I dented house edge with math (not trolling)
lets say the house edge of this dice game is 1% meaning you as the player have a 49.5% chance to double your money and therefore the house has a 50.5% chance to take your money.
if you place wagers of $0 until you get a desired loss streak then set your bet to $1 or whatever you want the base bet to be then you have basically stretched the odds of actually winning or losing any money to a much smaller amount it seems.
for example: lets say you place $0 bets until a 3 loss streak that means you have 5 possible outcomes in a run of this method.
1. 49.5% chance to WIN
2. ~24.99% chance to LOSE+WIN
3. ~12.62% chance to LOSE+LOSE+WIN
4. ~6.37% chance to LOSE+LOSE+LOSE+WIN
5. ~6.5% chance to LOSE+LOSE+LOSE+LOSE
and remember youre only betting money when you lose 3 in a row so only outcomes 4 and 5 actually gain or lose you money.
now all you have to do to calculate the house edge is add up the probability of all your winners and subtract them by the probability of all your losers. in this case you have a 6.37% chance to win $1 and a 6.5% chance to lose $1 so the equation would look like this
((0.0637)(1)) + ((0.065)(-1)) = -0.0013 or 0.13% house edge which is much lower than the default house edge of 1%
you can increase the amount of losses required before you place your base bet to lower the house edge even further so for example if you hypothetically waited for loss streaks of 10 before you placed your base bet your house edge would be extremely small but it would never flip into your favor it would just infinitely approach 0.
now i wouldnt be surprised if there was a simple oversight that explained all this but if there isnt i think its pretty neat.
Quote: wawoim not sure what you mean can you explain how this doesnt effect the house edge
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ChatGPT 4 gave a good explanation -
Quote:Betting systems do not change the house edge but can change the variance of a game due to the way they manipulate bet sizes and patterns. It's important to understand the distinction between the two concepts: house edge and variance.
House Edge:
The house edge is the built-in advantage that the casino has in a game. It is a long-term measure that represents the average percentage of each bet that the casino expects to win over an extended period of play. The house edge is determined by the rules and probabilities of a particular game and cannot be changed by betting systems.
For example, in American roulette, the house edge is 5.26% because there are 38 possible outcomes (1 to 36, 0, and 00), but the payout for a single-number bet is 35 to 1. This discrepancy between the true odds and the payout odds gives the casino a long-term advantage.
Betting Systems:
Betting systems, such as the Martingale or Fibonacci system, are strategies that players use to manage their bets. These systems usually involve adjusting bet sizes based on previous outcomes, often with the goal of recouping losses or capitalizing on winning streaks. However, they do not change the underlying probabilities or house edge of a game.
Variance:
Variance refers to the short-term fluctuations in a player's results, which can deviate significantly from the expected long-term outcome (house edge). Betting systems can change the variance by altering the bet sizes and patterns, leading to different sequences of wins and losses.
For example, the Martingale system involves doubling your bet after every loss. This can create higher short-term variance, as a player may experience a series of losses followed by a big win that recoups all previous losses. However, it does not change the house edge, and in the long run, the player is still expected to lose at the same rate.
In summary, betting systems can change the variance of a game, leading to fluctuations in short-term results. However, they do not impact the house edge, which remains constant and represents the casino's built-in advantage over the long term. It is essential to remember that no betting system can overcome the house edge, and players should be cautious when using these strategies.
lets say your base bet is $1 and again you have 49.5% chance to double your bet and 50.5% chance to lose your bet and lets also say for simplicity's sake you only have a balance of $7 so youre only allowed 3 losses that would give you 4 outcomes
1. 49.5% chance to WIN
2. ~24.99% chance to LOSE+WIN
3. ~12.62% chance to LOSE+LOSE+WIN
4. ~12.88% chance to LOSE+LOSE+LOSE
outcomes 1, 2, and 3 are a $1 profit and outcome 4 is a $7 loss so to find the house edge you do this:
((0.495)(1)) + ((0.2499)(1)) + ((0.1262)(1)) + ((0.1288)(-7)) = -0.0305 or 3.05% house edge
when you plot out the common betting systems like this you find that the house edge gets bigger not smaller but the one ive proposed seems to be an exception so i would like an explanation why im mistaken.
Quote: wawook but can you explain specifically why this specific system is flawed?
