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Roulette Reducing average house edge vs an EC

Discussion in 'Roulette Forum' started by TwoUp, Jun 5, 2022.

  1. TwoUp

    TwoUp Well-Known Member

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    I have claimed previously that a series of straight up bets pays less house edge on average than a single large bet on an EC. The series of bets pays even less house edge on average than an EC with La Partage (LP) where one gets half their EC bet back on zero.

    To be absolutely clear the house edge is always a percentage of what is risked, my claim is the amount risked is, on average less when using a series of bets on a single straight up number than with a single large EC bet. It follows then that the average amount of house edge paid in absolute terms is lower as a result. The house edge in percentage terms will remain 2.70% of the total risked. I am not claiming anything magical or any leaps of faith.

    The conventional wisdom suggests that you cannot do better than risking 100% of your bankroll on a single EC bet. But I show this is not necessarily so, especially when we are not recycling our action. We are always better to incrementally or progressively risk on a higher/longer odds bet when the house edge is the same as an EC. Roulette offers this opportunity.

    Here is a scenario where one attempts to make $1000 before losing $1000.

    The house edge on an EC is $1000 ×1/37 = $27.03

    The house edge in an EC with LP is half the above and 1.35% or $13.51 on a $1000 bet.

    Understand that house edge is always paid on what we risk, win or lose. An EC should pay $1055.55 based on the true odds but only pays $1000. The fact we contracted to risk $1000 to make less than the actual true odds, we just paid 2.7% (1.35% for LP) house edge right there win or lose.

    Every dollar risked pays house edge as determined by the expected value.


    The expected value (E.V.) is simply the probability of winning multiplied by the payout minus the probablity of losing multiplied by the amount risked. For roulette bets this is always 1/37 or 2.70% and for EC with LP (1.35%).

    So with all that out of the way I present a series of bets on a single straight up number and compare the effective house edge vs what one pays with an equivalent EC.

    The average amount risked on a series of bets is less than an EC and that is why the average house edge (the E.V) is less in absolute value terms and will always be 1/37 of what is risked.

    I have used $1000 as the basis in the table below but you could just as easily substitute 1000 units.

    Screenshot_20220530-213335_Drive.jpg


    As we see in the table the probability of losing all 24 bets is very close to an EC (51.81%). The series of bets is expected to lose $982 51.81% of the time (expected loss $508.77). This is not house edge, just the expected loss. House edge is 2.7% of total risked or $26.54 as the table shows)

    An EC is expected to lose $1000 51.35% of the time (expected loss $513.50). This is not house edge just the expected loss. House edge is 2.7% of total risked or $27.03 (1.35% or $13.51 for LP).

    The real difference is in the case of a win and that is where house edge is saved with the series of bets vs the EC. The series can win on earlier bets thus terminating the progression and reducing the total risked and total house edge paid. This is the only thing I am claiming: that a straight up bet can win and that playing up to 24 of them is 48.19% likely to win and 51.81% likely to lose. When we do win we pay less total house edge on average than an EC.

    I would like to keep the thread clean and on the topic of the math as hopeful as that might sound.

    I am more than happy to answer genuine questions and explain the math. I will not feed trolls, attention seekers and those who are incapable of logic or reason and just want to kick up dust. If you are one of these people you will be ignored after exhausting your chance to be reasonable.

    Please keep the discussion to the math. If you think a calculation is wrong. Identify the columns you agree with first. Then identify the column and row you have a problem with.

    Please ask a question to clairfy your understanding first before ranting based on faulty assumptions. It may be a simple case of not understanding what the particular column represents.

    Where you genuinely believe a calculation is still wrong, please provide your forumula, and your calculated values and we will work it out together.
     
  2. Rustyshackleford

    Rustyshackleford Active Member

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    wait im confused, are you saying betting black/red is better or betting straight up numbers is better for the player?

    in normal gambling going in with a 1,000 unit and betting 1 unit out of that basket is the trick.
    if you win and how you handle and manage that 1 unit win reflects the fate of your over all bankroll of a 1,000 unit
    betting all 1k unit in 1 turn is worst thing you could do but gives best chances/odds of double.
    betting 1 unit and using a positive system (upper after a win) is your only option.

    in reality negative/positive/oscars/cancel/martingale they will all fail before doubling.
    i read a post on wizard of odds forum someone in queens name richbailey played electronic baccarat and waiting for a 3 player or banker streak and betted against it and won x2 his roll
    10min bet so (two rolls of 320 total up 640)now hes gambling w/ house money.

    i walk by my casino where they have real shoes of baccarat stadium and i would have gotten clipped for 150$ per 6hr, however he played electronic and dont really know the bust out rate of a 30 unit martin gale, i do believe the electronic seems to show less streaks of over 8ish
     
    Last edited by a moderator: Jun 5, 2022
  3. Nathan Detroit

    Nathan Detroit Well-Known Member Founding Member

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    There is no such an animal as the " casinos money " once the money is paid over to you it is

    YOUR money .


