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Phil 148 Chances

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The importance of understanding chances: A great many injustices are perpetrated upon people who have a poor understanding of mathematical truths by those who do. People have a financial incentive to fool you out of your money. Since money comes in numerical increments, a lack of skill with numbers inevitably translates into a lack of skill with money.

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Common fallacies of probability: The Gamblers Fallacy – Is assuming that the odds of a single truly random event are affected in any way by previous iterations of the same or other truly random event. Ignoring the Law of Large Numbers – Is assuming there must be other explanations for very improbable events. In many cases there may be other explanations than luck, but remember that it might just be like the truckload of pennies mentioned by the text.

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Heuristics (the common language of probability) People, in order to quickly make sense of a vast array of information, rely on simple decisional gimmicks known in the research as heuristics. These heuristics are generally useful, but are prone to break down in systematic ways. The tendency of a heuristic to supply the wrong answer is referred to as a bias. Some examples follow:

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Representativeness Which is the more likely hand in poker? Hand A : (A, K, Q, J, 10 ) Hand B : (3, J, 2, 7, 9 )

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Representativeness bias Both hands are actually equally likely, though most say that hand B is more likely. People very quickly recognize that Hand A represents a successful hand, which is a much less common outcome than what Hand B represents, which is a nothing-hand. The heuristic that selects patterns that indicate success has then contributed to misjudging probability, and thus is a bias.

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Other Examples: See the text, pp. 280-281, for discussion questions.

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a priori probability: We generalize the probability of any random event in the following way: The probability of any hypothesis Pr(h) = # of outcomes that confirm the hypothesis/total # of possible outcomes.

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a priori probability and other chance events: It is easy to misapply a priori probability. It is important to limit its usage to events that are actually random. For one example, consider the case of a batter in a baseball game: – Only two outcomes are available in each at bat: an out, or the batsman on base. This means that the batter has a 50% a priori chance of getting on base. – However, this a priori probability is of little interest because the batter getting on base is not a random event in the same way that a coin toss is a random event. It is based on many interdependent skills and actions and mental states of both the batter and the pitcher. In real practice, batters fail to get on base much more than 50% of the time. – It is true that luck of some kinds has some influence in baseball, but a priori probabilities dont tell much of a useful story.

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Rules of Probability (1) 1.Negation: The probability that an event will not occur is 1 minus the probability that it will occur Pr (not-h) = 1 – Pr(h)

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Rules of Probability (2) 2i. Independent Conjunction: Given two independent events (events that do not affect each others a priori probabilities) the probability of their both occurring is the product of their individual probabilities. Pr(h1 & h2) = Pr(h1) * Pr(h2)

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Rules of Probability (2G) 2d. Dependent Conjunction Given two events that affect one anothers a priori probability, the probability of their both occurring is the probability of the first occurring times the probability of the second occurring, given that the first has occurred. Pr(h1 & h2) = Pr(h1) * Pr(h2|h1)

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Rules of Probability (3) 3e. Exclusive Disjunction: The probability that at least one of two mutually exclusive events will occur is the sum of the probabilities that each will occur. Pr(h1 (exclusive)or h2) = Pr(h1) + Pr(h2)

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Rules of Probability (3G) 3i. Inclusive Disjunction The probability that at least one of two events will occur is the sum of the probabilities that each of them will occur, minus the probability that they will both occur. Pr(h1 (inclusive)or h2) = [Pr(h1) + Pr(h2)] – Pr(h1 & h2)

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Rules of Probability (4) 4. Series with Independence: The probability that an event will occur at least once in a series of independent trials is 1 minus the probability that it will not occur in that number of trials. Pr(h at least once in n trials) = 1 – Pr(not-h) n

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You may have noticed: The previously discussed rules of probability involved each of the logical operators: negation, disjunction, and conjunction, except for conditional. Bayess Theorem is a theorem of conditional probability. Youll notice that we are now progressing beyond a priori probability, and into statistical probability.

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An Example: Wendy has tested positive for colon cancer. Colon cancer occurs in.3% of the population (.003 probability) If a person has colon cancer, there is a 90% chance that they will test positive (.9 probability of a true positive) If a person does not have colon cancer, then there is a 3% chance that they will test positive (3% chance of a false positive) Given that Wendy has tested positive, what are her chances of having colon cancer?

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Answer: The correct answer is 8.3% Most people assume that the chances are much better than they really are that Wendy has colon cancer. The reason for this is that they forget that a test must be absurdly specific to give a high probability of having a rare condition.

