Is it valid to apply large scale statistics to a single case? I have a small black box that displays a 0 or a 1 every time I press its button. 
I can tell you, truthfully, that over the past 20,000,000,000 times it has been pressed, 19,999,999,990 times it has shown a 1.
I ask you to make a prediction of what number will appear when I press the button next.
For the sake of simplicity, I won't attach any form of consequences to the results. You make a guess, and you are either right or wrong. 
How do you make your choice, and in what way do you find that choice to be valid?

Here's some other constraints that you can feel free to apply or discard if you think the result would be interesting:
1. The temporal distribution of the recorded results is unknown.
2. The display is blank until the button is pressed, and the digit will appear for a short amount of time afterwards. It is blank when you see it. 
3. The button being pressed causes the display of the digit.

 A: First I would like to say that I'm not sure if I fully understand what you ask in your actual question. The whole concept of statistics (or more rigorously inference theory) is to draw conclusions from the given data, so yes it is possible to use statistics to make predictions about a single case as long as you have any data to base you prediction on. You would not need statistics to predict the outcome if you know all the variables and relations in the the process exactly.
In your example you are asking for a prediction of the next outcome in some unknown random process. We have no knowledge of the dependence between the trials you describe. What we do know is that we have a dichotomous outcome ($0$ or $1$) with an unknown probability $p$. 
Essentially what we are looking for is the conditional expectation of the next state given all previous values. 
$$
  E[X_{i+1}|X_i, X_{i-1}, \ldots, X_0]
$$
Here we need to make an assumption on the dependence between trials in order to continue. Lets investigate both possible assumptions.
1. Independent trials
Assuming the trials are independent, i.e the expected outcome is independent of previous outcomes. If this is the case you describe something which is known as a Bernoulli trial. The best estimator of the probability of success (to get $1$) is the sample mean. 
$$
 \hat p = \frac{1}{N}\sum_{i=0}^N x_i,
$$
where $x_i$ are each of our previously recorded outcomes. Each outcome is however not given in the problem but instead only the count of events displaying $1$ (lets call this sum $T$). Luckily $T$ is a sufficient statistic for the probability parameter $p$ of the Bernoulli distribution. By sufficient statistic we mean a summary statistic containing all information about the sought parameter.
$$
 \hat p = \frac{1}{N}T = \frac{19999999990}{20000000000} \approx 1.
$$
This gives us a probability very close to 1 of getting a $1$ in a trial. The expectation of the next trial is thus
$$
 E[X_{i+1}] = p \approx 1
$$
And our best estimate of the next trial is $1$ with a very high probability.
2. Conditional trials
Here we assume the expected outcome is dependent on previous outcomes. This seems very reasonable, since as the process is unknown we cannot rule out that we have a dependence on previous values. However, dependence is something we cannot really address as we have only been given a summary statistic which is not sufficient for models making use of this dependence. (At least I do not know of any such model)
Say for example that the process is a Markov process with a large probability of staying in the same state. In this case our prediction in the independent model might be very wrong if the last observation was $0$. However, as the summary statistic ($T$) given in the problem statement is not sufficient for the parameters in this model.
Conclusion
This leaves us with only one option. In order to make a prediction given the data at hand we have to assume independence. If we were to use a more complex model without the support of data we would have to start guessing.
The best prediction given the data is therefore $1$.
I hope this answers your question.