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Quote:The gambler's fallacy, also known as the Monte Carlo fallacy or the fallacy of the maturity of chances, is a cognitive bias or erroneous belief that occurs when a person mistakenly assumes that the probability of a certain outcome in a random event is affected by the outcomes of previous events.
The fallacy stems from a misunderstanding of the concept of independence in probability theory. In independent events, like coin flips, dice rolls, or roulette spins, the outcome of one event has no bearing on the outcome of subsequent events. The probabilities remain constant and are not influenced by past outcomes.
For example, in the case of a fair coin toss, the probability of getting heads or tails is always 50%. If a person witnesses a series of 5 heads in a row, the gambler's fallacy might lead them to believe that the next toss is more likely to result in tails, as a way to "balance out" the previous outcomes. However, the reality is that the probability for the next flip remains 50% for both heads and tails, regardless of prior outcomes, because each flip is an independent event.
The gambler's fallacy often affects people's behavior in gambling situations, leading them to make irrational decisions based on the belief that a certain outcome is "due" or "overdue" to occur after a series of other outcomes. This erroneous belief can result in significant losses, as it ignores the fundamental principles of probability and randomness.
Quote: wawoif you compare this system to a martingale for example you might see my point a little. a simplified martingale run would look like this.
lets say your base bet is $1 and again you have 49.5% chance to double your bet and 50.5% chance to lose your bet and lets also say for simplicity's sake you only have a balance of $7 so youre only allowed 3 losses that would give you 4 outcomes
1. 49.5% chance to WIN
2. ~24.99% chance to LOSE+WIN
3. ~12.62% chance to LOSE+LOSE+WIN
4. ~12.88% chance to LOSE+LOSE+LOSE
outcomes 1, 2, and 3 are a $1 profit and outcome 4 is a $7 loss so to find the house edge you do this:
((0.495)(1)) + ((0.2499)(1)) + ((0.1262)(1)) + ((0.1288)(-7)) = -0.0305 or 3.05% house edge
when you plot out the common betting systems like this you find that the house edge gets bigger not smaller but the one ive proposed seems to be an exception so i would like an explanation why im mistaken.
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You're betting system flaw is the fact you are calculating the house edge based on streaks as if each dice roll is not an independent event.
The house edge on the next roll does not change based on the outcome of previous rolls. No matter how many "streaks" you wait until you bet, the house edge on the next roll in your hypothetical example is going to be 0.5%, no matter what, the thing with any of these betting systems is that they "work" until they don't. What ends up happening if you say....just keep getting a losing roll after you do your bet after 3 loss streaks? Even after 10 in a row being your entry point, what happens if you keep getting a losing roll?
"Well that won't happen" you might say, I'm sure the people that have lost all their hands in a blackjack shoe have said the same thing, if you keep playing long enough, the best of the best streaks will happen to you, and the worst of the worst will happen as well.
Quote: wawothe intention isnt for the system to "work" i already explained how you cant beat the house edge i just want to know if i successfully lowered it or not and at first glance i had the same thought that the streaks of $0 wagers wouldnt matter and id be left with exactly 1% house edge but thats not what happened. i plugged in the variables into a known equation and this is the result how do you explain that?
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You are calculating 5 independent events as if they were a single independent event
Quote: wawothe intention isnt for the system to "work" i already explained how you cant beat the house edge i just want to know if i successfully lowered it or not and at first glance i had the same thought that the streaks of $0 wagers wouldnt matter and id be left with exactly 1% house edge but thats not what happened. i plugged in the variables into a known equation and this is the result how do you explain that?
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You want to know if you've lowered the house edge (or not).
Do you have more ways to win?
Does the house have fewer ways to win?
Quote: wawoi guess we both have fewer ways to win? im not exactly sure id just like to know whats supposedly wrong with the math. is house edge NOT calculated by summing the probability of each outcome multiplied by the return of each outcome???
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The problem with the math is that dice have no memory.
The dice can't remember that they just won for the house three times in a row, and now the players are due for a win.
((0.495)(0)) + ((0.2499)(0)) + ((0.1262)(0)) + ((0.0637)(1)) + ((0.065)(-1)) = -0.0013
what am i getting wrong explain with math please do not explain with words i know it seems to make no sense when i just tell you the system but it works on paper.
.495
.2475
.12375
.061875
.0309375
... and past posting is a no-go.
edit: i have some errors too. Wait a moment please.
(0.505)^4 = 0.0309375??????? what?????? how are you an admin???????
