Critical And Creative Thinking Lecture 5 PDF

Critical And Creative Thinking Lecture 5 Probability and Frequency IMPORTANT NOTICE There is substantial evidence from research in cognitive psychology and neuroscience that distraction can negatively affect learning. Distractions interfere with one’s ability to process and retain information, l...

Critical And Creative Thinking Lecture 5 Probability and Frequency IMPORTANT NOTICE There is substantial evidence from research in cognitive psychology and neuroscience that distraction can negatively affect learning. Distractions interfere with one’s ability to process and retain information, leading to decreased focus, reduced comprehension, and impaired memory consolidation. For this reason, laptops and tablets are strictly prohibited in this class (unless otherwise instructed). If your laptop or tablet is visible you will be marked as having an unauthorized absence from the class. Each unauthorized absence will reduce your overall grade. Four or more unauthorized absences mean an automatic grade of 0 for the class. All slides will be made available on Canvas after the class. If you like to take notes, you may use pen and paper. 2 Probability and Frequency  Please download and install the Slido app on all computers you use Suppose you are playing a lottery, where you pick 6 numbers between 1 and 59, then 6 number- balls between 1 and 59 are chosen at random. Which sequence of balls is more likely to be chosen? ⓘ Start presenting to display the poll results on this slide. 4 Representativeness Heuristic Both sequences of numbers are equally (un)likely. But when people without training in critical thinking are asked this question, they will typically choose 12,50,6,4,32,47. This is due to a System 1 heuristic or cognitive shortcut: the representativeness heuristic. We tend to judge as more likely examples we regard as more typical. The sequence 12,50,6,4,32,47 more closely resembles a typical example or stereotype of a sequence of lottery numbers than 1,2,3,4,5,6. 5 Representativeness Heuristic Like other heuristics or cognitive shortcuts, the representativeness heuristic doesn’t always give the wrong results. For example, four-legged dogs are more typical than three-legged dogs, and they are more common. However, it often leads to bad probabilistic reasoning. 6 Representativeness Heuristic A famous example is the gambler’s fallacy. If, for example, a roulette ball lands red ten times in a row, people typically judge it more likely to land black the next time. This is because a sequence with a mixture of black and red more closely resembles a typical random sequence than a sequence of eleven reds. However, each roulette result is probabilistically independent of other results. The previous results don’t make it any more likely to land black or red this time. 7 Representativeness Heuristic Another example of the representativeness heuristic distorting probabilistic reasoning is the conjunction fallacy. Consider the following scenario: There has been a shooting in a suburban neighbourhood known for gang violence. Which is more likely? 1. The shooting was committed by a man. 2. The shooting was committed by a male gang member. 8 Representativeness Heuristic People typically judge that 2 is more likely. But this is mathematically impossible. Men If the shooting was committed by a male gang member, it must also be true that the shooting was committed by a man. Male gang members So the probability of the shooting being committed by a male gang member cannot be higher than the probability of the shooting being committed by a man. 9 Base Rate Neglect Think back to our puzzle in Lecture 1 about the rare disease X. Almost everyone (no matter how intelligent or educated) tends to overestimate probabilities in cases such as this. This is because of something called the base rate fallacy or base rate neglect: we fail to factor in the base rate or prior probability of something being the case. Although having symptoms associated with disease X makes it more likely you have disease X, the prior probability of having disease X is so low, it is still very unlikely you will have it – even with the symptoms. 