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ThomasK
ThomasK
Joined: Jul 4, 2015
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June 10th, 2022 at 9:49:20 AM permalink
Dear Wizard,

Thank you for your videos on betting systems, instructions, and challenges. It inspired me to design a (hypothetical) experiment which might show that "Betting systems are ..." neither "... equally ..." nor "... completely worthless."?

This (hypothetical) experiment would involve you, Heather Ferris, and Angela Marie Wyman in the setup below. I'm not sure whether this was suitable for another video.

Angela Marie would be the financial controller watching over the wagers during the experiment and would be the final judge, deciding who performed better - you or Heather.
Angela Marie might be placed in a comfortable lounge with music, drinks, and snacks and enough table space to manage the two stacks of chips for Heather's and your wagers and another two stacks for both your returned chips.

PROCEDURE
1) Angela Marie will hand seven chips to each of you, asking you to play one single game (rules see below) and ask you to then come back with the chips you have left from this game.
2) On your return Angela Marie will put these chips (or none, if you lost) on the second stack and start over with step 1), handing you another seven chips from the stack of wagers.
3) After a number of games (the more, the better) Angela Marie will compare the two stacks of returned chips an announce if and who performed better.

RULES
The games are played at a roulette table, preferably single zero, but double zero will do as well. To simplify things, only bets on red are made. Both players will bet on red on the same rolls (first, second, third) as long as they are qualified for a roll according to their individual rules.

Heather Ferris will bet one chip on red on the first roll.
* If red wins, she collects the two chips from the table and is done for this game, now holding eight chips. She will have to wait until the Wizard's rolls are resolved.
* If red loses, Heather now bets two chips on red for a second roll. If red wins, she collects the four chips and waits with her eight chips for the Wizard to finish his rolls.
* If red loses again, she bets her last four chips on red. After this roll she leaves the table together with the Wizard.
* She returns to Angela Marie with either no chips at all or with eight chips.
(Betting system is Martingale with max. three steps.)

The Wizard will bet all his seven chips on red on the first roll.
* If red loses, he will have to wait until Heather's rolls are resolved.
* If red wins, the Wizard bets all fourteen chips (his initial seven plus the seven chips won) on red for a second roll. If red loses he will wait for Heather to finish her rolls.
* If red wins, the Wizard bets all twentyeight chips for a last roll. After this roll he leaves the table togehter with Heather.
* He returns to Angela Marie with either no chips at all or with fiftysix chips.
(Betting system is Paroli with max. three steps.)

PREDICTION
My guess is that Angela Marie will announce Heather as the winner (girl's power?).
To substantiate the experimental result, Angela Marie might calculate the house edge she experienced on the two games she played.


Angela Marie on average invested seven units in Heather's game.
Heather returned either nothing (loss: 7) or eight units (win: 1).
The probablity of returning nothing was q=19/37*19/37*19/37 approx. 13.54%.
The proablity of returning eight units was p=1-q approx. 86.46%.
The expected value is approx. 0.8646 * 1 - 0.1354 * 7 = 0.8646 - 0,9479 = -0.0833.
The house edge is the expected value divided by the average wager, approx. -0.0833 / 7 = -0.0119.
Remark: Only 44% of the house edge of a single roll on single zero roulette.



Angela Marie on average invested seven units in the wizard's game.
The Wizard returned either nothing (loss: 7) or fiftysix units (win: 49).
The probablity of returning chips was p=18/37*18/37*18/37 approx. 11.51%.
The probablity of returning nothing was q=1-p approx. 88.49%.
The expected value is approx. 0.1151 * 49 - 0.8849 * 7 = 5.6417 - 6.1940 = -0.5523.
The house edge is the expected value divided by the average wager, approx. -0.5523 / 7 = -0.0789.
Remark: Nearly three times (2.92) the house edge of a single roll on single zero roulette.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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June 10th, 2022 at 10:24:33 AM permalink
That's not entirely surprising, considering that Wizard is expected to bet far more than Heather, so he is also expected to lose far more.

