## Friday, July 11, 2014

### Unizor - Probability and Prediction

Yet another viewpoint to probability is its direct connection to prediction.
Why do we need probability after all? To make a prediction.
Without using the probabilities for predicting of occurrence or non-occurrence of certain events we would not need the probabilities. So, ability to predict is the main purpose of introducing probabilities.

So, from the viewpoint of prediction, the statement that "the probability of a dice to have number 3 on top is equal to 1/6" means that in the course of many random independent experiments of throwing a dice we expect the number 3 to be on top, approximately, in 1/6 cases. It does not mean that in every series of 6 experiments the number 3 will be on top exactly once. It does not mean that in every series of 600 experiments the number 3 will on top exactly 100 times. What it means is that if we conduct N experiments and the number of times the number 3 is on top equals to K(N) then a fraction P(N) = K(N) / N tends to 1/6 as N increases to infinity.

Can the number 3 be on top of a dice 10 times in a row? Yes. It's still possible. But this is a different experiment with different elementary events and different probabilities. In this case a single experiment is throwing a dice 10 times in a row as a series. An elementary event is a series of 10 numbers (each from 1 to 6) with a probability of 1/6^10 for each such series. That very small value is the probability of having number 3 on top 10 times or number 2 on top 10 times or any other series of 10 numbers, 1 to 6 each.

But, if we want to approximately evaluate the count of times when number 3 is on top among 10 throwings, we have to add all the probabilities of elementary events above that contain no number 3's, 1 number 3, 2 number 3's etc. up to 10 number 3's to know the probability of number 3 for each count. Let's evaluate it. For number 3 to be on top K times out of 10 we have to pick K positions out of 10 (there are C(10,K) ways to do it). These K positions must be occupied by number 3 (so, it's only one choice to have it), while the other 10−K positions can be occupied by any other number (that is, 1, 2, 4, 5 and 6) - 5 choices for each position, which makes 5^(10−K) combinations. Therefore, the number of combinations of 10 numbers from 1 to 6 each with K numbers 3 among them equals to C(10,K)·5^(10−K). That is the number of elementary events in our experiment with 10 rolls that contain number 3 exactly K times.
So, the probability of having number 3 exactly K times equals to a probability of a single elementary event multiplied by this number, that is
P(K)=C(10,K)·5^(10−K)/6^10

The approximate values of these probabilities for different K are:
P(0)=0.16; P(1)=0.32; P(2)=0.29; P(3)=0.15; P(4)=0.05; p(5)=0.01
and for greater K the probabilities are less than 0.01.

So, if we want to predict the results of 10 random experiments of throwing a dice, we can say that "it's quite probable to have the number 3 on top somewhere from 0 to 4 times from each series of 10 experiments". Narrower prediction (like from 1 to 2 times) has lower probability.
Of course, "quite probable" has its numerical equivalent as we tried to demonstrate above. That's exactly what Theory of Probabilities is studying.