    ND
     
  4. TwoUp

    TwoUp Well-Known Member

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    I am saying betting up to 24 times on a straight up number is better than the conventional wisdom of a single EC like red/black.

    It is better because the average risk is less for the equivalent odds of winning $1000. When it loses, your downside is similar to the EC (losing 24 bets back to back $982) but on a win it does not pay as much house edge on average because you don't risk all of the $1000 up front like an EC bet (the average win is higher also)
     
    Last edited: Jun 5, 2022
  5. Nathan Detroit

    Nathan Detroit Well-Known Member Founding Member

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    Of no value .
     
  6. Junket King

    Junket King Well-Known Member Compulsive Liar

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    I haven't gone through your attachment above, one things strikes me however, why aren't you comparing apples with apples. When you bet an EC you are in-effect backing 18 numbers, yet your quote above references 24 times?

    I will review your attachment later on :)
     
  7. TwoUp

    TwoUp Well-Known Member

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    That's a good question. The probability of losing 24 times on a single number is roughly the same as losing an EC so they are comparable from a win/loss perspective:

    (36/37)^24 ≈ 19/37
    51.8% ≈ 51.4%

    This means the probability of winning is also comparable: 48.2% vs 48.6

    So it's an Apples to Apples comparison from a probability perspective and the worst case risk in $ terms.
     
    Last edited: Jun 6, 2022

  8. TwoUp

    TwoUp Well-Known Member

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    Thanks for your troll remark.
    Any further posts from yourself will be ignored.
     
  9. Dr. Sir Anyone Anyone

    Dr. Sir Anyone Anyone Well-Known Member Lineage to Founders

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    breakingbad-mike.gif


    NO. Just NO.
     

    Attached Files:

    Last edited: Jun 6, 2022
  10. TwoUp

    TwoUp Well-Known Member

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    Yes. Just Yes. I like breaking bad.

    Are you able to explain why "No"?

    House edge is only paid on what one actually risks.

    Do you you say no to that premise?
     
  11. Junket King

    Junket King Well-Known Member Compulsive Liar

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    You can't really compare betting a single number X number of times vs betting an EC. There lies the problem.
     
  12. TwoUp

    TwoUp Well-Known Member

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    The basis of the comparison is we risk up to $1000 to win $1000 in both cases with a probability of success of 0.412 (series) and 0.416 (EC) to win.

    It is all about which bet risks the least on average and therfore pays the least house edge on average.

    The target, probability, and worst case risk are all fixed, which leaves us with average risk and house edge to compare which is the point of the post.

    Same with trading strategies, they can be very different in their execution but comparable based on factors like profit target but have very different trading costs and drawdowns.
     
  13. 6probability9

    6probability9 Member

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    column 10 x column 11 = column 14

    column 14 you use -$26.54 × (36/37)²³. -$26.54 is the sum E.V. of 24 levels = (36÷37)²⁴, how can you multiply 23 losses by the -$26.54 total of all your losses to get a -$14.13 on the 24th "cum. Level" E.V.?

    -$1.59 × (36/37)¹ is not the E.V for level 2.
    -$0.81 is the E.V. for level 2.

    -$26.54 × (36/37)²³ is not the E.V. for level 24
    -$1.51 is the E.V. for level 24

    -$8.10 is not the E.V. of wagering $982
    -$26.54 is the E.V. of $982
     
  14. 6probability9

    6probability9 Member

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    48.2% is the Pr that you don't always reach level 24, overall Pr 48.2% of events you win. 51.8% is the Pr that you reach 24 levels and still lose $982.

    That's worse Pr than EC + LP that has 48.65% Pr of Win or lose (1 or -1).

    Your Pr of Win @ levelⁿ is 2.7%
     
    Last edited: Jun 6, 2022

  15. 6probability9

    6probability9 Member

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    column 10 × column 11 = column 14

    20220531_185102.jpg

    The meat of your "-$8.10" average E.V. is false.
     
  16. 6probability9

    6probability9 Member

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    "$8.10" E.V. which isn't the true value or average of your losses.
    Bet levels 2-24:

    W = (36÷67)ⁿ × (1÷37)ⁿ

    L = (36÷37)ⁿ × (36÷37)ⁿ

    @ bet level 1 You lost. $29 × (36÷37) that would expect negative 0.78 value. If recycled
    @ bet level 2 You lost. $30 × (36÷37) that would expect negative 0.81 value. If recycled

    The cumulative level E.V. (column 14)
    On bet level 2 you sum the cumulative bet E.V. at level 1 & 2 . aka You've lost twice (36÷37)² that's (-0.78 + (-0.81 =) -1.59 E.V. and multiply this by the probability of losing one level prior, (36÷37)¹ to arrive at -$1.55 = -$1.59 × (36÷37)¹..

    Why are you multiplying the cum. bet E.V with the Pr of losing -1 level prior?
    You do this to all the subsequent level E.V. in column 14. Miscalculating an average -$8.10 E.V. that doesn't exist.

