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In the American Roulette wheel, the winning odds of betting colours(black/red) is 47.36%. Consider this method, if the minimum single bet is \$10 and if I were to bet \$10 and double my next bet of what my previous bet is $10 if I happen to lose my previous bet for a total of 3 times going forward making the total bet count of 4, will this be a profitable play in the long run.

Like this way:

Initial bet: \$10, if I win I put \$10 again for the next bet. If I lost I put double the amount of my previous bet for my next bet, which is \$10 making it \$20. If I win this bet of \$20, I go back to betting \$10, but if I lost I will double this amount for my next bet making it \$40. I'm going to do this for a total of 3 consecutive times in the event of a lost, so it's 10 + 20 + 40 + 80 = \$150. So if each single bet gives me 47.36%, 0.4736 * 4 = 1.8944 * 100 =189.4% meaning if I were to do this doubling for 3 consecutive times for a total of 4 bet counts, each single bet technically gives me 189% chance of winning. Am I right, is this the correct way of calculating it?

I was also thinking about the application of this same method to the dozen bets. For a dozen bet, the winning odds is 31.57%, but the payout is 2 to 1 vs. only 1 to 1 for colour bets. If the same method is applied to betting, which of the two types of bets will be more profitable in the long run?

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Hi Theo. I think you were having trouble with the text displaying correctly because of all of the dollar signs. The editor reads a dollar sign as the beginning of 'math mode', which is a mode allowing you to write using math characters. Text written in this mode will automatically italicize, and have no spaces. This is what you were seeing. To fix this, just add a \ in front of all of the dollar signs (click edit to see that I've done this). Good luck with Roulette! – Jared Mar 12 '13 at 16:42
@Jared Thanks :) – Theo Mar 12 '13 at 16:45
For almost every bet at the roulette table, the house edge is $\frac 2{38}$ so each dollar bet returns on average about $0.9472$. Your expected losses only depend on the total dollars bet, not on the pattern of betting or which bets you choose. – Ross Millikan Mar 12 '13 at 17:17
up vote 5 down vote accepted

The error in your calculation is that the chance of winning at least once in four tries is $1-(\frac {20}{38})^4 \approx 0.9233$. You can't keep adding the $\frac {18}{38}$s-the fact that the sum exceeds $1$ should tip you off. This is because you might win more than one-the events are not mutually exclusive.

Your expected win from one series is then $10 \cdot 0.9233 + (-150) \cdot 0.0767=9.233-11.51=-2.28$

The important truth is that no series of losing bets can be winning. You can have a high probability of profit, as here, but the expectation will be negative.

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I was just looking at the second equation, how do you get the numbers 150 in the bracket, 0.0767, and 11.51? because I wanted to try calculate the result of a series of 8 tries instead of 4. – Theo Jun 6 '13 at 19:24
@Theo: $150$ is the total amount you lose on four bets. $0.0767=1-0.9233$ is the chance that you lose them all. Generally, when trying to calculate the expected value, you multiply the probability of each result by the value of that result and add them up. The first sentence says you win $10$ with probability $0.9233$, so you must lose $150$ the rest of the time. – Ross Millikan Jun 6 '13 at 19:45
What about the 11.51? – Theo Jun 6 '13 at 20:05
@Theo: It is the indicated product: $(-150)\cdot 0.0767$ – Ross Millikan Jun 6 '13 at 20:32
oh I get this now. I'm so bad at this -.- – Theo Jun 6 '13 at 21:03

You haven't calculated the odds correctly. See binomial probability for an explanation of how to calculate the correct probability.

If you had an unlimited amount of time, money, and chances then the strategy would win. The problem is you don't have that. The most practical matter (besides actually having the money to front the operation) is the casino won't let you bet as high as you would need to in order to cover your loses. If you lose $n$ times in a row then you will have lost

$$\sum_{k=1}^n 10(2^{k-1})$$

and you will need to bet $10(2^{n+1})$ just in order to make \$10. The risk reward isn't worth it. If you've lost 6 times in a row (which is quite possible) then you will need to bet \$640 on top of the \$630 you lost from the first 6 bets. So you will have to front over \$1,000 just to make \$10. If you lose 8 times in a row (its certainly not impossible) then you'll need to front \$2,560 for your next bet (on top of the \$2550) you've already lost just to make \$10.

