Binomial probability

Life is usually complicated. But sometimes you can simplify some specific problems to yes/no or good/bad answers. If you have a sequence of independent repetitions of an event, with two possible outcomes, and the probabilities of those outcomes does not vary between repetitions, we call that a Bernoulli trial.

If all the above conditions are fulfilled, we can denote p as the probability of success and q = (1 – p) the probability of failure for one such event. For a sequence of n events we have the probability to be successful in k of them:

P = C(n, k) \cdot p^{k} \cdot q^{n - k} (1)

While this may seem a bit boring, we can get use the above relation for various probability estimates, as long as we apply it correctly, and the events are Bernoulli trials.

Example 1

Let’s say you have to pass a test. There are 20 multiple choice questions, each one with 4 choices and one correct answer. What is the probability to pass if you do not study? That would be the probability to answer correctly to 12 questions from 20, with p being 0.25 (1 out of 4 possible choices):

P = C(20, 12) \cdot 0.25^{12} \cdot 0.75^{8} \approx 0.000752

A probability of 0.075% is really low, so you will probably fail without study. However this particular test also has other rules like: 3 questions are mandatory to answer correctly, or you fail regardless of the other answers. So let’s suppose you study somewhat so you’ll correctly answer those 3 mandatory questions. You probability to pass becomes:

P = C(17, 9) \cdot 0.25^{9} \cdot 0.75^{8} \approx 0.009284

which is 120 times larger that if you don’t study at all, but still low.

In the end the best idea to pass any test is to study for it, but in case you don’t feel like it at least you can estimate how much you should study based on how lucky you feel.

Example 2

Let’s suppose we’re playing a game where you can win 1 dollar with probability p = 0.5 and lose 1 dollar with q = 1 – p. The expected value of the game is:

E = p \cdot 1 - q \cdot 1 = 0

so in the long run we would expect to neither gain, nor loose from repeatedly playing this game.

However let’s calculate the probability to win at least 4 dollars after 20 games. This is the probability to win 11 games (this means 11 vs. 9 so we gain 2 dollars) or the probability to win 12 games (this means 12 vs. 8 so 4 dollars). We can add the probabilities because the 2 events are mutually exclusive. According to what we discussed this means:

P = C(20, 11) \cdot 0.5^{11} \cdot 0.5^{9} + C(20, 12) \cdot 0.5^{12} \cdot 0.5^{8} \approx 0.28

This looks like a good enough probability to win. For a run of 80 games we have:

P = C(80, 41) \cdot 0.5^{41} \cdot 0.5^{39} + C(80, 42) \cdot 0.5^{42} \cdot 0.5^{38} \approx 0.167

So although the probability decreases with the number of runs it is still looks significant.

So how does this work? On one side we have the expected value of the game 0, so you expect to win 0 dollars on the long run but if we actually calculate the probability to win a couple of dollars in a series of repeated games we get big numbers. The answer is simple actually, there is a probability to lose at least 4 dollars and in this case it is equal with the probability to win, so you will probably get nothing in the end of a long series of games.

Always think of all the probabilities or you will miss important ones.


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