Fibonachi Dozen Strategy on Double Bonus Spin Roulette

I am curious if using Fibonachi Dozens Strategy on the online casino game Double Bonus Spin would change the expected return?

I was playing through a Fibonachi Dozen sequence starting at $0.10 (as this all my "frugal" nature will allow). On the 11th progression of $8.90 the ball started the bonus round. Both Bonus numbers hit my Dozen letting me win an extra $17.60.

Standard profit on the $8.90 is $3.30, as it has to cover the previous $14.30 in losses.

To be honest, I was just going by the excellent info I found on your Wizard of Odds site.

It shows an expect return of -0.022893. I hope there is a math solution for the variable return the bonus round permits when using a negative progression system like Fibonachi.

Thank you for your hard work across all the sites you have. The knowledge contained has certainly helped my bankroll. Enjoy your day. 🖖

Anyone have any idea on this?

Like all progressions, it's nonsense. Progressions are like calories, the more you consume, the more weight you gain. You can't lose weight eating slices of pizza, even if you eat them in the sequence 1, 1, 2, 3, 5, 8, ...

Quote: teliot

Like all progressions, it's nonsense. Progressions are like calories, the more you consume, the more weight you gain. You can't lose weight eating slices of pizza, even if you eat them in the sequence 1, 1, 2, 3, 5, 8, ...
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Read the OP again. He's not asking if a progression works, he's asking if there is an effect on the expected return when hitting a bonus round used in conjunction with a progression.

Quote: Krispy519

I am curious if using Fibonachi Dozens Strategy on the online casino game Double Bonus Spin would change the expected return?

I was playing through a Fibonachi Dozen sequence starting at $0.10 (as this all my "frugal" nature will allow). On the 11th progression of $8.90 the ball started the bonus round. Both Bonus numbers hit my Dozen letting me win an extra $17.60.

Standard profit on the $8.90 is $3.30, as it has to cover the previous $14.30 in losses.

To be honest, I was just going by the excellent info I found on your Wizard of Odds site.

It shows an expect return of -0.022893. I hope there is a math solution for the variable return the bonus round permits when using a negative progression system like Fibonachi.

Thank you for your hard work across all the sites you have. The knowledge contained has certainly helped my bankroll. Enjoy your day. 🖖
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Can you, in simple words, consistently explain (describe) the betting strategy step by step?

There is no "variable return." You have to understand that, in effect, you are playing an entirely different game from regular roulette.

The expected return on a bet of 1 is 481,752 / 493,039, so the expected value is -0.022893, as you, and The Wizard, said. (This assumes double-zero roulette.)

Playing a progression won't change the "expected return" in this game any more than it would in pretty much any other. Your current bet's ER is unaffected by the number and amounts of your previous ones.

"How did you calculate the ER?"

Fibonachi Dozens Strategy is a negative progression system that follows this sequence on any pays 2 to 1 bet. 1,1,2,3,5,8,23,21,34,55,89,144,233,377...

To calculate the new bet add the 2 previous losses together. This leads to increased profits as you progress up the system.

Here is a table of profit at each level.
1 wins 2
1wins 1
2 wins 2
3 wins 2
5 wins 3
8 wins 4
13 wins 6
21 wins 9
34 wins 14
55 wins 22
89 wins 35
144 wins 56
233 wins 90
377 wins 145

To calculate the win at any level take the number 2 steps back and add 1. Example: a win for a bet of 377 is 144 + 1.

On a single 0 wheel it is still an expected loss of 2.7%. Where my question lies is in the online game of Double Bonus Spin Roulette. The table on this site shows an expected loss of 2.2893% on any pays 2 to 1 bet.

Here is a table for IF you should hit a "JACKPOT" of winning on both bets on the bonus spin play any pays 2 to 1 bet:

1 wins 4
1 wins 3
2 wins 6
3 wins 8
5 wins 13
8 wins 20
13 wins 32
21 wins 51
34 wins 82
55 wins 132
89 wins 213
144 wins 344
233 wins 556
377 wins 899

To calculate a "Jackpot" win take the number 2 back in the sequence and add 1 plus double the current bet. Example: a bet of 377 is 144 + 1 + 754 = 899.

So this leads to my original question of: Does using a negative progression system affect the expected loss of 2.2893%?

I made a program in Basic to simulate 10000 spins. It shows "Jackpots" to be quite rare at about 1 in 320 spins.

Hopefully someone can do the REAL math and calculate if this does affect the expected loss ratio.

Quote: Krispy519

So this leads to my original question of: Does using a negative progression system affect the expected loss of 2.2893%?

I made a program in Basic to simulate 10000 spins. It shows "Jackpots" to be quite rare at about 1 in 320 spins.

Hopefully someone can do the REAL math and calculate if this does affect the expected loss ratio.
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I did the math, and you are still expected to lose 2.2893% of the total amount that you bet. This is true regardless of what system you use.

