chance for the bank to lose

[Toby]

How do you calculate the chance to lose when you have 2,7% edge in a single zero european wheel?

How do you build a formulae to calculate any of this events?

For instance I'd like to know the way to calculate the chance that the bank loses in a 100 spin session, in a 200 spin one or in a 500 one.

Does it matter if there are several players playing all over the layout or if there is only one playing 1 or 2 numbers to calculate the bank's result?

[Bayes]

Hello Toby,

I assume that what you're asking for is the so-called 'risk of ruin'; the chance that you will lose your entire bank. This depends on the size of your bankroll relative to the bankroll of your 'opponent'. From the casino's point of view, this means that their risk of being wiped out is essentially nil (since their bankroll is so much larger than any individual player). This would apply even if their advantage was zero, the fact that they have 2.7% only lessens the chance that they will go broke.

This is another reason why casinos set house limits. If there was no maximum stake you could win with a martingale (if you were Bill Gates). 

Here's a simulation you can play around with:
link:://math.ucsd.edu/~anistat/gamblers_ruin.html

[VLS]

Thanks for your explanation Bayes.

[Toby]

Thanks for your answer.

I mean what is the chance to lose a session if for instance there are 2 players against the bank.

The bank has 500 units and players have 200 each.

Player#1 plays 6 numbers, from #9 to #14, 1 unit each flatbetting.

Player #2 playes from #15 to #36, 22 numbers 1 unit each, flatbetting.

What happens after 100, 200, 300 or 1000 spins?

What % of loss has the bank and every player in those sessions?

[Bayes]

For any particular scenario you could work out the average result, but I don't think this is what you're looking for because it wouldn't take into account the variance (and therefore the risk of ruin).

To use your example, in the case of 100 spins:

player #1 covered 6 numbers âÅ"• 1 unit/number âÅ"• 100 spins = 600 units.
player #2 covered 22 numbers âÅ"• 1 unit/number âÅ"• 100 spins = 2200 units.

Profit for the bank would then be, on average:

0.027 âÅ"• 2800 = 75.6 units, so the final bank after 100 spins would be 575.6 units.

Player #1 would have lost 0.027 âÅ"• 600 = 16.2 units, leaving 183.8 units.
Player #2 would have lost 0.027 âÅ"• 2200 = 59.5 units, leaving 140.6 units.

Rinse and repeat for 200, 300, 1000 spins.

You could work out a formula for this calculation, but like I said, it doesn't give you the whole picture. It doesn't take into account that the bank might be lost in the event of a run against it. What you really want to know is how often this might happen for various parameters. The easiest solution would be to write a simulation in which you could enter the size of bank, no. of players and their bets, length of session etc and then run multiple sessions to find the chance of losing a session (both for the bank and the player(s).)

This would be an interesting little project. Give me a few days and I'll throw some code together.

Stay tuned!

[Toby]

"You could work out a formula for this calculation, but like I said, it doesn't give you the whole picture.  It doesn't take into account that the bank might be lost in the event of a run against it"

This is what I want.

I know tha math involved here about the 2. 7% edge of the total waged in bankÃ,´s favour.

But, we know the long run.

Playing against those 2 players can give different results after 50 sessions of 100.

In average, how many 100 spin sessions does the bank wins and loses after 100 trials?(10k trials).

And, what about 200 spin sessions?

Do you get what I want?

[Bayes]

I know what you want, but I'm not aware of any formulas which will take into account all the factors you mention, which is why a computer simulation is the best (if not the only) option.

[Toby]

Ok, thank you.

[Ekis]

Don't know if this answer will help.   .   

If one player is flatbetting you can use the binomial distribution to calculate the probabillity to lose or win.    The formula is not very complicated, but I think it is easiest to use excel.   



An Example in excel:
One player is flatbetting an even chance for 100 spins.   
If he wins 49 or less spins he lose, if he wins 50 spins he breaks even and if he wins 51 or more spins he wins the session.   

In order to calculate the probabillity for the player to lose the session you write
=Binomdist(49;100;18/37;TRUE)
The first number,49, is the number of wins.   
the second number,100, is number of trials.   
the third number, 18/37, is the probabillity.   
and TRUE means that we want the cumulative value for "at most" 49

The answer will be 0,567916304 which is the probability that the player will win 49 or less spins in a 100 spin session => the player lose


If player win 50 spin he will break even.   
=Binomdist(50;100;18/37;FALSE)
We choose FALSE because we want the exact probability for 50 wins and not the cumulative value

The answer is 0,076733818 which is the probability to break even

the probabillity for the player to win is 1-0,5679-0,0767=0.   3554
(=1-BINOMdist(50;100;18/37;TRUE)  = 0,355349879)


Flatbetting even chance 1000 spins
P(x=<499)=  0,79486466  (Player lose)
P(x=500)=   0,01750475  (even)
P(x=>501)=  0,18763058  (Player win)

[Toby]

Thank you very much EKIS.

I tested the formulae and found some outcomes.

IÃ,´d like to apply the formula to this:

a 9-number-sector that hits 0. 26%(9/34. 5)

to test what happens to this situation IÃ,´d try this:

=binomdist(50;200;0. 26;true) to see the % n a 200 spin session

the % is 0. 398 to hit at least 50 times in 200 spins, is thst ok?

what is the % to hit 51 times in 200 spins?0. 4618? or 0. 6345%

What happens in more or less spins?

IÃ,´d like to know how to break even at least with one hit with this formulae and its % in a couple of spins.

thanks in advance

[Ekis]

If you are betting 9 numbers the chance of hitting in one spin will be 9/37=0. 24
(If I understand you correctly.   Where does the 34,5 come from?)

=binomdist(50;200;9/37;true) gives the answer 0,6247.  This is the probability to hit 50times or less (50,49,48,. . . ,1,0)

The probability to hit exact 51 times is =binomdist(51;200;9/37;FALSE)  which gives the answer 0,05998
The probability to hit 51 times or more is 1-0,6247 = 0. 3753


The formula is btw:

P(x)=(n!/(x!*(n-x)!))*(p^x)*(1-p)^(n-x)

In the formula above x=wins n=spins, p is probability for one hit.  This formula give the probability for exact x hit in n spins.

Not sure if I answered your question. .

[Toby]

Yes, you answered my question.

I wanted to know in the example the chance to hit 51 or more times in 200 spins.

We take this "=binomdist(50;200;9/37;true) gives the answer 0,6247" and substract from 1.

0.  3753 is the % to hit 51 or more time in 200 spins.

Other scenario:

Suppose you are  AP(VB, DS, Computers, Biased wheels) and have the skill to have an edge.

So, the numbers you play do not hit 9/37, they hit 9/34.

How do you build the formula for a 200 spin session then?.

=binomdist(51;200;9/34;FALSE)=0,061530055
1-0,061530055=0. 93847

Does it mean that you have over 93% probability to hit 51 or more time in 200 spins when your numbers hit 9/34?

[Ekis]

No, you have to use true instead of false.   6.  15% is the probability of exact 51hit. 

And you have to use 1-the probability of 50 or less hit. 

=1-binomdist(50;200;9/34;TRUE)=0,6482

So you have 64.  82% chance to hit 51 or more  in 200 spins when your numbers hit 9/34. 

[Toby]

IÃ,´ve got 58. 66% to hit 51 or more times, sorry.  =1-binomdist(51;200;9/34;true)=0,586646146

64. 51% for 76 hits in 300 trials
68. 78% for 101 hit in 400 trials
82. 79% for 250 hits in 1000 trials


More trials the merrier.