# Real-world interpretation of probability

I'm confused about the real-world interpretation of a certain probability.

Say that in 1000 observed cases a certain event occurs 10 times. We can then say that the probability is 10/1000. Now, of course, this only tells us how often something may occur, not when (in a series of 1000 observations) it will occur. So it's perfectly possible that the first event occurs after 200 observations, and the second event after only 260 observations, etc. Therefore, if we say that 10/1000 = 1/100, there is no reason why in a new sample of 100 observations an event should occur. On the other hand, it's perfectly possible that a third sample finds all 10 occurrences of an event in the first 100 observations. In reality then, and contrary to ordinary intuition, a probability of 1/100 is in perfect keeping with no occurrence of an event in 100 observations, as well as with 10 occurrences of an event in 100 observations. Is this reasoning correct, or is there some flawed assumption I'm being blind to? Thanks a lot for your help!

• You seem to have answered yourself, so it is not really clear what is your question?
– Tim
Commented Jan 29, 2020 at 15:45

1) If in 1000 cases a certain event occurs 10 times, we wouldn't say that the probability (of the event occurring) is 10/1000=1%. More correctly we would say that this is an estimator of the true probability, and that 1% is actually the observed relative frequency. Chances are that the true probability (assuming that it exists, see below) is not too far away from 1%, however it can be different; it is easy to imagine that in the next 1000 observations you will find, say, 9, or 12 occurrences without a change of the underlying probability.

2) The standard assumption that is made here is that the observations are identically and independently distributed (i.i.d.), meaning that in every case the probability is the same (otherwise it isn't clear what probability you're talking about putting together the 1000 cases), and that the outcome of certain cases isn't influenced by the outcome of other cases (for example those that were observed directly before in time).

3) Obviously, you cannot be sure that exactly the same happens in the future that has happened in the past. So indeed you're right. In one instance you may observe just one occurrence in 100 observations (and/or 10 in 1000), in the next one you may observe ten in 100. This does not necessarily mean that the underlying true probabilities (which cannot be directly observed but only estimated) are different.

However, huge variation of the relative frequency of observed occurrence under the i.i.d. assumption is very unlikely, so you shouldn't expect to observe 10 occurrences in the first 100 observations and then no further one in the following 900. (It can be made mathematically precise how much variation can be expected.) If such a thing happens, chances are that something is wrong with the assumption, i.e., either the probability has changed over time, or there is a positive dependence between outcomes.

• as we are talking about probabilities, yes chances are something is wrong but the chance is still there. Commented Jan 29, 2020 at 16:33
• Thanks Lewian! Could you also tell me how the amount of expected variation is calculated? Not necessarily any specific formula, but is there a technical term for this which I could look into further? Thanks again! Commented Jan 29, 2020 at 18:36
• For a given probability $p$ the variance for the relative frequency in $n$ i.i.d. observations is $p(1-p)/n$ (standard result). The distribution of this relative frequency is approximately normal (for $n$ large enough). In particular this can be used to construct tests or confidence intervals for the difference between two observed relative frequencies for independent data, without knowing the underlying true $p$. There's also Fisher's exact test for comparing two probabilities, and corresponding confidence intervals. Hope this helps. Obviously not all details are given. Commented Jan 30, 2020 at 13:39
• The binomial distribution allows to compute the actual probabilities. For example for $p=0.01$ I get that the probability for having 10 or more occurrences in 100 observations is very precisely zero (so small that my computer won't tell the difference), on the other hand, the probability for having 0 occurrences in 900 observations is 0.0001, very, very small, so $p=0.01$ is too small for the first result and too large for the second one, and so in practice you wouldn't expect to see them both in succession, unless the i.i.d. assumption is violated. Commented Jan 30, 2020 at 13:47

this reasoning is correct because we are only estimating the situation.

You might want to use a dice and roll it x times. You won't find a perfect 1/6 distribution per side because of imperfections on the dice/material and you might be able to influence the result by throwing the dice in a specific way.

However, you will find when you roll the dice 6 times that chances are very low you encountered all sides exactly once. Maybe you will encounter the following result:

1,2,2,3,6,2
Chances now are:
1: 1/6
2: 3/6 (50%)
3: 1/6
4: 0/4
5: 0/6
6: 1/6

But when you roll the dice 1000 times you will get a more even distribution and the more often you roll the dice the more accurate the estimation for this specific dice and your throwing technique will get.

It is important to realize this. I actually made a dicing bot and realized that even if that chance of winning is 80% eventually you will have a loosing streak of 1000 games in a row.