Re: Need help with a probability problem
Donald Burrill wrote: Then, again, you are asserting that this is not a probability problem but a measuring-skill problem. Your postulate that the subsequent executioners must have reduced probabilities (or success rates) only makes sense if all executioners use the same method of execution: a condition you have not heretofore required. Surviving a fencing match with the first executioner needn't imply anything about one's ability to survive hand-to-hand combat with the second; except insofar as the Yes. There is no reason to suppose that such fencing ability is strictly monotonic. In fact anecdotal evidence suggests otherwise. For example, the best executioner might be left handed, but have his handedness advantage removed when fighting a left-handed prisoner etc. Thom = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Need help with a probability problem
On 20 Nov 2001, J. Peter Leeds wrote: The problem actually breaks down to a rather simple analogy: Imagine that a man has been sentenced by court to run a gauntlet composed of four club-wielding executioners. (ill-defined, and thus insoluble, problem omitted) and Donald Burrill responded: Easier it may be, but one can't help suspecting that some aspects of the inanities evident are not paralleled by structures or relationships in whatever your real problem is... I agree with Donald...reading EDSTAT-L, I am continually reminded of the cliche about the patient at the doctor's clinic who has a friend who thinks (s)he might have VD. It seems as if many correspondents posting real (non-homework) problems go to great lengths to anonymize their problems, usually to the extent of making sensible advice impossible. Some possible explanations: 10) They don't want their colleagues to know they've consulted EDSTAT-L. 9) We might steal their data. 8) Statisticians are like mushrooms - they do best when kept in the dark and fed well-composted BS. 7) EDSTAT-L is the favorite consulting service for researchers on witness protection programs. 6) There is actually no research project as described; rather, this is part of an elaborate psychological study of mailing list subscribers. 5) They know that editors don't like putting the entire list of EDSTAT-L posters as coauthors. 4) They reckon we wouldn't understand the difficult stuff they're working on don't have time to explain it. 3) Some researchers realize that they are working on subjects of truly embarrassing stupidity, but they need the grant money. 2) They want to do as much of the work as they can for us and think that this is the first step. And the top reason: 1) These posters are from the NSA: they could tell us but then they would have to kill us. Robert Dawson = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Need help with a probability problem
I'm working on a formula for measuring decision making skill and am trying to estimate the probability that a person of known skill can distinguish among different response option contrasts and avoid a type II error. The problem actually breaks down to a rather simple analogy: Imagine that a man has been sentenced by court to run a gauntlet composed of four club-wielding executioners. The court places the best execution at the beginning of the gauntlet followed by the second, third and fourth best. Based on past performance the first executioner has a .90 probability of striking the man, while the remaining executioners have .50, .30, and .20 respectively. What is the man's probability of being struck by at least one of the executioners and how is this calculated? Notice that the events are not independent because if the man is fast enough to make it past the first executioner his odds of making it past the rest are improved since he will have survived the best executioner. What is this sort of problem called? (e.g., conditional probability, joint probability, Baysian probability, etc.). Please excuse the inanity of the example but it is much easier than trying to explain my research. Peter = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Need help with a probability problem
On 20 Nov 2001, J. Peter Leeds wrote: I'm working on a formula for measuring decision making skill and am trying to estimate the probability that a person of known skill can distinguish among different response option contrasts and avoid a type II error. One effective way of avoiding a Type II error is to reject the hypothesis being tested. Of course, this entails a non-zero probablility of making a Type I error... :-) Seriously, though, I believe it is not possible to _avoid_ a Type II error in the process of accepting the hypothesis being tested; one can only [attempt to] control the probability of such an error. Perhaps this is what you meant, but it isn't exactly what you wrote. The problem actually breaks down to a rather simple analogy: Imagine that a man has been sentenced by court to run a gauntlet composed of four club-wielding executioners. The court places the best execution You mean executioner, surely? at the beginning of the gauntlet followed by the second, third and fourth best. Based on past performance the first executioner has a .90 probability of striking the man, while the remaining executioners have .50, .30, and .20 respectively. What is the man's probability of being struck by at least one of the executioners and how is this calculated? Notice that the events are not independent because if the man is fast (or lucky, or skillful?) enough to make it past the first executioner his odds of making it past the rest are improved since he will have survived the best executioner. In other words, the probabilities associated with the other three executioners are NOT .50, .30, and .20 as advertised, but some (presumably) smaller values? In other words, the probability of being struck by the second executioner is .50 only if one has already been struck by the first executioner? This doesn't seem very sensible... And what model have you (if any) for recalculating the other three probabilities for those who manage to escape the first (and then the second, and then the third) executioner? I do not see why you quote values of alleged probabilities, only to say in the next breath that those probabilities are false. Nor do I quite believe your assertion of non-independence: seems to me they might very well BE independent, if only one knew what the REAL probabilities were. No? What is this sort of problem called? (e.g., conditional probability, joint probability, Bayesian probability, etc.). Please excuse the inanity of the example but it is much easier than trying to explain my research. Easier it may be, but one can't help suspecting that some aspects of the inanities evident are not paralleled by structures or relationships in whatever your real problem is; which rather vitiates the underlying (if unstated) assumption that analysis of the inane example will be in some way helpful in analyzing the real circumstances. Or, to put it another way, the inane example may be wholly inadequate as a model for whatever phenomenon you're really trying to deal with. -- DFB. Donald F. Burrill [EMAIL PROTECTED] 184 Nashua Road, Bedford, NH 03110 603-471-7128 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
