0
$\begingroup$

Firstly, I totally get that a single bet on red and black is 50% regardless of anything, period.

However, check out this scenario.

I consider betting 8 times. I want to start with the first spin and double up after each loss as I walk up to the roulette wheel. God is punishing me from above and unknowingly, 8 losses will occur and I will lose the lot, but I change my mind on the first 4 spins and don't put the chips down so don't lose any money. I start betting from the 5th spin with my initial amount.

Surely I have saved the losses from those first 4 spins and this is the advantage of the Martingale system, right?

I get to chose( not randomly) from a series of numbers where the probability of an event happening is small. I'm not just betting on a single 50/50 ball, I'm betting on an event that has small chance and where the reward outweighs the risk because of that picking up of the first 4 chips.

I have other tips, but have to keep them to myself for now. Will let you guys know if the ideas fail. hehe.

Open for comment or correction.

$\endgroup$
  • $\begingroup$ A single bet on red or black is less than 50%. 0 and 00 are neither black, red, even or odd. The casino wants to win, not just to break even. $\endgroup$ – kviiri Dec 15 '15 at 13:31
  • $\begingroup$ In addition to the green 0, casinos have a maximum bet. So even if you have enough money to double each time, the casino won't allow you to bet this much money $\endgroup$ – stity Dec 15 '15 at 13:40
  • $\begingroup$ Is it your theory that God manipulates the wheel for the explicit purpose of making you win or lose, and that God's disfavor lasts for only eight spins of the wheel, after which God will relent and give you a winning spin? That's not how roulette wheels are generally considered to work. Or if you mean that you initially decided to quit after 8 bets but then predetermined that you would only bet at most 4 times, you reduce the maximum loss (the risk) but you also reduce the chance of winning (the reward). $\endgroup$ – David K Dec 15 '15 at 14:34
0
$\begingroup$

(for this answer I'm assuming a double-zero Roulette table - 38 pockets with 0 and 00 which lose for all outside bets of red, black, even and odd)

I consider betting 8 times. I want to start with the first spin and double up after each loss as I walk up to the roulette wheel.

How many times you consider betting a priori has no effect on the outcome. But to the question itself...

God is punishing me from above and unknowingly, 8 losses will occur and I will lose the lot, but I change my mind on the first 4 spins and don't put the chips down so don't lose any money. I start betting from the 5th spin with my initial amount. Surely I have saved the losses from those first 4 spins and this is the advantage of the Martingale system, right?

This just doesn't work. I'll start with the intuition, and follow it up with the math.

First of all, let's assume for a moment it makes sense to stop doubling your bet after four losses. But why does it have to be exactly four to tip the scales in your favor? Why not three? And if three, why not two? And if two, why not one? At which point you're back at the second simplest gambling system there is - simple fixed bet. (the simplest is not to play, and has a better expected outcome than other systems in typical games)

The mathier part: So after four losses, you've made a loss of $15x$, where $x$ is the size of your initial bet. You argue that a fresh start after four consecutive losses limits your losses enough for you to have a decent probability of recovering. Since every win, regardless of the amount of losses it follows, yields exactly $x$ in profit, we only need to count how many wins we're likely get before landing a streak of four consecutive losses.

Probabilities of possible outcomes for a series of at least 0 losses followed by a win, up to at least 4 consecutive losses:

$P(W) = \frac{18}{38} \approx 47\%$

$P(L, W) = \frac{20}{38} \times \frac{18}{38} \approx 25\%$

$P(L, L, W) = (\frac{20}{38})^2 \times \frac{18}{38} \approx 13\%$

$P(L, L, L, W) = (\frac{20}{38})^3 \times \frac{18}{38} \approx 7\%$

$P(L, L, L, L) = (\frac{20}{38})^4 \approx 8\%$

Now, if we model the situation as a geometric distribution where each trial is the sequence of rolls up to a win or four losses, we can calculate the expected number of wins before landing a four-loss streak:

$E(wins) = (1 -(\frac{20}{38})^4) / (\frac{20}{38})^4 \approx 12$

So you're expected to win $12x$ before hitting the first series of four consecutive losses... which will cost you $15x$, so you're likely to be making a loss. Feel free to try this strategy if you ever happen by my casino, though!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.