I am unsure how to do the advanced math that my question asked. Could you break it down for me?

Maybe it's "Gambler's Fallacy" but it seems that hitting a high value "Jackpot" should vastly skew the the standard expected loss of 2.2893%.

Right now I have been using an online site called Replit to make programs in Basic that run simulations. Kind of a brute force way to get results.

Perhaps to understand the math I'd need a PHD in probability but I'd sure like to learn the process.

Quote: Krispy519

I am unsure how to do the advanced math that my question asked. Could you break it down for me?

Maybe it's "Gambler's Fallacy" but it seems that hitting a high value "Jackpot" should vastly skew the the standard expected loss of 2.2893%.

Right now I have been using an online site called Replit to make programs in Basic that run simulations. Kind of a brute force way to get results.

Perhaps to understand the math I'd need a PHD in probability but I'd sure like to learn the process.
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Not at all.

However, before I try to explain, I would like to know what you mean by "jackpot." I would also like to know how many losses in a row it would take before you run out of money or reach some house betting limit.

The possibility of a double win is factored into the 2.2893%; remember, without it, double-zero roulette has an expected loss of 5.2631%.

I thank you kindly for taking an interest in my dilemma.

I'll use the Draftkings version as an example min bet is $0.10 to max of $600 on the pays 2 to 1 bets.

This allows for a max bet of $600 which is not the true Fibonachi number of $674.40 but is still a profit of $108.90. This gives a max of 20 bets before you are bust. By my calculation you have a 0.0505586% chance of failing 20 times or about 1 in 2000.

I nicknamed "JACKPOT" after the instance when you hit the bonus spin and win with both numbers.

I'll have to do further testing but at 10000 spins the results look fairly promising.

How many spins would you think are needed to prove it viable? I have been using a bank of $1000 trying to cash out at $1250, is that reasonable?

Here is a table showing how much you stand to win if you lose 0, 1, 2, ..., 19 bets before winning
The columns are:
Round - what bet number this is
Bet - the amount bet
Prev Losses - the total amount lost up to that point
Profit on Win - the total profit made (winnings minus previous losses) on a "single" win, either from the first spin or a single win on the second spin
Prob of Win - the probability on a "single" win
Profit on Jackpot - the total profit made of a jackpot (double win)
Prob of Jackpot - the probability of a jackpot
ER - the expected return of a win at that point

Round	Bet	Prev Losses	Profit on Win	Prob of Win	Profit on Jackpot	Prob of Jackpot	ER
1	1	0	2	0.31986111	4	0.0008762	0.64322701
2	1	1	1	0.21642893	3	0.00059287	0.21820753
3	2	2	2	0.1464432	6	0.00040115	0.29529331
4	3	4	2	0.09908846	8	0.00027143	0.20034839
5	5	7	3	0.06704663	13	0.00018366	0.20352749
6	8	12	4	0.04536604	20	0.00012427	0.18394958
7	13	20	6	0.03069621	32	0.00008409	0.18686801
8	21	33	9	0.0207701	51	0.0000569	0.18983259
9	34	54	14	0.01405376	82	0.0000385	0.19990942
10	55	88	22	0.00950925	132	0.00002605	0.21264199
11	89	143	35	0.00643428	213	0.00001763	0.22895418
12	144	232	56	0.00435366	344	0.00001193	0.24790729
13	233	376	90	0.00294583	556	0.00000807	0.26961157
14	377	609	145	0.00199325	899	0.00000546	0.29392998
15	610	986	234	0.0013487	1454	0.00000369	0.32096791
16	987	1596	378	0.00091258	2352	0.0000025	0.35083381
17	1597	2583	611	0.00061748	3805	0.00000169	0.38371688
18	2584	4180	988	0.00041781	6156	0.00000114	0.41984052
19	4181	6764	1598	0.0002827	9960	0.00000077	0.45947345
20	6765	10945	2585	0.00019129	16115	0.00000052	0.50292073

The sum of the ERs on wins = 6.01196163
However, the ER of a total wipeout (20 losses in a row) is -7.16631970, so the total ER of a round of this system is -1.15435807.

Calculating how likely you are to end up 250 ahead with a starting bankroll of 1000 is something I think is best done with Monte Carlo simulation (i.e. do it millions of times, and see how often you win).
If you let your initial bankroll be 1094.5, you can play until you lose 11 in a row. You will reach 250 before losing 11 in a row about 30% of the time.

Just so I understand, the playing of a negative progression DOES affect the ER in favour of the player?

With the 1094.5 the last is beyond max allowed. Still profitable at least. Would the losses be 21 in a row?

I assume Monte Carlo simulation is the exact name for an app/program?

Quote: Krispy519

Just so I understand, the playing of a negative progression DOES affect the ER in favour of the player?