On any given spin, the probability of red winning (assuming a double-zero wheel) is 9/19, and the probability of red losing is 10/19.
Over 19^3 = 6859 sets of three spins, the following possibilities occur:

Wizard will bet 7 on the first spin of each set; this is 6859 spins
Wizard will bet 14 on the second spin where the first spin wins; this is 9 x 361 = 3249 spins
Wizard will bet 28 on the third spin where the first two spins win; this is 9 x 9 x 19 = 1539 spins.
The total amount Wizard is expected to bet is 7 x 6859 + 14 x 3249 + 28 x 1539 = 136,591.

Heather will bet 1 on the first spin of each set; this is 6859 spins
Heather will bet 2 on the second spin where the first spin loses; this is 10 x 361 = 3610 spins.
Heather will bet 4 on the third spin where the first two spins lose; this is 100 x 19 = 1900 spins.
The total amount Heather is expected to bet is 6859 + 2 x 3610 + 4 x 1900 = 21,679.

Wizard is expected to bet 6.3x what Heather does, so he is expected to lose 6.3x what Heather does.
DeMango
DeMango
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June 10th, 2022 at 10:44:31 AM permalink
D'Alembert and La Bouchere. Conspicuous in their absence!
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
ThomasK
ThomasK
Joined: Jul 4, 2015
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June 10th, 2022 at 11:36:46 AM permalink
Quote: ThatDonGuy

That's not entirely surprising, considering that Wizard is expected to bet far more than Heather, so he is also expected to lose far more.

[ ... ]

Wizard is expected to bet 6.3x what Heather does, so he is expected to lose 6.3x what Heather does.
link to original post



For each game, or set, both the Wizard and Heather only receive 7 units from Angela Marie. So that is all they are able to bet and, in the bad case, to lose. Angela Marie certainly will be able to confirm this.
ThatDonGuy
ThatDonGuy
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June 10th, 2022 at 12:21:50 PM permalink
Quote: ThomasK

Quote: ThatDonGuy

That's not entirely surprising, considering that Wizard is expected to bet far more than Heather, so he is also expected to lose far more.

[ ... ]

Wizard is expected to bet 6.3x what Heather does, so he is expected to lose 6.3x what Heather does.
link to original post



For each game, or set, both the Wizard and Heather only receive 7 units from Angela Marie. So that is all they are able to bet and, in the bad case, to lose. Angela Marie certainly will be able to confirm this.
link to original post


Except that, whenever Wizard wins his first bet, he now has 14 to bet, and whenever he wins his second bet, he now has 28. If you're going to give 56 chips back when you win three bets, you do have to take this into account.

Look at it another way:
Wizard has a 6130 / 6859, or about 89%, chance of losing the original 7.
Heather has a 1000 / 6859, or about 15%, chance of losing the original 7.

To paraphrase George Orwell, "All systems are worthless...but some lose your money much faster than others."
ThomasK
ThomasK
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June 10th, 2022 at 1:11:46 PM permalink
Quote: ThatDonGuy


Except that, whenever Wizard wins his first bet, he now has 14 to bet, and whenever he wins his second bet, he now has 28. If you're going to give 56 chips back when you win three bets, you do have to take this into account.

Look at it another way:
Wizard has a 6130 / 6859, or about 89%, chance of losing the original 7.
Heather has a 1000 / 6859, or about 15%, chance of losing the original 7.

To paraphrase George Orwell, "All systems are worthless...but some lose your money much faster than others."
link to original post



Hmm, you're focusing on the Wizard and on Heather and on the very details of what they are doing at the roulette table.

How about Angela Marie?
She invests the same amount in two games in order to find out which one is better. She doesn't know anything about the Wizard betting 7, 14, and 28 units.
She sees 7 units going away and, when lucky, receiving back 56 units from the Wizard.
And, when lucky, receiving back 8 units from Heather.
What she also can do, is estimate the probability of these lucky events, when playing long enough.

Since we know that the events are produced at a roulette table, we are able to name these probablities a priori.

Knowing the amounts and their probabilities, Angela Marie is now able to calculate the expected value for both games and compare them accordingly. And compare them to other games as well, for example to the expected value of a single roll at roulette.

