this post was submitted on 16 Jan 2024
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No Stupid Questions

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Ok so here's the rules

  • I just bet on red every time
  • I start with 1 dollar
  • every time I lose, I triple my previous bet
  • every time I win I restart

I'm going to simulate 10 games

  • Game 1 - Bet $1 Lose
  • Game 2 - Bet $3 Lose
  • Game 3 - Bet $9 Win $18
  • Game 4 - Bet $1 Lose
  • Game 5 - Bet $3 Lose
  • Game 6 - Bet $9 Win $18
  • Game 7 - Bet $1 Lose
  • Game 8 - Bet $3 Lose
  • Game 9 - Bet $9 Lose
  • Game 10 - Bet $18 Win $36

In this simulation I'm losing at a rate of 70%. In reality the lose rate is closer to 52%. I put in $54 but I'm walking away with $72, basically leaving the building with $18.

Another example. Let's pretend I walk in with $100,000 to bet with. I lose my first 10 games and win the 11th.

  • 1 lose
  • 3 lose
  • 9 lose
  • 27 lose
  • 81 lose
  • 243 lose
  • 729 lose
  • 2187 lose
  • 6561 lose
  • 19683 lose
  • 59049 win $118098

$88573 spent out of pocket, $118098 won

Walk out with roughly $29525.

I get most casinos won't let you be that high but it's a pretty extreme example anyway, the likelyhood of losing 10/11 games on 48% odds is really unlikely.

So help me out here, what am I missing?

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[–] DScratch@sh.itjust.works 148 points 10 months ago (4 children)

This is the Martingale system, it will 100% work if you have infinite money and can bet infinite amounts.

[–] teft@lemmy.world 62 points 10 months ago* (last edited 10 months ago) (2 children)

It won’t work on any table that has a 0 and/or 00. The reason those green numbers exist is to give the house an edge over high rollers who can sustain a martingale strategy.

[–] bluGill@kbin.social 62 points 10 months ago

It works even with 0 and 00 - but only if you have infinite money and there is no limit to bets. So long as there is finite money, or limited bets you will - statistically - hit a run of losses that exceeds your money.

Maximum bets and the fact that money is limited is what breaks this, once you hit the maximum bet odds are slightly against you winning on that next turn because of the 0 and 00; and so eventually you will lose as you hit running up to the max bet and then lose on that turn.

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[–] Valmond@lemmy.mindoki.com 38 points 10 months ago

Me: "bets 1 infinity dollars"

Loses

Me: "bets 3 infinities of dollars"

[–] A_A@lemmy.world 3 points 10 months ago

You can't have an infinite amount since past a certain mass it will collapse onto itself and create a black hole. (or other practical limits) So eventually you lose anyway.

[–] take6056@feddit.nl 2 points 10 months ago

I thought this was a pretty good in-depth explanation of the infinite case and the finite case: https://www.youtube.com/watch?v=zTsRGQj6VT4

[–] dhork@lemmy.world 105 points 10 months ago

It's been tried before, the House always wins.

https://en.m.wikipedia.org/wiki/Martingale_(betting_system)

Roulette in particular has no skill involved, only various table bets, all of which have a House edge. The only guarantee in Roulette is that if you play long enough, you will lose all of your money. The only thing you are in control of is how much money you bring to the table, and when you stop.

[–] m0darn@lemmy.ca 76 points 10 months ago (1 children)

Another way of thinking about it is betting your entire bankroll for 99.9....% certainty that you will win $1.

Say you go into the casino with $1000.

Bet:

$1 lose.
$3 lose.
$9 lose.
$27 lose.
$81 lose.
$243 lose.
$729 oh wait you can't bet that much, you only have $457 left. Dang, do you bet $457 or find another $272? 

Bet $457 and you win $914! Congrats you're now only down $86!
Or maybe you lost and are down $1000.

Or maybe you scrounged up $272 so you could keep playing
Bet 729 and lose. Now you're down $1272.
Or
Bet 729 and you win 1458. Pay back the $272 you borrowed from your buddy, you're still up $186. 
You just bet $729 dollars for a %50 chance of winning $186.

But what are the chances of getting 6 or 7 losses in a row? 1 in 64, or 128 respectively, actually worse because roulette wheels aren't 50/50, they're 18/19 (18 wins and 19 losses in 37 plays on average) or worse. So losing 6 times in a row will happen 1 in 54 plays, 7 losses is 1 in 106.

Google says roulette wheels spin 55 times per hour so with your strategy you will lose your bank roll in about one hour assuming your starting bet is 0.1% of your bank roll.