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Formal statement of Bayess Theorem: BT: Pr(h | e) = ___Pr(h) * Pr(e|h)___ [Pr(h) * Pr(e|h)] + [Pr(~h) * Pr(e|~h)] h = the hypothesis e = the evidence for h Pr(h) = the statistical probability of h Pr(e|h) = the true positive rate of e as evidence for h Pr(e|~h) = the false positive rate of e as evidence for h

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Formal Statement of Bayess Theorem:

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The Table Method: h~hTotal eTrue Positives False Positives Pr(e)*Pop. ~eFalse Negatives True Negatives Pr(~e)*Pop. TotalPr(h)*Pop.Pr(~h)*Pop.Pop. = 10^n n = sum of decimal places in two most specific probabilities.

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The Table Method: h~hTotal e= Pr(e|h) * [Pr(h)*Pop.] = Pr(e|~h) * [Pr(~h)*Pop. ] Total of this row ~e= below - above Total of this row TotalPr(h)*Pop.Pr(~h)*Pop.Pop.

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The Table Method for Wendy: h~hTotal e= Pr(e|h) * [Pr(h)*Pop.] = Pr(e|~h) * [Pr(~h)*Pop. ] Total of this row ~e= below - above Total of this row TotalPr(h)*Pop.Pr(~h)*Pop.Pop.

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The Table Method for Wendy: has CC~ have CCTotal e= Pr(e|h) * [Pr(h)*Pop.] = Pr(e|~h) * [Pr(~h)*Pop. ] Total of this row ~e= below - above Total of this row TotalPr(h)*Pop.Pr(~h)*Pop.Pop.

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The Table Method for Wendy: has CC~ have CCTotal tests positive = Pr(e|h) * [Pr(h)*Pop.] = Pr(e|~h) * [Pr(~h)*Pop. ] Total of this row ~ test positive = below - above Total of this row TotalPr(h)*Pop.Pr(~h)*Pop.Pop.

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The Table Method for Wendy: has CC~ have CCTotal tests positive = Pr(e|h) * [Pr(h)*Pop.] = Pr(e|~h) * [Pr(~h)*Pop. ] Total of this row ~ test positive = below - above Total of this row Total.003*Pop..997*Pop.100,000

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The Table Method for Wendy: has CC~ have CCTotal tests positive = Pr(e|h) * [Pr(h)*Pop.] = Pr(e|~h) * [Pr(~h)*Pop. ] Total of this row ~ test positive = below - above Total of this row Total30099,700100,000

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The Table Method for Wendy: has CC~ have CCTotal tests positive = True Positive Rate (.9) * 300 = Pr(e|~h) * [Pr(~h)*Pop. ] Total of this row ~ test positive = below - above Total of this row Total30099,700100,000

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The Table Method for Wendy: has CC~ have CCTotal tests positive 270= Pr(e|~h) * [Pr(~h)*Pop. ] Total of this row ~ test positive = below - above Total of this row Total30099,700100,000

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The Table Method for Wendy: has CC~ have CCTotal tests positive 270= Pr(e|~h) * [Pr(~h)*Pop. ] Total of this row ~ test positive 30= below - above Total of this row Total30099,700100,000

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The Table Method for Wendy: has CC~ have CCTotal tests positive 270= False positive rate (.03) * 99,700 Total of this row ~ test positive 30= below - above Total of this row Total30099,700100,000

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The Table Method for Wendy: has CC~ have CCTotal tests positive 2702,991Total of this row ~ test positive 30= below - above Total of this row Total30099,700100,000

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The Table Method for Wendy: has CC~ have CCTotal tests positive 2702,991Total of this row ~ test positive 3096,709Total of this row Total30099,700100,000

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The Table Method for Wendy: has CC~ have CCTotal tests positive 2702,9913,261 ~ test positive 3096,70996,739 Total30099,700100,000

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The Table Method for Wendy: has CC~ have CCTotal tests positive 270 (true positive) 2,991 (false positive) 3,261 ~ test positive 30 (false negative) 96,709 (true negative) 96,739 Total30099,700100,000

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What are Wendys chances? has CC~ have CCTotal tests positive 270 (true positive) 2,991 (false positive) 3,261 Wendys Chances given that she tests positive are the true positives divided by the number of total tests. That is, 270/3261, which is.083 (8.3%). Those who misestimate that probability forget that colon cancer is rarer than a false positive on a test.

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How about a second test? Note that testing positive (given the test accuracy specified) raises ones chances of having the condition from.003(the base rate) to.083. If we use.083 as the new base rate, those who again test positive then have a 73.1% chance of having the condition. A third positive test (with.731 as the new base rate) raises the chance of having the condition to 98.8%

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Another example: I highly recommend reading the discussion question that runs from p.299-302. See also this excellent Wikipedia write-up that contains an update to the Sally Clark case: http://en.wikipedia.org/wiki/Prosecutor's_fallac y

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Copyright © 2010, 2007, 2004 Pearson Education, Inc. Chapter 14 From Randomness to Probability.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. Chapter 14 From Randomness to Probability.

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