10 Frequency Trees Frequency Trees Frequency trees are System 2 method of calculating probabilities, that make clear both base rates and conditional probabilities. Using frequency trees avoids relying on the representativeness heuristic and rules out base rate neglect. Studies also suggest that using frequency trees trains and improves our intuitions about probabilities. 12 Frequency Trees Suppose you have an important flight to catch tomorrow, but it will be delayed if it rains that day. You have the following information: Probability of rain = 20% Probability of a delay if it rains = 75% Probability of a delay if it does not rain = 10% How likely is it that the flight will be delayed? 13 Frequency Trees We can represent a 20% probability of rain as a frequency. A 20% probability of rain corresponds to a frequency of 20 out of every 100 days being rainy. We can represent this using the frequency tree on the right. (Note, there is nothing special about starting with 100 days; you can start with a different number. But 100 is a convenient number to use here.) Frequency Trees The probability of a delay, given rain, is 75%. 75% of 20 is 15. We represent this in our frequency tree like this. Frequency Trees The probability of a delay, given no rain, is 10%. 10% of 80 is 8. We represent this in our frequency tree like this. Frequency Trees By looking at the frequency tree, we can now see that out of every 100 days there will be 15 + 8 = 23 where the flight is delayed. So there is a 23% probability of a delay. Frequency Trees Frequency trees are also a helpful way to understand the probabilities involved in medical diagnostics. Suppose a male in a low-risk group tests positive for HIV. 0.1% of men in low-risk groups have HIV. The test is highly accurate. If someone has HIV, there is a 99.9% chance he will test positive. And if someone does not have HIV, there is a 99.99% chance he will test negative. What is the probability this man has HIV? 18 Frequency Trees A false positive is when a test gives a positive result for something that is not present. A false negative is when a test gives a negative result for something that is present. In HIV tests, there is a 0.01% chance of a false positive result 0.1% chance of a false negative result 19 Frequency Trees There is a 0.1% prevalence of HIV in low-risk groups. This is also known as the base rate or prior probability of having HIV (for those in low- risk groups). We can represent the base rate in a frequency tree this way. Frequency Trees The one male (out of 10,000) in a low risk group with HIV has a 99.9% chance of testing positive. So approximately 1 out of 1 male with HIV will test positive. This is represented in the frequency tree this way. Frequency Trees Out of 9,999 men in low risk groups without HIV, 99.99% will test negative. So approximately 9,998 out of 9,999 men without HIV will test negative. This is represented in the frequency tree this way. Frequency Trees By looking at the frequency tree, we can see that (approximately) out of every 2 people who test positive, 1 will have HIV. So (for a male in a low risk group), a positive test for HIV means approximately a 50% chance of having HIV. Base Rate Neglect Even though the test for HIV is extremely accurate, a positive result only means a 50% chance of having HIV (for those in low-risk groups). Psychologists have shown that people (including doctors and other medical professionals) tend to be very bad at making probabilistic judgements in cases like these. This is because System 1 thinking conflates the chance of diagnostic error (the false positive rate) with the overall chance of not having the illness given a positive result. This is the fallacy of base rate neglect: ignoring the role of base rates in calculating probabilities. 24 Base Rate Neglect Frequency trees make it clear that, in order to know the probability of something, given some probabilistic evidence, we must know: the base rate the false positive rate the false negative rate Unless we can know all three of these, we cannot know the probability of something, given some probabilistic evidence. For example, we cannot know the probability of a disease, given a test, unless we know all three rates. 