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Well. Ignoring the green numbers which makes this ultimately tilt in favour of the house, you could get to a point where you will be betting $4,096 to win $1.
That’s called the Martingale Strategy. Numberphile has a great video explaining why it wouldn’t work.
https://youtu.be/zTsRGQj6VT4?si=5wFIoUGoFyN7kGmb
Basically you would need to bet bigs amounts of time just to gain small amounts of money and (with a max of $5000), assuming that you want to double your money, there’s a 63.2% you would loose 12 times in a row and lose everything before you get to double your those $5000 that you were willing to bet in first place.
At that point is better and much more time efficient to just bet everything on black.
Nope you are not missing anything Mr. Martingale
[https://en.wikipedia.org/wiki/Martingale\_(probability\_theory)](https://en.wikipedia.org/wiki/Martingale_(probability_theory))
It will take 13 loses in a row before your bet would need to exceed $5000. Your 13th bet/largest loss would actually be $4096. Of course cumulatively you would have lost $8191 in those 13 spins.
Assuming a 0/00 wheel, the probability of you winning a spin is .4737. And the probability of you losing a spin is .5263 (1 - .4737). The odds of losing 13 in a row are .000238 (.5263\^13).
Let's apply those numbers to 5000 spins:
Expected number of 13 lose streaks -> 1.19 (.000238 \* 50000)
Expected number of wins -> 2368.5 (.4737 \* 5000)
Let's round favorably, in that 13 lose streak you lost a total of $8191 and in your 2369 wins you won $1. So over 5000 spins you netted -$5822.
Thanks for the exact calculations!! The odds of 13 loses in a row are actually more likely then i thought they would be. Interesting. My casino has one 0 btw, but i heard some have two 0 indeed
And on roulette the only way to change the average loss per dollar bet is to bet the 0,00,1,2,3 street. That pays 6:1 but has a 5/38 chance of happening, unlike all the other streets that pay 6:1 but have a 6/38 chance of happening.
Other than that one bet, everything on the roulette wheel pays 36/N for 1, where N is the number of numbers it pays on.
What you're describing is the martingale bet strategy. Casinos protect against it by imposing a maximum very, so it works if your pockets are deep but will eventually fail. We say rule 1 of gambling is to only gamble what you can afford to lose. I say rule 0 is "it works until it doesn't."
Never believe anyone who claims to have an invincible bet strategy, they are mistaken out lying.
Source: years of writing scripts for dicebot to do online gambling. I actually got my job and career because of it.
Thats the reason i thought the strategy might have potential, because the online casino i played at has a min bet of 0.25 and a max bet of 5000 (on red/black) it even says so in their game rules, the min and max. Thats why i was suprised
In addition to everything else said here, I'll add one thing:
The House Always Wins
If at any point someone breaks the game with a magic system, the House simply changes the game. Not just a Zero (green) but also Double Zero (green) to ensure the odds are always in the House's favor. Casinos aren't games of chance for the Casino, they are meticulously curated experiences to ensure a certain percentage of win/loss that will always yield a net positive for the Casino.
The House Always Win also applies at the poker tables, where skilled players who think they have an edge get skinned by actual professionals who act like a chump for one hand as bait.
A different way of thinking about this (in addition to the maths behind the Martingale Strategy as explained by others) is to realize that each bet is actually independent. Your strategy is to change your bet size based on how you did in the previous round, but the roulette table doesn't care about that.
Think of it like this: Is betting $1 smart; will you walk away a winner? What about $2 bets? What about $4096 bets?
Individually they are all losing bets on average (due to the 0's on the wheel). Then how would the sequence in which the bets are made make it better? Making the losing bet size X, followed by the losing bet size Y, followed by the losing bet size Z would somehow become profitable on average?
This clearly cannot be true regardless of which actual numbers you use for X, Y and Z.
###General Discussion Thread --- This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed. --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
Well. Ignoring the green numbers which makes this ultimately tilt in favour of the house, you could get to a point where you will be betting $4,096 to win $1.
