Suppose I have a random event with k possible outcomes of equal probability. What distribution (if any) describes the probability of obtaining a specific sequence of length m after n events?

#probability #probabilitydistribution #statistics

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@faelif @UlrikeHahn

the probability of observing a single specific sequence is (1/k)^m.

@jerlich @faelif what are k and m? the number of outcomes and the sequence length? because if yes that is answering a less general question than the one asked

@UlrikeHahn @faelif

E.g. if you are flipping a fair coin then p(h)=p(t) = 0.5 (k==2). The probability of observing some specific sequence htththhthtis just 0.5^m where here m = 10.

@jerlich @UlrikeHahn

The question is about finding the sequence in a larger string, though - for example, HHTTH contains TH

See also here:
mathstodon.xyz/@faelif/1143041

@faelif @jerlich you are right- it‘s not: which is the whole point of the paper I linked too (and renewal processes) the wait time for HHHH is completely different than that for HHTT ….

@faelif @jerlich Jeff is confusing the probability of a sequence of length m occurring in n flips of a coin where m=n with that probability when n > m (ie ocurrence as a subsequence). Subsequence probabilities are not equal across sequences. The references I gave are precisely about that! d1wqtxts1xzle7.cloudfront.net/

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