### Author Topic: When is the risk of ruin greater than the HA!  (Read 3499 times)

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#### Jesper

##### When is the risk of ruin greater than the HA!
« on: June 05, 2016, 08:28:53 AM »
All must agree it is better chance to survive a session if the HA is zero than it is 2.7 or greater. But we have the variance, and the risk of losing the bankroll is larger if we have a short bankroll.

It is a lot of wheels in the world, but few offering zero HA.

If somebody has 10 Euro available for playing, and he has to chose between tables, one offering 1 cent min and has an HA of 2.7%, the other table has no HA but a minbet of 5 cent.

Could the fact one has 1000 units overcome the HA at 2.7%  compare to have no HA but only have 200 units, make it a better choice play at the table with the 2.7% HA.

Would be nice if we could calculate the limits when the bankroll size make one of the tables a better choice.

Short bank is a common losing factor.

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#### UK-21

##### Re: When is the risk of ruin greater than the HA!
« Reply #1 on: June 05, 2016, 10:43:36 AM »
Two considerations for the mix, and missing from the above, are how long (how many spins?) one intends to play for and what methodology, "system", range of numbers to be covered each spin will be used.

What three standard deviations south of the EV represents will vary with the above, and therefore the amount of bankroll needed to keep from tapping out if unlucky enough for results to be in the southerly tail of the bell curve will vary as well.

If you bet in a way that negates all variance, and intend playing for 30 spins, then the bankroll required to not tap out will be 66 units (37+(30-1)); 37 units bet per spin, expect to lose 1 each time (EV) and so after 29 consecutive losses, there would be 37 in hand for the final spin. No need to calculate the standard deviation as there would be no variance present.

#### Jesper

##### Re: When is the risk of ruin greater than the HA!
« Reply #2 on: June 05, 2016, 12:15:22 PM »
Two considerations for the mix, and missing from the above, are how long (how many spins?) one intends to play for and what methodology, "system", range of numbers to be covered each spin will be used.

What three standard deviations south of the EV represents will vary with the above, and therefore the amount of bankroll needed to keep from tapping out if unlucky enough for results to be in the southerly tail of the bell curve will vary as well.

If you bet in a way that negates all variance, and intend playing for 30 spins, then the bankroll required to not tap out will be 66 units (37+(30-1)); 37 units bet per spin, expect to lose 1 each time (EV) and so after 29 consecutive losses, there would be 37 in hand for the final spin. No need to calculate the standard deviation as there would be no variance present.

#### UK-21

##### Re: When is the risk of ruin greater than the HA!
« Reply #3 on: June 05, 2016, 12:25:05 PM »
True . . . but part of the answer's in there if you look for it.

#### Jesper

##### Re: When is the risk of ruin greater than the HA!
« Reply #4 on: June 05, 2016, 12:31:33 PM »
The good answer should be if by example like.
If you bet on an EC for 1000 spins using 200 units with no HA, the risk of loss all is : %
If you bet on an EC for 1000 spins using 1000 units and a HA of 2.7 the risk of lose all is: % (small must lose all).

#### UK-21

##### Re: When is the risk of ruin greater than the HA!
« Reply #5 on: June 06, 2016, 06:21:52 PM »
I'll do one (assuming an adverse edge of 2.7%), you do the other (no edge) . . .

For a thousand spins, betting one unit per spin (on an even-payoff bet), the EV for losing spins will be 1000 x 19/37 (=~514), and the EV for winning spins must threfore be 1,000 less this figure (486). To calculate the variance / standard deviation, one uses:

SQRT ( number of  trials x probability x (1-prob) ). So the standard deviation for the number of losing spins is:

SQRT ( 1,000 x 19/37 x (1-19/37) )   =  15.81.      3 x Std Devs would therefore be 47.43.

The worst case scenario for betting 1,000 spins on an evens-payoff bet would therefore be 561 losing spins (514+47). Three StdDevs either side of the EV covers 99.73% of outcomes, so there is a relatively tiny amount of room for worse results than this.

So assuming the worst case scenario is losing 561 units, and therefore winning 439 units, there's the potential to end up with a net loss of 122 units. So having a bankroll of 122 units would mean you'd have only a 0.14% chance of tapping out (only half of the 0.27 not covered by 3 x Std Devs as the other half would be at the "winning" side of the bell curve.

---------------------

Repeating the exercise with 1 x standard deviation . . .

1 x Std Dev south of the EV would be 530 losses (514+16) - so losing 530 units, and therefore winning 470 units, the net loss would be 60 units. 1 Std Dev either side of the EV covers 68.26% of occurances, and so the half south of the EV (losing side) covers 34.13%. Having a bankroll of 60 units would therefore leave you with a risk of tapping out of 15.87% (50.00%-34.13%).

Two standard deviations covers 95.44% of occurances, so half of this - 47.72% - that is south of the EV, gives a risk of tapping out of 2.28% (50%-47.72%).

I found this useful tool on the web which allows you to see what multiples of the standard deviation any particular percentage of occurances is covered by, ie 0.5 Std Devs either side equates to 38.30% of the population. Very handy for us degenerate gamblers.

https://www.mathsisfun.com/data/standard-normal-distribution-table.html [nofollow]

=====================================================

This is a fairly simple set of calculations, but once you start introducing progressions, inconsistent betting patterns and similar, it becomes a lot more complicated, and a degree of over-estimating can save on the aspirin. I apologise if this comes across as patronising in any way - it's not intended to.

Best of luck at the felt.

