"There are N elements in a set, the probability of randomly selecting any element is the same (so there is no restriction) meaning that for every single element, the probability is 1/N.

Now suppose that there are K elements (K < N) with a given characteristic, then the probability of randomly selecting one of the K elements out of the sett with N elements is equal to K times the probability of each element, this is:

P = K*(1/N) = K/N"

Then if a dice has 6 possible outputs, and we want to find the probability of getting a 3 or a 5, we need to find the quotient between the outcomes where we get a 3 or a 5 (2) and the total number of outcomes (6)

P = 2/6

Now let's go to the case in the problem:

We know that out of the last 20 games, your team won 18.

Then yes, your team won (18/20)*100% = 90% of their played games.

Does this mean tath the probability of winning the next game is 90%?

Well no, because this is not a random selection, there are a lot of other things needed to analyze, as the win rate of the other team (suppose that the other team won 14 of their last 20 games, then their win rate is (16/20)*100% = 70%

Does this mean that the probability where our team wins is 90%, and the probability of the enemy team winning is 70%?

Well no, this makes no sense at all, and this is why this method is not a good way to estimate probability in complex cases like this one.

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3 years ago