[–] Anticorp@lemmy.world 5 points 10 months ago

I tried the Martingale Strategy in my early 20's, thinking I had invented a system that would actually beat the house. I was playing $1 roulette, and had a budget of $100. I did manage to win $400ish, and then I lost 12 times in a row. I busted on the 9th loss, and went back to betting $1 just to see how long it would take to win. I would have needed $4096 to stay in the positive, and I would have won $1 on that last bet.

[–] patatahooligan@lemmy.world 50 points 10 months ago (4 children)

So help me out here, what am I missing?

You're forgetting that not all outcomes are equal. You're just comparing the probability of winning vs the probability of losing. But when you lose you lose much bigger. If you calculate the expected outcome you will find that it is negative by design. Intuitively, that means that if you do this strategy, the one time you will lose will cost you more than the money you made all the other times where you won.

I'll give you a short example so that we can calculate the probabilities relatively easily. We make the following assumptions:

  • You have $13, which means you can only make 3 bets: $1, $3, $9
  • The roulette has a single 0. This is the best case scenario. So there are 37 numbers and only 18 of them are red This gives red a 18/37 to win. The zero is why the math always works out in the casino's favor
  • You will play until you win once or until you lose all your money.

So how do we calculate the expected outcome? These outcomes are mutually exclusive, so if we can define the (expected gain * probability) of each one, we can sum them together. So let's see what the outcomes are:

  • You win on the first bet. Gain: $1. Probability: 18/37.
  • You win on the second bet. Gain: $2. Probability: 19/37 * 18/37 (lose once, then win once).
  • You win on the third bet. Gain: $4. Probability: (19/37) ^ 2 * 18/37 (lose twice, then win once).
  • You lose all three bets. Gain: -$13. Probability: (19/37) ^ 3 (lose three times).

So the expected outcome for you is:

$1 * (18/37) + 2 * (19/37 * 18/37) + ... = -$0.1328...

So you lose a bit more than $0.13 on average. Notice how the probabilities of winning $1 or $2 are much higher than the probability of losing $13, but the amount you lose is much bigger.

Others have mentioned betting limits as a reason you can't do this. That's wrong. There is no winning strategy. The casino always wins given enough bets. Betting limits just keep the short-term losses under control, making the business more predictable.

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[–] Melobol@lemmy.ml 50 points 10 months ago* (last edited 10 months ago) (2 children)

To make things worse, Vegas now have triple 0 roulette tables. Three zeros in a row. In the same place. Do not play Roulette if you want to win. The only bet in a casino isn't to the casinos favor is on Craps and it's called ODDS. But to get to able to play it you have to win against 1.41% house edge first. And it is capped about 10x or 5x of original bet.

If you want to have the best odds against a casino that is hard work and not solid strategy. Counting cards can make money but honestly the amount of work it requires - you are better on the stock market or even in crypto.

Do not gamble against the casino.

The only reason to gambe is to have fun and safe risk with money you decided to spend and to never see again.

[–] ryathal@sh.itjust.works 8 points 10 months ago (1 children)

Haven't auto shufflers killed counting cards now anyway?

[–] LKPU26@lemmy.ml 5 points 10 months ago

The House will use multiple decks, I think up to 8 but I haven't gambled for a few years, + the auto shuffler.

They'll usually deal about 1 decks worth of cards and then pass all cards through the shuffler.

This means that you have to count 8 decks worth of cards, and they reset that before you can really get much of an advantage.

[–] mack7400@lemmy.world 4 points 10 months ago

Vegas now have triple 0 roulette tables

Stupid inflation!

[–] ZagamTheVile@lemmy.world 40 points 10 months ago (1 children)

Remember, betting red isn't a 50/50 chance of winning. 0 and 00 are green. So your odds at winning are about 48% with 0, or 47% with 00. Same with odd/even bets. There's a reason casinos add the 0's. They aren't in the business to give you money.

[–] davidgro@lemmy.world 4 points 10 months ago (1 children)

That's a reason OP went with tripling the bet instead of the more common doubling when people independently discover this system.

[–] ZagamTheVile@lemmy.world 2 points 10 months ago (1 children)

Yes but a 45% win chance is the same long term losing plan no matter how much you bet. I could bet 10 times the amount every time I lose and I'll still have a 45% chance of winning.

[–] davidgro@lemmy.world 1 points 10 months ago

Yes that is true... if you end at a "random" point (running out of money for example). But if you can always end at a win (particularly a large one after a string of many losses) then that changes things, the odds become much less relevant because you are manipulating them.*

The problem is that consistently doing so would require genuinely unlimited money and no betting limits, since the number of losses-that-can-happen-in-a-row is itself unlimited. With that little detail the Martingale system would actually work.
Then again if you have unlimited money then why bother to gamble?

