# 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?

• Will it $\big(1-\frac{18}{37}\big)^{6} \sim 0.01833628123$ – L.K. Jun 24 '14 at 9:30
• Probability of loss is $6$ times to loss game, so it is $(19/37)^6\approx 0.018336281$, and probability of win is $1-(19/37)^6\approx 0.981663718$. – Oleg567 Jun 24 '14 at 10:23

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.

• I understand the high amount risked for the low profit but I thought a 98% chance, taken only once in my life, of winning $€30$ was worth it to "beat" the casino. Now I have lost $€210$ I am wondering what the best course of action is. If I continue the next three bets I believe I have at least an 86% chance of winning. – WhatHaveIDone Jun 24 '14 at 10:30
• Simply, it is your choice :) But if play multiple times, then "average win:average loss" is $$Average Win : Average Loss = (1\times 53.537) : (63\times 1) = 53.537 : 63,$$ and it is not profitable. – Oleg567 Jun 24 '14 at 10:35
• If I continue are my chances 86% or still 98% though? – WhatHaveIDone Jun 24 '14 at 10:38
• I think your chances will 86% - because you've done a "half of way" already. 98% is on the start, when all results are unknown. – Oleg567 Jun 24 '14 at 10:39

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?

• More like initially hoping to flip HH or HT or TH, flipping once and getting T (which leaves TH as now possible but not H) – Henry Aug 24 '15 at 13:44