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From: John O
Remote Name: 69.208.140.49
Date: 18 Mar 2005
Time: 21:34:18 -0700
If you have a fair die and you regard a success as rolling a "1", what is the expected number of rolls it will take you to achieve 4 successes? Here is what I think: You would expect 6 rolls to get the first success (geometric with p = 1/6). You would expect another 6 for you second success (hence, 12 tosses for your first two succeses). You would then expect 6 tosses again for your third success and another 6 for your fourth and final success (brining your total to 24 tosses.) Because the tosses are all independent, the geometric r.v. would apply. But according to the negative binomial expected value ( =r(1-p)/p) you would expect 4(5/6)/(1/6) = 20 tosses to get four successes. So does it take 20 tosses or 24 tosses? My logic, I believe, makes sense but does not follow conventional wisdom of the negative binomial expected value. Thanks