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?

  • patatahooligan@lemmy.world
    512·
    10 months ago

    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.

    • TooPoor@lemmy.world
      57·
      10 months ago

      All your math above is irrelevant and your last paragraph is completely wrong. Table limits are the reason this strategy won’t work…

      If you have a large enough bank roll and continuously double your bet after a loss, you can never lose without a table limit.

      Source: data analyst.

      • patatahooligan@lemmy.world
        91·
        10 months ago

        If you have a large enough bank roll and continuously double your bet after a loss, you can never lose without a table limit.

        Unless your bank roll is infinite, you always lose in the average case. My math was just an example to show the point with concrete numbers.

        In truth it is trivial to prove that there is no winning strategy in roulette. If a strategy is just a series of bets, then the expected value is the sum of the expected value of the bets. Every bet in roulette has a negative expected value. Therefore, every strategy has a negative expected value as well. I’m not saying anything ground-breaking, you can read a better write-up of this idea in the wikipedia article.

        If you don’t think that’s true, you are welcome to show your math which proves a positive expected value. Otherwise, saying I’m “completely wrong” means nothing.

      • RvTV95XBeo@sh.itjust.works
        61·
        10 months ago

        This is just factually incorrect. Losing is always an option, even if you have a billion dollars so you can double your bet ~30 times, you could still lose the weighted coin toss 30 times. On top of that, all the money that’s ever existed still only gets you like 50 spins on the roulette table, exponential growth is a removed. OPs strategy of tripling their bet only goes for about 13 rounds before OP is slinging around a million dollars. Even if they just won the lottery, round 19 puts them out a cool billion.

        Also this strategy is foolish in so many ways because you’re just playing to hopefully break even, and like everything in a casino, the odds of you breaking even before running out of cash ALWAYS favors the house.

        Source: Statistics.