290

99.7%= 219 to 305


1200(4 wheels), average 348, break even 334, 1sd for 10 numbers=16.5

68%=332 to 365

95%=315 to 380

99.7=298 to 397

We can continue adding wheels but we have a real situation of what a AP can face on his work.

I donÃ,´t know how to calculate a day-byday outcome playing 1, 2, 3 or 4 wheels.

[Toby]

LetÃ,´s see another situation, you hit 10/34.5(4% edge)(an Advantage-play) playing 10 numbers in a 300-spin-session.

The break even point is the same 83.3 hits, the average now is 3000/34.5=87 hits instead of 81(regular player).

1SD for this situation is 8.25.

68% will hit from 79 to 95 times, winning 430 or losing 150.

95% will hit from 70 to 103 times, winning 700 or losing 470.

99.7% will hit from 62 to 112, winning 1040 or losing 750.

It changed a lot, what a difference.

To undestand the scenario, having +4% edge we  will be +14% to -5% at the 68% of the times. So it is 10 points more or 9 point less than 4% edge.

In the 95% of the times we have +23% to - 15.65. It is  19 points more or 19,6 points less.

In 99.7% bof the times you have 29 points more or 29 points less.

It seems that every SD ads 10% to the best or the worst situation.

[nottophammer]

Supose you play on an European single-zero wheel.

You decide to play on 27 numbers each spin, 1 unit each flat bet  with -2.7% edge.

You have a 500 bankroll to invest.

Your goal is to double your BR or leave the casino with 0 dollars.

1)What is the chance to achieve the goal(reach 1000 units) or ruin(lose the 500) after 100, 200, 300 or 500(or more) spins?

2)Another scenario: you decide to split the BR and play on 2 wheels at the same time, 27 numbers with the same conditions.
What are the chance to win 500 between 2 wheels or to lose your entire BR? Same trials.

3)You play on 3 wheels at the same time, you divide your 500 BR in 3(166.6) and try to get 500 profit playing on the 3 wheels.

4)How do you calculate the odds against?

Thanks in advance
toby

Found this by accident as looking for some help on a math question. But bayes comments

There's no simple way to calculate the answers to these 'risk of ruin' type questions. The easiest way is to write computer simulations. I might get around to doing it... or not!   

Well i'm no math person, but Toby my answer is you wont do it mate.

[nottophammer]

So the bank is 500 and double the Bank
500/27 = 18.51 potential stakes, a win +9, 500 profit 500/9= 55.55 wins.

Does this mean the 18.51 potential stakes, needs to win 3 times minimum, 18.51*3 = 55.53

[nottophammer]

This is from the document. 370 games up to today. So 10th is in making the 11th non-hit our stake for 27.

Total spins is 504 for the 370 games.
504*27= 13608
370*36= 13320

so -288 units  is that right, if so, dont do it Toby

[nottophammer]

What i was looking for is how many spins could a 13 pocket wide section of the wheel miss. I'll take the math answer, but would rather have an actual seen situation answer

[nottophammer]

lets assume 0-13 on the wheel thats 13 pockets. #2 hits, how long is that section likely to miss for?

[Toby]

There is not way to know it. You could find 40 spins gaps if you test 1.000 millions.
A dozen can sleep up to 40.
You mean a 13-number-section of a wheel. It should be the same as any 13 isolated numbers.

[nottophammer]

The above is for 364 games max taken for the remaining 13 is 11 spins, the 25 is the 25th non-hit, so the 13 #'s here are spread around the wheel, but like said to Pri, 9 remaining non-hit does not replicate 9 random.
So the section of 13 might not react the same as the remaining 13#'s that are spread. Better to have info based on actuals not the math answer to me