Roulette paradox - martingale probability after leaving table My plan was to bet the following amounts on red/black until I win and then stop (or accept losing everything):
$€30, €60, €120, €240, €480, €960$
My odds of winning one bet on red/black are $\frac{18}{37}$.
By my calculation, my probability of winning once was:
$1 - (19/37)^6 = 0.98$ or $98\%$
Q1. Is that probability correct?
I'm not a gambler and recognise gambling is stupid but with such a high chance of winning I thought I could play once, "beat" the casino and then never play again.
After losing the first three bets, a loss of $€210$, and being unaccustomed to the feeling, I decided perhaps I wasn't willing to risk that much money and left.
Q2. If I were to return to the casino to complete the sequence, would my overall chances still be $98\%$?
I can't see how that would be correct but equally I don't see why a gap in play would affect my original odds. If I were to complete the sequence would I be correct in stating "I had 98% chance of winning but was incredibly unlucky"?
Q3. What actually happened was, on my third bet, I made an error that meant the bet wasn't placed. I would have won if it was placed. Does this affect anything?
 A: It is classic trap for beginners.
Each time you risk to loss $€30+€60+\ldots+€960=€1890$. 
And you can win only $€30$ each time.
Yes, total probability to win is $\approx 98.16\%$. 
But your (probable) profit is very small in comparison with (probable) loss.
Their ratio is $$\text{Win}:\text{Loss} = €30:€1890=1:63.$$
But probabilities ratio is
$$
P(Win):P(Loss) = 1-(19/37)^6 : (19/37)^6 = 0.981663718 : 0.018336281 \approx 53.537 : 1. 
$$

$1$st play. $€ 30$.
If win ($P=\frac{18}{37} \approx 0.486486$), then amount of win is $€30$. Otherwise you'll loss $€30$ and start to play $...$
$... 2$st play. $€ 60$.
If win: $(P=\frac{19}{37}\cdot \frac{18}{37} \approx 0.249817$), then amount of win is $-€30+€60=€30$. Otherwise you'll loss $€30+€60=€90$ and start to play $...$
$... 3$rd play. $€ 120$.
If win: $(P=\left(\frac{19}{37}\right)^2\cdot \frac{18}{37} \approx 0.128284$), then amount of win is $-€30-€60+€120=€30$. Otherwise you'll loss $€30+€60+€120=€210$ and start to play $...$
$\ldots$ etc.
A: Suppose you plan to flip a coin twice, hoping to flip HH.  Your probability of winning is 1/4.
You flip once, and the coin comes up T.  Do you believe that if you flip again, your probability of getting HH is still 1/4?
