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# Advanced Math/Time Series and Conditional Probability

randy patton wrote at 2015-07-07 06:19:39
Perhaps this rendition of Bayes' Rule will help. You can always start with the equality from the axioms of set and probability theory

P(B|A)P(A) = P(A|B)P(B)

where B would represent the 4-day streak event and A the 3-day streak. This can be rearranged by dividing by P(A) to get

probability of 4-day given 3-day = P(B|A) = [ P(A|B)P(B) ] / P(A)

As you say in your follow-up, a 4-day streak can only happen if a 3-day streak has already happened, which means P(A|B) = 1, so

P(B|A) = P(B)/P(A)

just as the radio guy said.

As far as the other streaks go, if for example, the probability of an 8-day given a 7-day would have the same form as the 4 and 3-day case above. The probabilities of the longer streaks do in fact go down quickly, but the ratio of consecutive streaks stays the same (the respective probabilities decrease at the same rate).

You comment that, after a 7-day streak, it should be less probable to get an 8-day streak, meaning an up day occurs, than for a down day to occur ("reverting to the mean"). however, if the occurrence of a single up or down day is 50-50 and is independent of what has happened before, then both possibilities have equal probabilities. This whole problem seems to be a rehashing of the well known example of asking what the probability of getting a heads in a fair coin toss after lots of heads have just been tossed in a row. The answer is 1/2 since the coin doesn't know or care what happened before, it will land on heads or tails with equal probability.

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