Roulette Probability With a Coin toss

Question

Suppose you are playing roulette and determining to bet on red or black (Evenly 50% chance of occurring) by flipping a coin. You lose 59 times in a row. From this experience, it is most rational to conclude that:

a) Using a coin toss to determine whether to bet on red or black is in general a very bad strategy for playing roulette

b) The game is somehow rigged against you and the casino or its employees are cheating you

c) You are very likely to win on your next bet if you continue this coin flip based strategy

d) The roulette game is broken, but there is no reason to assume that it was broken intentionally

e) You were merely very unlucky

f) One cannot reasonably conclude which of the above options is more likely

So I went ahead and found the probability of losing 59 times and found it to be 1 in 5.764 x10^17. I am struggling to determine which answer is correct and how I should go ahead eliminating answers. Going along with that; how many times would you need to lose to suspect something is going on?

Answer 1

We will assume the variant of roulette where there are two green spaces numbered "0" and "00", neither of which win when choosing Red or Black. This gives $38$ spaces.

Suppose $X \sim \operatorname{Binomial}(n,p)$ is a random variable that counts the number of successes (wins) in $n$ independent Bernoulli trials with success probability $p$, and let $\hat p = X/n$ represent the proportion of successes. If we have the following hypotheses $$H_0 : p = 9/19, \quad H_a : p \ne 9/19,$$ then the distribution of $\hat p$ under the null hypothesis is approximately normally distributed; i.e., $$\hat p \mid H_0 \sim \operatorname{Normal}\left(\mu = p = \tfrac{9}{19}, \sigma = \sqrt{\tfrac{p(1-p)}{n}} = \sqrt{\tfrac{90}{21299}}\right).$$ Then the observed proportion of success, $\hat p_{\text{obs}} = 0$, would correspond to a test statistic $$Z \mid H_0 = \frac{\hat p_{\text{obs}} - \frac{9}{19}}{\sqrt{\frac{90}{21299}}} \approx -7.28697.$$ This is a $z$-score, and the corresponding $p$-value for this two-sided test is about $3.17 \times 10^{-13}$. The exact $p$-value for this outcome is $7.15431 \times 10^{-17}$. In any case, the null hypothesis is rejected with such overwhelming evidence against it: this would mean that the lack of any wins in $59$ trials is strong evidence that the true probability of winning is not $9/19$, which is what we would expect if the roulette wheel were truly fair.

This, however, does not suggest that the cause of unfairness must be due to the wheel. Notwithstanding the possibility (though strange from a causal perspective) that the coin could be landing in a way that predicts the wheel's behavior and shows you the losing pick each time, the roulette wheel could in fact be fair and you could in fact have observed such an extreme result (i.e. $59$ losses or $59$ wins)--but such a probability would be at most $7.15431 \times 10^{-17}$.

Answer 2

Taking a Bayesian perspective, let's consider our priors for the various options:

a) This is in a sense trivially true, since from a statistical perspective every strategy for roulette is a bad strategy (in the sense of losing in the mean); but I'll take this option to mean that deciding your bets by coin tosses does worse than average, i.e. worse than a properly randomized strategy. With that interpretation, this option would require how coins and roulette wheels work to differ quite fundamentally from what we think we know how they work, so we should assign quite a low prior probability to this – let's say, $10^{-9}$.

b) This seems like something that is rather unlikely to happen, but on the other hand not as unlikely as a fundamental upheaval in our understanding of coins and roulette wheels; let's assign it a prior probability of $10^{-6}$.

c) This sounds like a classical error in statistical reasoning. Here, too, coins and roulette wheels would have to differ a lot from how we think they work for this to be true, so this is another $10^{-9}$.

d) Again, if the roulette game is merely broken without any intention of cheating you, it doesn't fit with our current understanding of coins and roulette wheels that your coin tosses would be perfectly correlated with the broken roulette game, so yet another $10^{-9}$.

e) According to you and heropup, the probability for this is something of the order of $10^{-17}$.

(We don't need to assign a probability to f), which is a sort of meta-option.)

For a complete answer, we'd also need to estimate, for each option, how likely the observed result was under that assumption; so far, we've only taken that into account in e). This could be complicated; we'd have to consider things like the fact that a cheating casino wouldn't be so obvious and let you lose $59$ times in a row. But for the sake of argument and of learning something about Bayesian probabilities, let's assume that the probabilities above are already the probabilities that the respective option is correct and the observed result of $59$ losses in a row occurs. Let's also assume for simplicity that the options are mutually exclusive.

Then a priori you'd have a probability of about $1.003\cdot10^{-6}$ to observe the result that you in fact observed: $10^{-6}$ for the cheating casino and $3\cdot10^{-9}$ for the three options that require a revolution in physics. Option e) is so much more unlikely than the others that it got lost in the rounding.

So a priori you'd find the observed result highly unlikely. But now you did observe it. A posteriori, you need to renormalize your probabilties by dividing them by $1.003\cdot10^{-6}$, the a priori probability for the observed result. Thus, a posteriori the probability that the casino is cheating you is

$$ \frac{10^{-6}}{1.003\cdot10^{-6}}\approx99.7\%\;, $$

with the remaining $0.3\%$ almost completely distributed over the three options that require a revolution in physics. The a posteriori probability for option e), that you were merely very unlucky, is

$$ \frac{10^{-17}}{1.003\cdot10^{-6}}\approx10^{-11}\;, $$

so for practical purposes you can disregard this possibility about as safely a posteriori as you could a priori, because, as unlikely as cheating casinos and revolutions in physics may be, they are still considerably more likely than $59$ consecutive truly random coin tosses that just happen to all correlate with a roulette wheel.

So, from this experience it is most rational to conclude option b). As Sherlock Holmes might put it:

When you have eliminated the impossible, whatever remains, however improbable, must be the truth.