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Precalculus Mathematics Homework Help
Guessing the number of throws required to get a 6 on a fair die
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[QUOTE="PeroK, post: 6473603, member: 493650"] 1) If you're betting on when the first six will come up, then your best chance is the first throw. Because all you need is to throw a six on the first throw. To get your first six on the nth throw you need ##n-1## throws that are not six [U]and[/U] a six on the nth throw. Hence the probabilities reduce quickly, as we can see from your data. 2) You reconcile it by calculating calculating the expected number of throws. Note that expected value has a precise meaning: $$E = \sum_{n = 1}^{\infty}np(n)$$ That's very different from the [I]mode[/I], which is the most common and in this case is ##1##, and the [I]median[/I] which is the midpoint of a distribution and in this case is ##4##. The expected value is also known as the [I]mean[/I]. If "expected value" carries too much linguistic baggage for you, then sick to using mean. [/QUOTE]
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Precalculus Mathematics Homework Help
Guessing the number of throws required to get a 6 on a fair die
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