The Probability of Long Runs
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suppose you’ve written a program that randomly assigns test subjects to one of two treatments, a or b, with equal probability. the researcher using your program calls you to tell you that your software is broken because it has assigned treatment a to seven subjects in a row.
you might argue that the probability of seven a’s in a row is 1/2^7 or about 0.008. not impossible, but pretty small. maybe the software is broken.
but this line of reasoning grossly underestimates the probability of a run of 7 identical assignments. if someone asked the probability that the next 7 assignments would all be a’s, then 1/2^7 would be the right answer. but that’s not the same as asking whether an experiment is likely to see a run of length 7 because run could start any time, not just on the next assignment. also, the phone didn’t ring out of the blue: it rang precisely because there had just been a run.
suppose you have a coin that has probability of heads p and you flip this coin n times. a rule of thumb says that the expected length of the longest run of heads is about
provided that n (1- p ) is much larger than 1.
so in a trial of n = 200 subjects with p = 0.5, you’d expect the longest run of heads to be about seven in a row. when p is larger than 0.5, the longest expected run will be longer. for example, if p = 0.6, you’d expect a run of about 9.
the standard deviation of the longest run length is roughly 1/log(1/ p ), independent of n . for coin flips with equal probability of heads or tails, this says an approximate 95% confidence interval would be about 3 either side of the point estimate. so for 200 tosses of a fair coin, you’d expect the longest run of heads to be about 7 ± 3, or between 4 and 10.
the following python code gives an estimate of the probability that the longest run is between a and b inclusive, based on an extreme value distribution.
def prob(a, b, n, p): r = -log(n*(1-p))/log(p) cdf = lambda x: exp(- p**x ) return cdf(b + 1 - r) - cdf(a - r)
what if you were interested in the longest run of head or tails? with a fair coin, this just adds 1 to the estimates above. to see this, consider a success to be when consecutive coins turn up the same way. this new sequence has the same expected run lengths, but a run of length m in this sequence corresponds to a run of length m + 1 in the original sequence.
for more details, see “the surprising predictability of long runs” by mark f. schilling, mathematics magazine 85 (2012), number 2, pages 141–149.
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