My Math Forum Confidence problem with dice

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 January 3rd, 2018, 06:10 PM #1 Newbie   Joined: Jan 2018 From: Seattle, WA Posts: 1 Thanks: 0 Confidence problem with dice In statistics, a confidence interval is a range of possible outcomes and an associated probability (confidence) that the actual outcome is part of that set. Let's start with rolling 2d6; the range of possible outcomes is [2..12] with 36/36 (100%) confidence. I'm interested in two questions: 1). If I lower the confidence interval to say [2..10] then my confidence in the outcome is also reduced. There is one chance to get a 12, and two chances to get an 11, so the new confidence in the outcome falling within the confidence interval is 1 - 3/36 = 33/36. 2). If I lower the confidence to some discrete probability appropriate to dice, perhaps 30/36, then what is the new range of values for that probability? There are three ways to get a 10, plus two chances for an 11, plus one chance for a 12. So there are 6/36 chances in the range [10..12] and 30/36 chances in the range [2..9]. I've used discrete dice problem to show the answers as I understand them, but eventually I'd like a solution to the continuous model, where confidence (probability) is a real in the range [0..1] and the confidence interval is a normalized range band [0..x] where 0 < x <= 1. Any suggestions or help on how to proceed?
 January 4th, 2018, 02:19 PM #2 Global Moderator   Joined: May 2007 Posts: 6,416 Thanks: 557 If you know the distribution function F(x), then you want to find a < b so that F(b)-F(a)= confidence. For the specific case you you mentioned, F(0)=0 and you want F(x)=confidence. In many situationss you will have a probability density function f(x) (=F'(x)}, so you are looking at $\displaystyle \int_{0}^{x} f(u)du$. Thanks from JeffM1

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