*(Actually I sorta take back my first comment in this thread, the odds aren't why to triple, just growing the pot faster each win is)

[–] MeatsOfRage@lemmy.world 38 points 10 months ago (1 children)

Thanks for all the responses. Out of curiosity I threw together a quick simulator to try out different variations.

https://roulette-test.netlify.app/

Needless to say, you usually blow everything you came in with:

[–] davidgro@lemmy.world 8 points 10 months ago (1 children)

This is good.
For example, with a million dollar pot and tripling the bet, usually less than 1000 games is safe, you'll win a small percentage over the initial amount.

But sometimes...

1000013402

[–] MeatsOfRage@lemmy.world 3 points 10 months ago

Lol, looks terrible on mobile, just updated it to look nicer

[–] rjthyen@lemm.ee 37 points 10 months ago (2 children)

I didn't get through all the comments, but most of the top ones and haven't yet seen the most basic casino check to stop these strategies is a table maximum. Same with doubling your bet each hand you lose at black jack, one bad run and you hit table max and you can't raise your bet far enough

[–] ExLisper@linux.community 2 points 10 months ago

What if I bet on black?

[–] Anticorp@lemmy.world 2 points 10 months ago (1 children)

That's easy though, you just go to higher max table. The hard part is having enough money to sustain the string of losses. And of course the casino sets minimum and max bets at tables, ensuring they don't align with this strategy.

[–] rjthyen@lemm.ee 1 points 10 months ago* (last edited 10 months ago)

That last part is key, higher max table comes with a higher minimum bet so the ratio stays roughly the same. I suppose you could move tables as your bet needed to increase and switch back to the low table if you manage to win one making cash reserves the main limiting factor, but I've always envisioned the table rules are set in a way to prevent someone with deep pockets from exploiting strategies

[–] zkfcfbzr@lemmy.world 36 points 10 months ago* (last edited 10 months ago)

Expanding a bit more on what everyone else says: This strategy works, as long as you never lose n times in a row, where n is the number of bets it takes to bet ALL of the money you currently have.

So the more money you bring with you, the longer you can make this strategy work - but the more devastating it'll be once you inevitably lose.

If you go with a doubling strategy instead of a tripling strategy, that means you have to lose floor(log₂(x+1)) times to realize an unrecoverable loss (you don't have enough to make your next bet), or one more than that to lose absolutely everything. With your tripling strategy the calculation is floor(log₃(2*x+1)). x is the amount of money you had after the last "reset".

So if you go to the casino with $100,000, your strategy will work as long as you don't lose 11 times in a row - once you do, you've suffered your devastating unrecoverable loss. Every time your money triples you can last one more loss. Tripling your money is very difficult with this strategy, as most of the time when you win, it's a small amount relative to what you're holding - you need large losing streaks to make a real difference, and large losing streaks make reaching the threshold of an unrecoverable loss easier, obviously.

Others have said it already, but - you can use this to win in the short term if you have a lot of money and only want to win a little bit more. If you use this strategy in the long term you will lose everything.

[–] bionicjoey@lemmy.ca 35 points 10 months ago

The fundamental theorem of gambling is a mathematically proven fact which states that as long as the house has an edge, there is no betting strategy that has a positive long-term expected value.

[–] solrize@lemmy.world 24 points 10 months ago

The limiting factor is called "Gambler's ruin".

https://en.wikipedia.org/wiki/Gambler%27s_ruin

[–] Cokeser@feddit.de 23 points 10 months ago

With the first rise of online casinos (pre 2010), a friend of mine thought he had rigged the system by more or less the same strategy. I think the usual approach is actually to double the bet.

Well..in the end he won a debt of several 10k€ before he had to quit. This put a huge strain on his early adulthood.

[–] lyth@sh.itjust.works 21 points 10 months ago (1 children)

Congratulations, you've invented the Martingale betting system and are well on your way to becoming an adept probability theorist

The short answer is that casinos account for this by changing the profit returns and the odds of making a profit at all so that catastrophic losses are much more likely

[–] foggy@lemmy.world 6 points 10 months ago (1 children)

Betting limits.

Also, n^3 grows fast.

[–] GamingChairModel@lemmy.world 5 points 10 months ago

It's 3^n, not n^3. n^3 is actually way slower.

[–] db2@lemmy.world 19 points 10 months ago

You're missing the house edge that means over time they'll effectively always win. Gambling is a suckers game.

