Can you create a method to produce seven integers with equal likelihood using a binary function with probabilities p and 1-p?

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Anonymous · Accepted Answer

to get 7 integers with equal probabilities , we could use a binomial (bernouli
distribution) and drop values till we get the required number. 2^3 = 8, so to
have 7 w equal probability we just drop one of them. the 8 possibilities here
would be 000, 001, 010, 011, 100, 101, 110, 111. To get 7, just exclude one of
these... ie. 111, and that's an answer: 000, 001, 010, 011, 100, 101, 110.

Anonymous · Answer

Flip = 1 (p) or 0 (1-p)
 Flip 1000 times to get a binomial distribution with mean 100p;  sd =10sqrt(pq)

Make it a normal dist by dividing by subtracting 100p and dividing by 10sqrt(pq)

Find cdf=1/7, 2/7, 3/7 and use that to produce numbers 1 to 7

Can you create a method to produce seven integers with equal likelihood using a binary function with probabilities p and 1-p?

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