« Last Edit: June 07, 2016, 04:39:52 AM by UK-21 »

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#### Jesper

##### Re: When is the risk of ruin greater than the HA!
« Reply #6 on: June 07, 2016, 06:53:05 AM »
The question is raised because the casino I use have a NOZ and there is fair odds, they have as well an ordinary EU-wheel which has 2.7% HA.

We can get 1000 bullits on the wheel with HA, using 10 Euro but only 200 on the NOZ due to minbet.

If somebody has 10 Euro in bank, would he benefit from the NOZ, as the bankroll is 5 times smaller?

So the question is at what bankroll  the wheel with HA will have less chance to lose.

I believe a too small bank can be worse than the HA.

The risk of ruin depends of both bankroll and HA, but a very small amount of chips, should increase the risk of ruin, more than the HA. At a stage the HA should be more in power.

Some says the HA  has very small effect on risk of ruin, if that's the case a wheel accepting  0.01 and has HA
should despite of that be better than a wheel accepting 0.05 and is fair, if the player has small money to play with.

#### UK-21

##### Re: When is the risk of ruin greater than the HA!
« Reply #7 on: June 07, 2016, 07:18:49 AM »
Do the sums for the "no edge" option (replace 19/37 with 18/37) and see what bank you'll need to avoid tapping out on the grounds of worst case scenario results - 3 standard deviations south of the EV.

From what I've done above you can see that 1,000 units will mean you won't run out of money when playing evens-payoff bets against a 2.7% house edge.

As the worst case scenario is when playing against an EV of -2.7% is a bank requirement of 122 units, then I should imagine that a bank of 200 will be more than adequate when playing against a wheel with no edge.

But for comparions purposes, ie what bank for each is required to leave  5% risk of ruin, you'll need to do the sums for the no edge wheel and compare. As to which is the better option to play - obviously the zero edge game, although this will depend on any "commissions" taken on withdrawn balances after any losses - commissions applied to withdrawn profits are immaterial, as despite it you'll still be up on the deal.

#### Jesper

##### Re: When is the risk of ruin greater than the HA!
« Reply #8 on: June 07, 2016, 12:23:27 PM »
"From what I've done above you can see that 1,000 units will mean you won't run out of money when playing evens-payoff bets against a 2.7% house edge."
[/size][/color]
[/size]No when have any player lost 1000 times in a row?[/color]

#### UK-21

##### Re: When is the risk of ruin greater than the HA!
« Reply #9 on: June 07, 2016, 05:54:43 PM »
What's losing 1,000 spins in a row got to do with it?

Your original question was what bank would be needed to avoid ruin if playing:

(a) 1,000 spins on a single zero wheel,
(b) 1,000 spins on a no zero, no edge, wheel.

You've got the tools here to calculate this. If you don't understand the background to quantifying variance by calculating the standard deviation I would recommend you set some time aside to do some reading.

#### Jesper

##### Re: When is the risk of ruin greater than the HA!
« Reply #10 on: June 07, 2016, 07:06:45 PM »
What's losing 1,000 spins in a row got to do with it?

Your original question was what bank would be needed to avoid ruin if playing:

(a) 1,000 spins on a single zero wheel,
(b) 1,000 spins on a no zero, no edge, wheel.

You've got the tools here to calculate this. If you don't understand the background to quantifying variance by calculating the standard deviation I would recommend you set some time aside to do some reading.

I can and understand it, but in the world of game we have, just to understand, we do not win using it.

A lot is about probability, which we all knows some or more of, but who wins knowing it?  On thousends of webpages the same math is repeated, and it is not wrong, but not a winner either.

#### UK-21

##### Re: When is the risk of ruin greater than the HA!
« Reply #11 on: June 08, 2016, 07:00:39 AM »
Players are supposed to lose (longer term), that's why there's hefty advantage/edge in favour of the party staking the game. If you understand all of the underlying numbers behind roulette, I'm surprised you're surprised - this is quite apparent.

The latest entry on my online blog has a chart recording recent wheel of doom results. You can see that even having experienced a fair degree of positive variance, the bank's running balance struggles to make it into the black and show a net profit. This is not a surprise to me - as the more one plays, the more the EV line heads south. The results of the final 22 spins (of almost 1,000) was an eye-brow raiser though.

This thread should be useful to anyone wishing to calculate what "bank" they need to avoid tapping out, based on applying a method that consistently covers the same number of numbers each spin - which was the original question.

#### ernroo1

##### Re: When is the risk of ruin greater than the HA!
« Reply #12 on: June 08, 2016, 08:48:25 AM »
I dont get everyone sharing their systems. Show us the money (charts of your profit) too, that's all that matters.

#### dobbelsteen

##### Re: When is the risk of ruin greater than the HA!
« Reply #13 on: June 08, 2016, 09:28:43 AM »
Rouletteplayers play Always short run sessions.The role of the house edge is minimal.It is useless to compare short run sessions of spins.
The succesful roulette player plays strategies. A  league between two strategies is much more interesting.

#### Harryj

##### Re: When is the risk of ruin greater than the HA!
« Reply #14 on: June 08, 2016, 05:00:09 PM »
Surely the important ratio is bet/Bankroll. The larger this ratio the more chance the punter has of staying in the game. The longer he stays in the game the better his chance of getting a favourable variance.

It is a mistake to accept the math attitude that every variance MUST favour the House. With a good bet selection and an adequate B/R. The unfavourable  variations can be "aborted", while those favourable are exploited.

Harry

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