[–] harrywrecker@kbin.social 17 points 10 months ago

This is what is known in the betting industry as "chasing your losses" and is a warning sign of a problem gambler who requires an intervention.

[–] kometes@lemmy.world 16 points 10 months ago (1 children)

I start with 1 dollar
Game 1 - Bet $1 Lose

You are done at this point, having lost all your money.

[–] MeatsOfRage@lemmy.world 2 points 10 months ago

I meant start the betting at $1. The pot was undefined for the first example.

[–] A_A@lemmy.world 10 points 10 months ago

What you are missing is that each time you are risking more and more. Eventually you risk something bigger than what you have and then you lose it. At this point you lose your life, your family and your friends.

[–] Anticorp@lemmy.world 9 points 10 months ago

It's the Martingale Strategy, except you're speed running the losses by tripling instead of doubling. I invented the Martingale Strategy on my own in my early twenties and thought I had figured out how to beat the house. I was pretty disappointed when I learned that 1. It already existed. 2. Would lose in the long run.

[–] Ainiriand@lemmy.world 8 points 10 months ago

Tables have a bet limit. But there is no limit to how much you can lose.

[–] JoBo@feddit.uk 7 points 10 months ago (1 children)

This isn't exactly the same as the Martingale system (where you double the bet after a loss so that you end up winning the starting bet when your colour does finally come up) but it's the same principle, just messier because you're tripling so it's harder to analyse (but you'll go broke quicker).

In principle, you will usually end up making a small profit while risking very big losses. And that's why it doesn't work. Either you run out of money for the next bet or you reach the house limit (which will almost always exist).

If you go to a casino with an amount of money you are prepared to lose, it's not the worst strategy (given that there are no good strategies with roulette). You will usually get a small win but you will sometimes get a big loss. You'll end up down in the long run.

[–] PopMyCop@iusearchlinux.fyi 3 points 10 months ago* (last edited 10 months ago) (1 children)

making a small profit

Right, this is what people always gloss over to just say that eventually the bet will be too big to sustain. Even if you win repeatedly, the bets you make after 3 or 4 losses are vast in comparison the amount you'll 'gain' per win. For the doubling (Martingale strategy), if your bet starts at $1, and you win $2 off of that, it doesn't matter how much you are eventually betting, you'll only make $1 for the whole cycle.

The tripling helps for the profit angle, somewhat. I ran the numbers for total amount of times betting before a win for net win. I wish the formatting let me make tables, but oh well.

Total Times Bet(bet amount): total of bet: net winning:

1 (1) ... 1 ... 1

2 (3) ... 4 ... 2

3 (9) ... 13 ... 5

4 (27) ... 40 ... 16

5 (81) ... 121 ... 41

6 (243) ... 364 ... 122

7 (729) ... 1093 ... 365

8 (2187) ...3280 ... 1094

9 (6561) ... 9841 ... 3281

10 (19683) ... 29524... 9842

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[–] eek2121@lemmy.world 7 points 10 months ago

In addition to what others have stated, many casinos have both minimums and caps on maximum bets. This means that you can only do this so many times. You WILL eventually lose it all.

I managed to do this once with an electronic machine and made money once because it had a minimum bet of $1.00, no maximum, and apparently wasn’t rigged. I made high 4 figures before I got bored. The next time I came in, the minimum was upped, a maximum was added, and the game was clearly changed to favor the house.

[–] earned_myself_a_gin@lemmynsfw.com 6 points 10 months ago

Assuming you don't have infinite money, and that there are limits at the table, this is a good example of a non-ergotic system, where the mean and the median diverge. Because the rules of the game you defined dictate that at some point, you run out of money and have to stop playing, the outcomes for the median dip below that of the lucky few, who win more than lose owing to random chance. This is a super cool concept in a lot of systems, not just gambling fwiw

Check out https://medium.com/incerto/the-logic-of-risk-taking-107bf41029d3 or https://taylorpearson.me/ergodicity/

[–] jet@hackertalks.com 5 points 10 months ago

Run your simulation to find The expected number of bets before you go bankrupt. That will be illuminating

[–] amio@kbin.social 4 points 10 months ago

Sure, it's a strategy. It will lose you a lot of money and make some casino very happy, but it is that. Sounds like Martingale, Wikipedia has an article that explains why it doesn't work.

[–] feedum_sneedson@lemmy.world 3 points 10 months ago
[–] Serinus@lemmy.world 3 points 10 months ago

You can save money by only betting one dollar on your second bet, since you always win on the third bet.

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