With the 1094.5 the last is beyond max allowed. Still profitable at least. Would the losses be 21 in a row?

I assume Monte Carlo simulation is the exact name for an app/program?
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NO, 'Monte Carlo simulation' is a generic term for arriving at probabilities and expectation values by performing a process that mimics the game in question.

This can be done by running an algorithm that simulates one game at a time with a random number generator. This can be done quite rapidly in computers compared to real world gambling games. There are millions of MC sim programs in existence. I have written dozens.

Before computers, MC sims were done by simulating a game physically. For example rolling out dice rapidly for thousands of rolls and recording results. The 'Monte Carlo' part of the MC sim name comes from the use of a randomizing method in the simulation.

MC sims are also done in science to simulate quantum or atomic processes. The behavior of spin glasses and roughening of crystal surfaces were investigated with MC sims and then compared to experimental results.

I wrote my simulation in Basic with a plethora of IF and THEN statements. It's from my grade 11 knowledge from 40 years ago.

Isn't an the total ER of a round of this system is -1.15435807 better than even the odds for French wheel using the La Partage rule?

Again many thanks for the quick and knowledgeable replies. 🖖

Quote: Krispy519

Isn't an the total ER of a round of this system is -1.15435807 better than even the odds for French wheel using the La Partage rule?

On the contrary - it is worse than playing 3-card monte or the shell game, or any other game where you lose every time, where the ER is -1 (you bet 1 and lose it 100% of the time). You are expected to lose more using this system because of the massive loss when you lose 20 bets in a row.

If your value for the French (single-zero) wheel with La Partage is -1.35135, that is a percentage; the actual ER is -0.0135135. The -1.15435807 means that if your initial bet is 1 and you increase it with a Fibonacci progression until either you win or you have 20 losses in a row, you are expected to lose an average of 1.15435807, or 115.436% of your initial bet, per progression started. Yes, you will only lose 20 in a row about once in every 1185 progressions, but also remember that you will lose over 10,000 units (assuming your initial bet is 1 unit) when that happens, whereas 1/5 of the time, you will win 1 unit, and another 1/3 of the time, you will win 2.

Just when I thought i could accept reality, I am unsure if your calculation takes into account that the Bonus well is 1.5 the size of a standard well. So i tried my hand at doing the Hard Math as follows:

Chance to Win first bet without bonus is: 12 / 38 = 31.57894737%

I may be a mistaken to assume that the Bonus actually acts like a push but gives 2 more spins. This leads me to what's the chance of a Bonus?

Bonus Chance is 1.5 / 39.5 = 3.79746835 % that gives 2 spins at 12 / 39.5 = 30.37974684%

Then chance to win in a Bonus is: 30.37974684% X 2 X 3.79746835% = 2.30732254%

Add the original chance to win and Bonus chance to win together: 31.57894737% + 2.30732254% = 33.88626991%

OR

Is it done like this:

Chance to win is 12/39.5 = 30.37974684% and then add the chance to win in Bonus spins resulting in: 30.37974684% + 2.30732254% = 32.68706938%

BUT

Your chart shows a 31.986111%

So, I am guessing my math is wrong OR didn't explain about the Bonus well being 1.5 the size of a standard for you to consider in your chart.

I hope you are at least amused by my inability to process probability in math. Thank you again.

Quote: Krispy519

Just when I thought i could accept reality, I am unsure if your calculation takes into account that the Bonus well is 1.5 the size of a standard well. So i tried my hand at doing the Hard Math as follows:

Chance to Win first bet without bonus is: 12 / 38 = 31.57894737%

I may be a mistaken to assume that the Bonus actually acts like a push but gives 2 more spins. This leads me to what's the chance of a Bonus?

Bonus Chance is 1.5 / 39.5 = 3.79746835 % that gives 2 spins at 12 / 39.5 = 30.37974684%

Then chance to win in a Bonus is: 30.37974684% X 2 X 3.79746835% = 2.30732254%

Add the original chance to win and Bonus chance to win together: 31.57894737% + 2.30732254% = 33.88626991%

OR

Is it done like this:

Chance to win is 12/39.5 = 30.37974684% and then add the chance to win in Bonus spins resulting in: 30.37974684% + 2.30732254% = 32.68706938%

BUT

Your chart shows a 31.986111%

So, I am guessing my math is wrong OR didn't explain about the Bonus well being 1.5 the size of a standard for you to consider in your chart.

I hope you are at least amused by my inability to process probability in math. Thank you again.
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In your bonus spin calculation, you forgot to take into account that, while one spin wins, the other loses.
The probability of winning the bet without the bonus is 12 / 39.5.
The probability of winning one bet (not both) with the bonus is 1.5 / 39.5 x 2 x 12 / 39.5 x 27.5 / 39.5.
When you add these two together, you get 31.9861106%.