    Your progession negates the expected value of your prior bet level and each loss will accumulate towards the toyal bet E.V: $982 × (-1/37) = -$26.54 E.V.
    Cum bet E.V. is theoretical as in if you recycled your progression over multiple times on average you will lose -$26.54.
     
    Last edited: Jun 7, 2022
  17. TwoUp

    TwoUp Well-Known Member

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    @6probability9 you already had a chance to understand on your own thread.

    I won't waste any more time on someone who has proven unable to count.

    You won't get any more replies from me.

    You can fuck off now.
     
    Last edited: Jun 7, 2022
  18. 6probability9

    6probability9 Member

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    No.

    You multiply the Pr before a level with the cumulative E.V. on that level.

    Simply answer the question to your own calculations.


    In column 14 example bet level 24 you use: -$26.54 × (36/37)²³.

    -$26.54 is the sum E.V. of 24 levels = (36÷37)²⁴.
    total bet $982 × (-1/37) = total E.V. -$26.54

    how can you multiply 23 losses by the $26.54 sum of 24 losses?

    to get -$14.13 on the 24th "cumulative level E.V".
    You done this on bet level 2:
    $1.59 ×(36/37)¹ = -$1.55

    You multiply the Pr before a level with the cumulative E.V. on that level.

    You done this on bet level 24:

    $26.54 × (36/37)²³ = -$14.13

    Column 14, Bet levels 2-24 is wrong.

    You're clearly not going to respond your incorrect formula. So I'm highlighting for the rest of the internet.
     
    Last edited: Jun 7, 2022
  19. TwoUp

    TwoUp Well-Known Member

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    Q. What decides that we bet at level 24 and realise the expected value at that level?
    A. 23 prior losses.

    Q. What is the probability of 23 losses in a row?
    A. (36/37)^23 = 53.24%

    Q. What is the probability of 24 losses?
    A. (36/37)^24 = 51.81%

    Q. What is the probability of making a bet after 24 losses?
    A. Zero. After 24 bets the series is complete and $982 loss has been realised.

    All as per the table.

    The table clearly shows the cumulative EV in column 10, and the total at the bottom of column 9. This represents the total house edge when ALL 24 bets are played.

    Similarly the $982 loss is shown at the bottom of column 3 and column 4. This represents the total when you lose ALL 24 bets. As per my original post.

    As per my original post there is little difference from a regular EC when all 24 bets lose. The series loses 51.8% of the time and wins 48.2%. Logically not all wins occur on the last bet at level 24 (with a probability of just 1.44%), wins do occur at the other levels.

    As per my original post, one does not always lose all bets and the difference in house edge is based on what is risked on average when we win. A win at any level the betting stops and the full $982 is not risked.

    Column 14 clearly shows the expected value based on the probability of reaching that level.

    The average of column 14 "Cum. level E.V." is the average house edge paid, taking into account the probablity of reaching evey level in the aeries and is shown at the end of the column.

    @6probability9 If you can't understand that a bet can actually win, and that continuing to bet is CONDITIONAL on the past bets losing no-one can help you. You have been indulged over and over again and are trolling at this point.
     
    Last edited: Jun 7, 2022
  20. 6probability9

    6probability9 Member

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    You have two outcomes that decides whether you bet at level 24.
    (36÷37)²³ = 53.25%
    (36÷37)²² × (1÷37)¹ = 1.48%

    If you win the later bet on level 23 this obviously influences the cummulative E.V. of the total series.
    W @ lvl23 = $926 × (-1/37) = -$25.03 E.V. the -E.V. of each session is calculated in column 10.

    After losing @ level 23 you don't multiply; -$26.54 × (36÷37)²³ = -$14.13. Wtf.

    Your system is a progression, to find the negative expected value at any level. You multiply the total bet × (-HE). The negative E.V. of all bet levels = -$982 × (-1/37) = $26.54.

    column 14 is honestly a load of shite.

    What is the probability of the first loss on level 1? It's not 100%.

    progression is risking more capital to chase profit without progression you could achieve 37 levels. that makes profit from bet levels 1-34.

    What the hell your series of Ls begins from 97.3% to 51.8% you only realise 48.2% Pr when the overall 51.8% is achieved. If you lost at level 24 in reality you had a 48.2% overall Pr of not losing 24 times in a row. If you lose @ lvl24 this meant your session fell in the 51.8% of the events. If you won before level 24 your overall Pr is less than 48.2%.
    Jesus.
    EC has a base 48.65% to win each consecutive bet.

    column 14 does not calculate the E.V. @ levels 2-24. you've obviously conflated bet E.V. as + & - values. column 10 is all negative this is the true values of your bet E.V.

    If you need to seek validation after the point I've repeatedly scrutinised your formula to calculate your "level E.V" and a non-existent -$8.10 average.
    column 10 × column 11 = column 14
    Go ahead.

    Your bet series still loses more than EC + LP.

    51.8 ÷ 100 × 982 = 508.76
    50 ÷ 100 × 1000 = 500
    50 ÷ 100 × 982 = 491
     
    Last edited: Jun 7, 2022

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