European Russian Roulette Casino Probability Help

Question

I employed a simple betting strategy while playing European Russian roulette, which features a single zero on the wheel. My approach involved exclusively betting on black, and in the event of a loss, I doubled my wager until I eventually won. Once I won, I would reset and place the minimum bet of $5$ again. The table I play at has a maximum bet limit of ${$300}$. Each day, my objective was to earn 100 and cease playing. However, if I encountered more than seven consecutive losses (resulting in the ball landing on red), I would be compelled to halt for the day and replenish my funds with another $300$ for the following day. if follow this precise strategy consistently for an entire year. I am curious to determine the number of winning and losing days, as well as the overall profit or loss incurred over the course of the year.

First, we need to determine the probability of winning or losing a single round of European Russian roulette. Since you only bet on black, the probability of winning a single round is $\frac{18}{37}$ (there are $18$ black slots out of a total of $37$ slots on the wheel).

To calculate the probability of losing seven times in a row, we multiply the probability of losing a single round by itself seven times:

Probability of losing $7$ times in a row = $(\frac {19}{37})^7$

Now, let's examine the potential outcomes over the one-year period:

Winning Day: To make $100$ in a day, you need to win a specific number of rounds. Let's assume you win n rounds in a day. Each win yields a minimum bet of $5$. Therefore, the equation to satisfy is:

$$n \times {$5}= {$100}$$ $$n = 20$$

Hence, you need to win $20$ rounds in a day to achieve your ${$100}$ goal.

Losing Day: If you lose more than seven times in a row, you are forced to stop playing for the day and reload another $300$. The probability of having a losing day is $(\frac{19}{37})^7$.

Next, we can calculate the expected number of winning and losing days over the year:

Overall profit or loss = (Expected number of winning days - Expected number of losing days)✖️ ${$100}$

Please note that this calculation assumes that the betting table always allows you to place the minimum bet of $5$ and reloads your funds to ${$300}$ each day. In reality, there might be additional factors or limitations that could impact the outcome.

Now could you go into Mathematical detail about the legitimacy of this strategy?

Answer 1

A losing day happens with much higher chance than you give it credit for. Also, the amount lost is much higher than you give it credit for. To lose seven times in a row, you will have lost your initial \$5 bet, as well as losing the \$10 bet, and the \$20 bet, and so on until having lost the \$320 bet for a total loss of $5+10+20+\dots+320=635$... Your total loss is not just the value of the final bet, but rather all lost bets until that point.

For a day not to be a losing day, then our gambler must have won 20 bets interspersed between runs of losses at most 6 in length.

To make things simpler to calculate, recognize that our gambler taking a break does nothing to affect the probabilities for the next time(s) they place more bets. Let us organize the day into 20 segments, and in each segment there is enough time for exactly seven bets. Our gambler will continue to bet until getting their first win in a segment after which they will take a break until the next segment begins where they repeat the process, or of course unless our gambler happens to lose every bet in the segment and must retire for the day.

To have a successful segment it must be that we did not experience 7 straight losses within that segment and will occur with probability $p=1-\left(\dfrac{19}{37}\right)^7$. To have 20 successful segments, and thus a "winning day" will occur with probability $p^{20}\approx 0.8276$

To have a "losing day" of any severity will occur so long as it was not a winning day and happens with probability $1-p^{20}\approx 0.1724$. With the way you have phrased it, by stopping if we lose seven times in a row... Note that on a losing day our losses are offset by how much we happened to have won until then so we could have lost $635$ if our seven straight losses occurred in the first segment, or $630$ if we happened to win the first segment but lost in the second segment, on up to having lost $540$ if we happened to win 19 segments and had our losing streak in the final twentieth segment.

The amount contributed to the expected winnings will be $100\cdot p^{20}\approx 82.76$.

The amount contributed to the expected losses will be, by breaking into cases based on $k$ the number of winning segments, $\sum\limits_{k=0}^{19}(-635+5k)\left(\frac{19}{37}\right)^7p^k\approx -101.5489$ (calculations) seen by taking the amount of the loss, multiplying by the probability of the first $k$ segments being successes, and the last segment being a loss.

The overall expected result is then:

$$\approx 82.76-101.5489\approx -18.79$$

That is, you are expected to lose approximately \$18.79 per day on average following this strategy. If you did this for a whole year, you will have lost approximately \$6858.35 in all.

Even if you were to modify this strategy to having stopped after the fifth loss in a row or however else since you didn't seem to realize how much you actually lose when you lose since the lost bets all add up, it will always end with a negative expected return for you. The house always wins. Maybe not all the time, but isolated cases of success do not outweigh the much greater number of people who lose money.

Answer 2

Lets say on a single day you keep doing this. If you lose $\ell$ rounds you reset to 5 bet again then, expected gains assuming you double your previous bet if you lose till $\ell$ rounds,

$$P(gain = $5) = \sum_{k = 1}^{\ell} (1-p)^{k-1} p = 1-(1-p)^{\ell}$$

$$P(losing = $5+$10+$20+...+$5 \times 2^{\ell-1} = 5 \times (2^{\ell}-1)) = 1-P(gain = $5) = (1-p)^{\ell}$$

Now if you repeat you replenishing bets continuously then you gain on an average $$= 5 P(gain = $5) -5 \times (2^{\ell}-1) P(losing = 5 \times (2^{\ell}-1)) = 5(1 -2^{\ell} (1-p)^{\ell})$$

So you need to check if $5(1 -2^{\ell} (1-p)^{\ell}) > 0$ to gain something out of this i.e., $\frac{1}{2^{\ell}} > (1-p)^{\ell} \implies p>1/2$

which is not true. So as you can see as $\ell$ increases your losses increases since $1 -2^{\ell} (1-p)^{\ell} \rightarrow -\infty$ as $\ell \rightarrow \infty$.

It seems in short term you might seem like winning but if you do it for a long time you will lose. So play quick and come out and do it less frequently. The clever part from the house is that they almost gave you an impression that $p \approx 1/2$ but still $p<1/2$.

So what if you bet at $n^{th}$ round an amount $$B_n = L+\sum_{i=1}^{n-1} B_i$$ Then $P(gain = L \ dollars) = P(winning \ in \ some \ round) \rightarrow 1$ as number of rounds goes to $\infty$.

Assuming you keep playing infinite number of times i.e., you have infinite money to put up with then probability of you winning $L$ dollars goes to $1$. Since this is continuous betting there is no expected value.

But this is how people lose all their money. Note that you need to be having cash to keep betting with infinite source of money. For this people take loans and double their bets and in the end due to finiteness of cash availability you lose and if you lose you lose everything.