Why would this martingale not work?

I am trying to understand the math and probability behind progressions and why they wont work in the ling run.

I have the follow (unrealistic) scenario;

I have a 1000 unit bankroll

I am a wizard and I can be in multiple casinos at the same time.

I love roulette and I like to bet one unit when an EC hasnt occured for 12 times (e.g. I bet on red after 12 blacks).

If my bet wins Ill start again and wait for next opportunity, while if it looses I ll implement a 6 step Martingale.
If my bet looses at the 18th spin I will start again from one.

My set max loss is 64 units for progression( 32+ 16 + 8 + 4 + 2 + 1). Thats 6.4% of my bankroll.

Putting aside the unfeasability and slowness of this, how long would it take before I go bust? I think 19 occurences in a row is like 1 in 200k possibility.

Quote: Ispendallinbitc

Putting aside the unfeasability and slowness of this, how long would it take before I go bust? I think 19 occurences in a row is like 1 in 200k possibility.

The chance of getting to 19 keeps decreasing after each spin. The chance of 19 reds in a row is indeed huge, 1 in 1.7 million (I think: 1 / (0.47)^19). But if you only wait until after one red has been hit to start counting, it drops to 1 in 800,000 ... if you wait until after two reds to start counting, the chance of seeing 19 in row is 1 in 375,000 ... and so on. If you watch every casino in the world and only start watching until red has been hit 12 times, you're expected to see a streak of 19 reds once every 197 times you start watching

(my numbers may be off a little because I was using 0.47 for the chance of red, not 0.473684...... )

I agree with what TomG said.

Remember that, if you are tossing a coin, while the probability of 20 heads in a row is 1 / 1,048,576, the probability of 19 heads in a row followed by a tail is also 1 / 1,048,576. The 20th toss is still just as likely to he heads as it is to be tails, regardless of how the first 19 turned out.

Similarly, in roulette, the probability of red coming up on a particular spin is the same as black coming up on the same spin, regardless of the results of the 19 previous spins.

Also, from another perspective, the martingale doesn't work because you're never changing the house edge and thus you'll always have a negative Expected Value (EV).

In general: EV = (NumTrials*AvgBet)*(HouseEdge)

In the martingale scenario when you lose you simply up your bet... upping your average bet. At no point in time do you ever change the HouseEdge which is negative.

Let's say your units are $10 and look at a few iterations...

EV($10) = (1*10)*(-0.0526) = -$0.53

lose

EV($20) = (2*15)*(-0.0526) = -$1.58

lose

EV($40) = (3*23.3333)*(-0.0526) = -$3.68

I hope you see how every time you up your bet you're upping your average bet and you're NEVER changing the negative house edge. So whenever you up your bet you're just simply LOSING MORE in expectation over the long run of the game.