25 Activity Use a frequency tree to calculate the following probability from the example used in Lecture 1. Your tongue is covered in blue spots. This symptom is associated with disease X, which afflicts 1 in every 100,000 people. Everyone with disease X gets this symptom, and 5% of people without disease X get the symptom. What is the probability you have disease X? 26 Activity Out of 5001 with symptoms, 1 has disease X. So the probability of having disease ! X given the symptoms is = "##! ~0.0002 The Probability Calculus The Probability Calculus Every mathematical truth about probability can be derived from just 3 axioms (fundamental principles): 1. 1 ≥ Prob(p) ≥ 0 2. If p is a logically necessary truth, then Prob(p) = 1 3. If p and q are incompatible (they cannot both be true), then Prob(p or q) = Prob(p) + Prob(q). 29 The Probability Calculus First, some points about notation: We use lower-case sentence letters p, q, r etc. to represent claims. We write ‘Prob(p)’ to mean ‘The probability of p’. We write ‘Prob(p|q)’ to mean ‘The probability of p conditional on q’/’The probability of p, given q’/’The probability of p, assuming q’. 30 The Probability Calculus AXIOM 1 1 ≥ Prob(p) ≥ 0 This tells us that, for any proposition p, the probability of p is a number between 0 and 1. Prob(p) = 0 means that there is no chance of p being true. Prob(p) = 1 means that there is no chance of p being false. 31 The Probability Calculus AXIOM 2 If p is a logically necessary truth then Prob(p) = 1. If p is logically necessary, then it must be true. As such, it has a probability of 1. Notice that AXIOM 2 doesn’t say that if Prob(p) = 1, then p is a logically necessary truth. Anything logically necessary is inevitably true, but things can be inevitably true without being logically necessary. 32 The Probability Calculus AXIOM 3 If p and q are incompatible, then Prob(p or q) = Prob(p) + Prob(q). Coin Coin Example: lands h lands t A coin landing h is incompatible with it landing t. Prob(h) = 0.5. Prob(t) = 0.5. So, Prob(h or t) = 0.5 + 0.5 = 1 33 The Probability Calculus AXIOM 3 If p and q are incompatible, then Prob(p or q) = Prob(p) + Prob(q). Notice that AXIOM 3 applies only when p and q p p&q q are incompatible. Say the probability of p (that it will rain tomorrow) is 0.7, and the probability of q (that there will be high winds tomorrow) is 0.6. It does not follow that Prob(p or q) = 1.3. 34 The Probability Calculus The problem with adding Prob(p) and Prob(q) in cases where p and q overlap (can be true together) is that we double count the cases where they overlap. p p&q q To handle this, we can use a General Disjunction Rule: Prob(p or q) = Prob(p) + Prob(q) – Prob(p & q) 35 Conditional Probability Say you wanted to find out the probability of wearing glasses, given that someone is a VinUni student. This is the probability of wearing glasses conditional on being a VinUni student. You could do this by finding out the proportion of VinUni students that wear glasses. You would find out the number of VinUni students find out the number of those wearing glasses, and divide the number wearing glasses by the number of students 36 Conditional Probability For instance, if there are 2000 VinUni students, and 500 wear glasses, then the probability of wearing glasses conditional on q:= VU Students "## ! being a VinUni student is or. $### % p & q:= VU In other words, we look at the number of p & students with q-cases (VinUni students & wear glasses) and glasses divide it by the number of q-cases VinUni students. 37 Conditional Probability More generally, Prob(p|q) (the probability of p conditional on q) is given by the proportion of q-cases that are p-cases. q:= VU Students That’s the number of p & q-cases, divided by the number of p-cases. This gives us the p & q:= VU students with Conditional Probability Rule: glasses Prob(p|q) = Prob(p & q) / Prob(q) 38 The Probability Calculus The Conditional Probability Rule tells us that Prob(p|q) = Prob(p & q) / Prob(q). Rearranging this equation gives us the: General Conjunction Rule Prob(p & q) = Prob(q) x Prob (p|q) Intuitively: if you want to know the probability of p and q together, you need to combine the probability of q on its own, and the probability of p when q is true. 39 The Probability Calculus Since p and not-p must be incompatible, AXIOM 3 tells us that Prob(p or not-p) = Prob(p) + Prob(not-p) ‘p or not-p’ is logically necessary. So Prob(p or not-p) = 1 So, Prob(p) + Prob(not-p) = 1. This gives us the Negation Rule Prob(not-p) = 1 – Prob(p) 40 Recap: The Probability Calculus AXIOM 1: 0 ≤ Prob(p) ≤ 1. AXIOM 2: If p is logically necessary, then Prob(p) = 1. AXIOM 3: If p and q are incompatible, then Prob(p or q) = Prob(p) + Prob(q). General Disjunction Rule: Prob(p or q) = Prob(p) + Prob(q) – Prob(p & q). General Conjunction Rule: Prob(p & q) = Prob(q) x Prob(p|q). Negation Rule: Prob(not-p) = 1 – Prob(p) Conditional Probability Rule: Prob(p|q) = Prob(p & q) / Prob(q). 