What would happen, if Heather and the Wizard put themselves in Angela Marie's shoes and simply ignored the details of what they are doing at the table?
ThatDonGuy
ThatDonGuy
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June 10th, 2022 at 1:41:40 PM permalink
I don't think that, when it comes to your original prediction, we disagree on anything. I am just explaining why this is the case.

If you want a true "black box" (i.e. the only things that matter are the initial bankrolls and the final results - how they got there is irrelevant), look at this:
Wizard has a 6130/6859 chance of ending up with zero and a 729/6859 chance of ending up with 56, so the expected result is 40,824 / 6859, or about 5.952.
Heather has a 1000/6859 chance of ending up with zero and a 5859/6859 chance of ending up with 8, so the expected result is 46,872 / 6859, or about 6.834.
I would probably declare Heather "the winner" as well.

Also note that, at least around here, "system" presupposes a guaranteed profit at some point. In this case, the Wizard's statement that all systems are equally and completely worthless is correct, as the probability of "guaranteed profit" is zero.
ThomasK
ThomasK
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June 11th, 2022 at 5:47:13 AM permalink
You're absolutely right. And we not only agree on the final judgement but on the way to explain it, as well: Calulating and comparing the expected values.

Imagine, you were forced to play roulette.
I guess I'm not mistaken to say that you would prefer Heather's way to the Wizard's way of playing.
How about simple bets on red?
* Angela Marie would hand you 7 chips from your own stack.
* You would bet all seven chips on red on the first roll together with Heather and the Wizard.
* After that roll you would have to wait until Heather and the Wizard have finished their games and then return back to Angela Marie altogether.

Would you want to change games with Heather, or even with the Wizard?

Angela Marie might calculate the house edge she experienced during your game.
And I suspect that she would recommend to you to change games with Heather.


Angela Marie on average invested seven units into this game.
The game returned either nothing (loss: 7) or fourteen units (win: 7).
The probablity of returning nothing was q=19/37 approx. 51.35%%.
The proablity of returning fourteen units was p=18/37 approx. 48.65%.
The expected value is approx. 0.4865 * 7 - 0.5135 * 7 = 3.4054 - 3.5946 = -0.1892.
The house edge is the expected value divided by the average wager, approx. -0.1892 / 7 = -0,0270.
Remark: Of course this is the house edge of a single roll on single zero roulette.


About your note: I understand that the term "betting system" commonly implies a method that overcomes negative expectation and turns a losing game into a profitable advantage play. My idea is to use them to reduce the disadvanage when playing negative expectation games. I am aware that it is impossible to switch such games to positive expectation, but approaching zero closer and closer should be doable. Following the general advice "When you have to play at a casino then chose the games with the lowest house edge."
ThomasK
ThomasK
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June 17th, 2022 at 9:02:29 AM permalink
I've set up a Monte Carlo simulation of the proposed experiment.
Here double zero roulette is assumed and bets on "1-18" instead of bets on red. These have the same probability as bets on red, but can numerically be selected much easier than the layout's red numbers.

In the result, one interesting thing stands out: Heather's play not only is the superior one, as expected, but it also has a slightly lower house edge (as experienced by Angela Marie) than that of a simple bet on "1-18" on a single zero roulette. So, if only double zero tables are available, this might be the way to go ...

I will give here all the details of my Monte Carlo simulation so that mistakes in the simulation or in the math can be found.

The simulation stretches over 380000 rolls. See here the first rows of the simulation. The last rows of the simulation are shown below in section "RESULT".
Since the jpegs tend to be blurry, I put magnified selections inside spoiler tags to make sure that the formulas are perfectly readable.



THE ROLLS
... are simulated in column C "number" with numbers 0 through 36 and -1 representing the double zero.

Column D "18" calculates the halves and the zero category:
0 - zero and double zero
1 - 1 through 18
2 - 19 through 36

Column E "win/lose" compresses this information into a pure binary value:
0 - lose on zeroes and on upper half "19-36"
1 - win on "1-18"



The Pivot table is used to verify the correctness of the frequency of the random numbers, as well as their assignment to the categories. Each number appears with a probablity of 1/38, random variation included, and is properly assigned to its half and to "win/lose".