The casio's aren't stupid though, if the table has a \$10 minimum bet it might have a maximum bet of \$2,500 for instance, so if you lose 8 times in a row your out a few thousand dollars. To make that back using your strategy you need to iterate the procedure 200 times to make back \$2000. But in those 200 trials its very possible you'll run into a losing streak of 8 or more.

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The loss after $n$ losses is $10(2^n-1)$ but the point still holds. The fundamental answer is that if each bet is losing, the whole series of them is losing as well. You can find series that win with probability greater than $50\%$, but they still lose on average-that just means the losses are larger than the wins. – Ross Millikan Mar 12 '13 at 17:04

If you stop after a finite number of bets, you will still make a loss in expectation. The probability of winning on any single bet is independent of all other bets and the expected value of any single bet is negative.

If you make a potentially infinite number of bets if you lose (doubling bet every time) and stop after the first win, this is a profitable strategy. But you cannot do this if your wealth is finite.

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When I see such a question, the first thing that comes to my mind is Gambler's Ruin. That is another model of betting. But let's cut to the chase for this question first.

Assume that you have $\$N$ worth of capital. This allows you to split it up into $a$ number of bets. This value of $a$ is determined by how many times you can split your capital.

So if $N\ = 10+20+40+80 = $, then $a=4$ as you can bet up to $4$ times consecutively before you go bust (for the evening). Similarly if $N = 630$, then $a=6$.

You are looking for the probability that you win at least once out of $a$ tries. Because, once the number of tries exceed $a$, then you would have gone bust and cannot place a new bet. To find this probability, use the binomial model. In school, I am always asked to define my random variable. Hence, here, I will define my random variable, $X$, as the number of wins I get out of $a$ tries. I will just need to score $1$ win despite the value of $a$. Hence, I always calculate $P(X \geq 1)$. I always use this expression to derive it.

$$P(X \geq 1) = 1-P(X=0)$$

Let's consider if $a=4$. $X$ takes parameters $p=0.4736$ as defined by your question and $n=4$. We will attempt to calculate $P(X\geq1)$, which is

$$P(X \geq 1) = 1-P(X=0) = 1- (p)^0(1-p)^4$$

One signing off comment here, as mentioned in other answers, each table has a maximum bet so you can't play this strategy (or this strategy expires quickly because $a$ max cap is low). Also, the probability of winning never exceeds $50\%$. Hence, casino games are never in your favour. If not, Las Vegas would not survive so many years.

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If I understand your method correctly, I believe this is exactly why casinos have maximum bets. Consider the following scenario.

Say you have an infinite amount of money, and are playing a game that pays 2 to 1 with a probability of $0<p<.5$. Let's employ a strategy similar to what you have described, by beginning with a bet of $\$1$, doubling the bet if we lose, and quitting if we win. With this strategy, if at any point in the game we win, we will have gained a profit of $\$1$ (check this in the case that we win after the first, second, and third times playing to see the pattern). Since $p>0$, we are guaranteed to win at some point, perhaps after playing the game many, many times, but we will always come out $\$1$ ahead.

By replacing your initial bet with a larger amount, say $\$10,000$, employing this strategy guarantees a profit of $\$10,000$.

Now suppose there is a maximum bet, i.e, we are not allowed to bet more than $N$ dollars, and find $n$ such that $2^{n-1}\le N<2^n$. If we haven't won by the time we've played $n$ games, we won't be able to double our next bet, and our strategy is shot. If $N$ is large enough and $p$ isn't too small, this should rarely happen, but when it does, your large loss in that case will outweigh all of your smaller gains in the cases that you win.

To see this, let's compute the expected value of our strategy with an imposed maximum bet of $N$ dollars. The chance that you lose $n$ times in a row is $(1-p)^n$ in which case you will have lost $\$2^n-1$ (check this). Notice that as $n$ increases, the chance that you lose decreases, but the hit you take for losing increases. Next, the chance that you win at some point within the first $n$ games is $1-(1-p)^n$, in which case you win $\$1$. Hence, your expected value is $$ 1\cdot(1-(1-p)^n)-(2^n-1)\cdot(1-p)^n=1-(2-2p)^n $$ Since $p<.5$ the above expected value is always negative, no matter how large the maximum bet is. Thus the only winning strategy is if you had the option of playing as many times as necessary to win, but this is only possible with infinite wealth and no maximum bet (which is not realistic).

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In roulette, each spin is independent (and not dependent on previous spin) and hence probability of winning at least once in a row of four is same as probability of winning when played.

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