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2009-02-09T22:46:14.330
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2009-02-09T23:00:27.727
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Body
EDIT: The below doesn't answer the question, sorry... Comment clarified that the real problem is about the probability of x consecutive 1s out of n bits, not just the simple thing I assumed. Had a quick look at this: <a href="http://www.mathhelpforum.com/math-help/probability-statistics/64519-probability-consecutive-wins.html" rel="nofollow noreferrer">http://www.mathhelpforum.com/math-help/probability-statistics/64519-probability-consecutive-wins.html</a> which may be what you are looking for - it seems to deal with working out the probability of a run of toin cosses out of a larger population of toin cosses, so sounds similar. But its late and I am tired so I haven't decoded the math :) OBSOLETE: It sounds like you are basically dealing with binominal probability - see <a href="http://en.wikipedia.org/wiki/Binomial_probability" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Binomial_probability</a>. I have to admit I haven't done the calculations for about 20 years, so somewhat rusty... Basically, binominal allows you to "add together" the probability of an event occuring multiple times, where there is only two possible outcomes each time. Order is significant in your case so it should be as simple as multiplying the probabilites; For 1 bit it is 50% For 2 bits it is 50%^2 = 25% For 3 bits it is 50%^3 = 12.5% Look at it another way; 1 bit only has 2 possible combinations, one of which is all 1s = 50% 2 bits have 4 possible combinations (10, 01, 11, 00), only one of which is all 1s - so 25% 3 bit have 2^3 = 8 possible combinations, only one of which is all 1s, so 1/8 = 12.5% So... probability of n bits all being 1 = 1/(2^n).
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1. POWhat's the probability that X *consecutive* bits in an array of N bits is set to 1?
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