In columns A and B the games and their individual rolls are numbered.
The games are either two or three rolls each. This is determined in column B "roll#" using the fact that the third roll is necesary, if rolls one and two yield the same result.

Column A "game#" counts the games simulated, incrementing on the first roll of each game.



THE CONTESTANTS
Columns F and G show the actions of betting and of waiting for the other party to finish their game.
Column F "Heather's action" is the Martingale procedure. She has to bet on the first roll unconditionally and will wait once she hit and won, or while she is waiting already. Otherwise she keeps betting.
Column G "Wizard's action" is the Paroli procedure. Again the first roll is mendatory. Once he loses or is already waiting, he will keep waiting. Otherwise he keeps on betting.



THE STACKS OF RETURNED CHIPS
Columns H and I represent the stacks where Angela Marie puts the chips returned from Heather and the Wizard.
As a monitoring and comparison reference, a third stack for pure roulette rolls is kept in column J. It accumulates the returned chips of the reference bets on "1-18". These bets, seven chips per bet, as with Heather and the Wizard, are made on roll 1 of each game, so that Heather, the Wizard and the simple single rolls play the exact same number of games.
Column H "Heather's returned chips" collects and adds 8 chips each time she bets and wins during a game. Otherwise she is waiting, adding 0 to her stack.
Column I "Wizard's returned chips" collects and adds 56 chips each time he is in his third roll and is qualified to bet and wins this third roll. Otherwise nothing is added, in particular while he is waiting.
Column J "simple single roll" collects and adds 14 chips on the first roll only when this roll is won.



RESULT
Row 380001 has the accumulated numbers of chips in columns H, I, and J. The differences between Heather's Martingale, the Wizard's Paroli are quite obvious. Even the pure rolls have their own sum.



To see the effect in detail, expected value and house edge (as experienced by Angela Marie) are calculated.
The loss produced in the Monte Carlo simulation is given in row 380004
Each player received seven chips per game. Multiplied by the number of games, given at the end of column A, results in the total chips wagered. This is the same for all players.
Subtracting this sum from the simulated number of returned chips results in the "actual loss".
Dividing the loss by the number of games produces the expected value, in row 380005.
Further dividing the expected value by the wager of seven chips results in the house edge as experienced by Angela Marie.

Rows 380009 through 380012 have the a priori calculation based on the known probabilities of a double zero roulette wheel.
The resulting figures of expected value and of the theoretical house edge are very close to the simulated results.



CONCLUSION
The experiment and the corresponding probability calculations suggest that ...

* A player playing a negative expectation game should definitely avoid using a betting strategy based on progression on winnings, like Paroli, because it increases the players disadvantage compared to the pure base game.

* A player playing a negative expectation game should favour a betting strategy based on progression on losses, like Martingale, even over the pure base game, because it reduces the player's disadvantage. Although, progression on losses, constructed properly, will converge towards zero expectation (i.e. the fair game) it will never reach it nor exceed zero and therefore will not turn the game into a winning, positive expectation advantage play.

FINAL REMARK
It can be shown that for a player facing a positive expectation game, the opposite advice is valid. Progression on losses decreases the player's advantage in direction of zero expectation (the fair game) while progression on winnings pushes the advantage further away from zero in positive direction.
ThomasK
ThomasK
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July 8th, 2022 at 11:14:50 AM permalink
The previous experiment showed that progression on losses and progression on winnings produce different effects in games of negative expected value. Two more experiments will show the effects of both types of progressions on games having an expected value greater than zero.

These two experiments might be performed at Heather's practice table. Heather would deal roulette according to the common rules. The only difference in the procedure would be that Heather would match the Wizard's wager before the roll, betting that the event won't happen.
(Remark: In general it should be good betting practice for both betting parties to put their wagers on the table. For obvious reasons casinos prefer to protect their bankroll and only expose the wagers, they actually lost to the players, after a roll.)
The convenient effect of Heather wagering in the experiments is that she now is a natural advantage player, playing at the houses positive edge.

Angela Marie again would be the financial controller, watching over the wagers, using separate stacks for the Wizard's and Heather's returned chips. She would also again calculate the house edges she experienced from both the Wizard's and Heather's play.

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