Quote: wawo?????????????????????????
(0.505)^4 = 0.0309375??????? what?????? how are you an admin???????
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I'm strictly here because of my looks and charm, not my keen ability to grind numbers.
In your scheme, you abstain from betting until the house has won 3 rounds in a row.
This means you sit out about 87% of the rounds, and play about 12.8% of the rounds.
In the rounds you play, I don't see how your chances improve beyond 49.5% chance of a win.
these are the same as 'never happened'Quote: wawo
((0.495)(0)) + ((0.2499)(0)) + ((0.1262)(0))
what we expect to see here is the probability of winning a single bet times amount won minus the probability of losing times the amount lost ... yet we see the wrong probability with no downside.Quote:+ ((0.0637)(1))
again describing something odd, and only a losing possibilityQuote:+ ((0.065)(-1)) = -0.0013
So I get it that you are looking at the sequence of bets as a whole ... but you can't place a bet on a sequence. You can only place a bet on one event at a time. Is this where you go wrong?
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I have grappled with the same issue as the OP
the house's chance of winning 4 in a row are very low
but once the house has won 3 in a row the chance of the house winning the 4th bet - meaning 4 in a row - are no longer anywhere near so low
it does seem kinna of strange - but the house's chances change when looking before any of the events happen to looking at it after 3 events
it's kinna like - an American male - when born has a life expectancy on average of about 73 years
but once he reaches age 70 his life expectancy is now about 84 years
.
Quote: lilredrooster.
it's kinna like - an American male - when born has a life expectancy on average of about 73 years
but once he reaches age 70 his life expectancy is now about 84 years
.
that's because he's survived the decrements of the first 70 years.
Quote: lilredrooster
but once the house has won 3 in a row the chance of the house winning the 4th bet - meaning 4 in a row - are no longer anywhere near so low
it does seem kinna of strange - but the house's chances change when looking before any of the events happen to looking at it after 3 events
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That’s not correct though, what you are describing is the gamblers fallacy exactly.
The house’s “chances” are the same for every single independent random event. Nothing that happened before can change that.
The error is so trivial I cannot believe there haven't been many correct responses already. You are not betting at all on sequence #5. You are betting one time after a sequence of three losses. Case #4 and case #5 represent the two outcomes of one wager. You are calculating the EV of the one wager but not dividing by the fraction of times you make the wager to get a house edge.Quote: wawoim trying to think of a reason why this doesnt work but it seems to check out on paper. it has no practical uses but if it is indeed the case you should be able to lower the house edge infinitely close to 0 on dice games with this method.
lets say the house edge of this dice game is 1% meaning you as the player have a 49.5% chance to double your money and therefore the house has a 50.5% chance to take your money.
if you place wagers of $0 until you get a desired loss streak then set your bet to $1 or whatever you want the base bet to be then you have basically stretched the odds of actually winning or losing any money to a much smaller amount it seems.
for example: lets say you place $0 bets until a 3 loss streak that means you have 5 possible outcomes in a run of this method.
1. 49.5% chance to WIN
2. ~24.99% chance to LOSE+WIN
3. ~12.62% chance to LOSE+LOSE+WIN
4. ~6.37% chance to LOSE+LOSE+LOSE+WIN
5. ~6.5% chance to LOSE+LOSE+LOSE+LOSE
and remember youre only betting money when you lose 3 in a row so only outcomes 4 and 5 actually gain or lose you money.
now all you have to do to calculate the house edge is add up the probability of all your winners and subtract them by the probability of all your losers. in this case you have a 6.37% chance to win $1 and a 6.5% chance to lose $1 so the equation would look like this
((0.0637)(1)) + ((0.065)(-1)) = -0.0013 or 0.13% house edge which is much lower than the default house edge of 1%
you can increase the amount of losses required before you place your base bet to lower the house edge even further so for example if you hypothetically waited for loss streaks of 10 before you placed your base bet your house edge would be extremely small but it would never flip into your favor it would just infinitely approach 0.
now i wouldnt be surprised if there was a simple oversight that explained all this but if there isnt i think its pretty neat.
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The probability that you get to make this one bet is 0.128787625. This is the sum of 0.06374987438 and 0.06503775063 but is also just equal to 0.505^3.
0.06503775063 / 0.06374987438 = 1.02020202 = 0.505 / 0.495
You have not changed the EV by one iota. You have just confused yourself by making the situation more complicated than you can handle mathematically. This is how most betting systems work.