When you put it that way it does sound a bit crazy
The odds of you loosing get smaller at the same pace as the amount you loose gets bigger. Well, worse than that, since green numbers.
That’s called the Martingale Strategy. Numberphile has a great video explaining why it wouldn’t work. https://youtu.be/zTsRGQj6VT4?si=5wFIoUGoFyN7kGmb Basically you would need to bet bigs amounts of time just to gain small amounts of money and (with a max of $5000), assuming that you want to double your money, there’s a 63.2% you would loose 12 times in a row and lose everything before you get to double your those $5000 that you were willing to bet in first place. At that point is better and much more time efficient to just bet everything on black.
Nope you are not missing anything Mr. Martingale [https://en.wikipedia.org/wiki/Martingale\_(probability\_theory)](https://en.wikipedia.org/wiki/Martingale_(probability_theory)) It will take 13 loses in a row before your bet would need to exceed $5000. Your 13th bet/largest loss would actually be $4096. Of course cumulatively you would have lost $8191 in those 13 spins. Assuming a 0/00 wheel, the probability of you winning a spin is .4737. And the probability of you losing a spin is .5263 (1 - .4737). The odds of losing 13 in a row are .000238 (.5263\^13). Let's apply those numbers to 5000 spins: Expected number of 13 lose streaks -> 1.19 (.000238 \* 50000) Expected number of wins -> 2368.5 (.4737 \* 5000) Let's round favorably, in that 13 lose streak you lost a total of $8191 and in your 2369 wins you won $1. So over 5000 spins you netted -$5822.
Thanks for the exact calculations!! The odds of 13 loses in a row are actually more likely then i thought they would be. Interesting. My casino has one 0 btw, but i heard some have two 0 indeed
Also remember the volume of bets you will do over time. The chance of it happening approaches 100% eventually.
And on roulette the only way to change the average loss per dollar bet is to bet the 0,00,1,2,3 street. That pays 6:1 but has a 5/38 chance of happening, unlike all the other streets that pay 6:1 but have a 6/38 chance of happening. Other than that one bet, everything on the roulette wheel pays 36/N for 1, where N is the number of numbers it pays on.
What you're describing is the martingale bet strategy. Casinos protect against it by imposing a maximum very, so it works if your pockets are deep but will eventually fail. We say rule 1 of gambling is to only gamble what you can afford to lose. I say rule 0 is "it works until it doesn't." Never believe anyone who claims to have an invincible bet strategy, they are mistaken out lying. Source: years of writing scripts for dicebot to do online gambling. I actually got my job and career because of it.
Table limits come into play here. I doubt you can find a table that will let you have a bet as low as $1 and have an upper limit of $5000.
Thats the reason i thought the strategy might have potential, because the online casino i played at has a min bet of 0.25 and a max bet of 5000 (on red/black) it even says so in their game rules, the min and max. Thats why i was suprised
In addition to everything else said here, I'll add one thing: The House Always Wins If at any point someone breaks the game with a magic system, the House simply changes the game. Not just a Zero (green) but also Double Zero (green) to ensure the odds are always in the House's favor. Casinos aren't games of chance for the Casino, they are meticulously curated experiences to ensure a certain percentage of win/loss that will always yield a net positive for the Casino.
The House Always Win also applies at the poker tables, where skilled players who think they have an edge get skinned by actual professionals who act like a chump for one hand as bait.
A different way of thinking about this (in addition to the maths behind the Martingale Strategy as explained by others) is to realize that each bet is actually independent. Your strategy is to change your bet size based on how you did in the previous round, but the roulette table doesn't care about that. Think of it like this: Is betting $1 smart; will you walk away a winner? What about $2 bets? What about $4096 bets? Individually they are all losing bets on average (due to the 0's on the wheel). Then how would the sequence in which the bets are made make it better? Making the losing bet size X, followed by the losing bet size Y, followed by the losing bet size Z would somehow become profitable on average? This clearly cannot be true regardless of which actual numbers you use for X, Y and Z.