OT stats discussion:
The choice of a binomial is okay, but note that by using it you are assuming that one day’s gains are independent from all the other days’ gains (assumption C on your link)
I think the question you answered was:
Q1) What is the probability that 6 out of 6 days will be up?
I think you wanted to answer:
Q2) Given I’ve already had 5 up days, what is the probability the next day will be up?
A1) 2% (either using binomial or simply 0.52^6)
A2) 52%
Note that since one day’s gains are assumed to be independent of the gains of all other days, A2 would be 52% no matter how long the string of up days that preceded it. Or you could calculate it by P(day 6 up) = P(6 of 6)/P(5 of 5) = 52%
A way to calculate it without assuming independence would be to find all the instances where there was a string of 5 up days and then see how many out of the total succeeded in printing a 6th up day. The only problem is the sample may be too small to say anything useful.
FWIW
Todd
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