Quote: gamerfreakQuote: lilredrooster
but once the house has won 3 in a row the chance of the house winning the 4th bet - meaning 4 in a row - are no longer anywhere near so low
it does seem kinna of strange - but the house's chances change when looking before any of the events happen to looking at it after 3 events
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That’s not correct though, what you are describing is the gamblers fallacy exactly.
The house’s “chances” are the same for every single independent random event. Nothing that happened before can change that.
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what I meant - which I guess I didn't make clear is
before any rolls - the house's chance of winning 4 in a row is very low
after the house has won 3 in a row - if you look backwards at that sequence - the house's chance of winning the next bet are much greater than their chance of winning 4 in a row before any rolls were played
.
Quote: lilredroosterQuote: gamerfreakQuote: lilredrooster
but once the house has won 3 in a row the chance of the house winning the 4th bet - meaning 4 in a row - are no longer anywhere near so low
it does seem kinna of strange - but the house's chances change when looking before any of the events happen to looking at it after 3 events
link to original post
That’s not correct though, what you are describing is the gamblers fallacy exactly.
The house’s “chances” are the same for every single independent random event. Nothing that happened before can change that.
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what I meant - which I guess I didn't make clear is
before any rolls - the house's chance of winning 4 in a row is very low
after the house has won 3 in a row - if you look backwards at that sequence - the house's chance of winning the next bet are much greater than their chance of winning 4 in a row before any rolls were played
.
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The chance to win one round should always be much greater than the chance to win 4 in a row, yes?
if you did a 10 stage martingale on an even chance bet trying to win $100 your tenth bet would be $51,200
if you lose that last bet you will have lost $102,300
in search of a $100 profit
.
Quote: lilredrooster.
if you did a 10 stage martingale on an even chance bet trying to win $100 your tenth bet would be $51,200
if you lose that last bet you will have lost $102,300
in search of a $100 profit
.
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It's not about the money, it's about the principal.
Quote: wawook but can you explain specifically why this specific system is flawed?
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Specifically, your betting scheme reduces the EV of the entire sequence. By not betting at all, you could reduce the EV to zero. The EV is an absolute number of bets you expect to win or lose during the sequence. The house edge is is the EV divided by the amount wagered. Percentage is a unitless metric, so the units of the numerator and denominator must cancel out. EV has units of bets or credits or chips or dollars. House edge has no units.
You used a known equation for EV. EV is not equivalent to house edge. Divide the EV by the amount bet per sequence (0.128787625 units) to get house edge.Quote: wawothe intention isnt for the system to "work" i already explained how you cant beat the house edge i just want to know if i successfully lowered it or not and at first glance i had the same thought that the streaks of $0 wagers wouldnt matter and id be left with exactly 1% house edge but thats not what happened. i plugged in the variables into a known equation and this is the result how do you explain that?
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Put this formula into a spreadsheet:
=(0.495*0.505*0.505*0.505*(1)+0.505*0.505*0.505*0.505*(-1)) / (0.495*0.505*0.505*0.505+0.505*0.505*0.505*0.505)
The result is -0.01 or a house edge of 1%. Dang, the magic is gone.
Quote: MentalYou used a known equation for EV. EV is not equivalent to house edge. Divide the EV by the amount bet per sequence (0.128787625 units) to get house edge.Quote: wawothe intention isnt for the system to "work" i already explained how you cant beat the house edge i just want to know if i successfully lowered it or not and at first glance i had the same thought that the streaks of $0 wagers wouldnt matter and id be left with exactly 1% house edge but thats not what happened. i plugged in the variables into a known equation and this is the result how do you explain that?
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Put this formula into a spreadsheet:
=(0.495*0.505*0.505*0.505*(1)+0.505*0.505*0.505*0.505*(-1)) / (0.495*0.505*0.505*0.505+0.505*0.505*0.505*0.505)
The result is -0.01 or a house edge of 1%. Dang, the magic is gone.
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You are neglecting the Tinkerbell Effect.
Quote: billryanQuote: lilredrooster.
if you did a 10 stage martingale on an even chance bet trying to win $100 your tenth bet would be $51,200
if you lose that last bet you will have lost $102,300
in search of a $100 profit
.
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It's not about the money, it's about the principal.
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billryan,
Oh... I thought it was about the interest! ;-)
Dog Hand