41 Activity Use the axioms and rules to calculate the following probabilities. State the answer and the rule used to obtain it. 1. Prob(p) = 0.4. Prob(not-p) = ? 2. Prob(p) = 0.2. Prob(q|p) = 0.4. Prob(p & q) = ? 3. Prob(p) = 0.3. Prob(q) = 0.3. Prob(p & q) = 0. Prob(p or q) = ? 4. Prob(p) = 0.6. Prob(q) = 0.6. Prob(p & q) = 0.3. Prob(p or q) = ? 5. Prob(p) = 0.2. Prob(q) = 0. Prob(p & q) = ? 6. Prob(p or q) = 0.7. Prob(not-p & not-q) = ? 7. Prob(p|q) = 0. Prob(q) = 0.6. Prob(p & q) = ? 42 Activity 1. Prob(p) = 0.4. Prob(not-p) = ? Negation Rule: Prob(not-p) = 1 – Prob(p) Answer: Prob(not-p) = 1 – 0.4 = 0.6 43 Activity 2. Prob(p) = 0.2. Prob(q|p) = 0.4. Prob(p & q) = ? General Conjunction Rule: Prob(p & q) = Prob(q) x Prob(p|q). Answer: Prob(p & q) = 0.2 x 0.4 = 0.08 44 Activity 3. Prob(p) = 0.3. Prob(q) = 0.3. Prob(p & q) = 0. Prob(p or q) = ? AXIOM 3: If p and q are incompatible, then Prob(p or q) = Prob(p) + Prob(q). or General Disjunction Rule: Prob(p or q) = Prob(p) + Prob(q) – Prob(p & q). Answer: Prob(p or q) = 0.3 + 0.3 = 0.6 45 Activity 4. Prob(p) = 0.6. Prob(q) = 0.6. Prob(p & q) = 0.3. Prob(p or q) = ? General Disjunction Rule: Prob(p or q) = Prob(p) + Prob(q) – Prob(p & q). Answer: Prob(p or q) = 0.6 + 0.6 – 0.3 = 0.9 46 Activity 5. Prob(p) = 0.2. Prob(q) = 0. Prob(p & q) = ? General Conjunction Rule: Prob(p & q) = Prob(q) x Prob(p|q). Answer: Prob(p & q) = 0 x whatever the value of Prob(p|q) is = 0 47 Activity 6. Prob(p or q) = 0.7. Prob(not-p & not-q) = ? Negation Rule: Prob(not-p) = 1 – Prob(p) Answer: This one is tricky. ‘not-p & not-q’ is equivalent to ‘not-(p or q)’. So, Prob(not-p & not-q) = 1 – 0.7 = 0.3 48 Activity 7. Prob(p|q) = 0. Prob(q) = 0.6. Prob(p & q) = ? General Conjunction Rule: Prob(p & q) = Prob(q) x Prob(p|q). Answer: Prob(p & q) = 0 x 0.6 = 0 49 Bayes’ Theorem Bayes’ Theorem Rev. Thomas Bayes (c.1701–1761) was an English philosopher and Presbyterian minister who is famous for formulating Bayes’ theorem. Bayes’ theorem is entailed by the three axioms of the probability calculus. Bayes’ theorem is a useful tool for calculating the probability of a hypothesis, given some evidence. 51 Bayes’ Theorem Let h be a hypothesis or theory, and e be some evidence. Then Bayes’ theorem is written: Prob(e|h) → Prob(h) AAACU3icbVHLSgMxFM2M9VVfVZdugkXQTZkRqW4E0Y3LCvYBnVIy6R0bzTxI7ohlnH8UwYU/4saFpg/Rtl4InJx7TnJz4idSaHScd8teKCwuLa+sFtfWNza3Sts7DR2nikOdxzJWLZ9pkCKCOgqU0EoUsNCX0PQfrob95iMoLeLoFgcJdEJ2F4lAcIaG6pbuPYQnVGFWU7F/2M+ff/aQH9Fz6gWK8WxKA7+avtF4KELQdPqYozyfMRmmWyo7FWdUdB64E1Amk6p1S69eL+ZpCBFyybRuu06CnYwpFFxCXvRSDQnjD+wO2gZGzAzSyUaZ5PTAMD0axMqsCOmI/evIWKj1IPSNMmTY17O9Iflfr51icNbJRJSkCBEfXxSkkmJMhwHTnlDAUQ4MYFwJMyvlfWZiRPMNRROCO/vkedA4rrjVSvXmpHxxOYljheyRfXJIXHJKLsg1qZE64eSFfJAvi1hv1qdt24Wx1LYmnl0yVfbGN7YntsA= Prob(h|e) = Prob(e) 52 Metaphysical vs Epistemic Probability So far we have discussed frequencies. Frequencies are a kind of metaphysical probability. Metaphysical probability concerns how objectively probable some claim is, due to facts about the world. Here, we will understand Bayes’ theorem in terms of epistemic probability. Epistemic probability concerns how subjectively probable some claim is for some person or group, due to the evidence they possess. Winning the lottery is objectively improbable. But a person might have excellent evidence that they have won the lottery, in which case it is epistemically probable (for them) that they have won the lottery. 53  Bayes’ Theorem “Likelihood”: how probable evidence e Prior probability: is, assuming the truth how probable the of hypothesis h. hypothesis h is, before we learn of Posterior evidence e. probability: the Prob(e|h) → Prob(h) AAACU3icbVHLSgMxFM2M9VVfVZdugkXQTZkRqW4E0Y3LCvYBnVIy6R0bzTxI7ohlnH8UwYU/4saFpg/Rtl4InJx7TnJz4idSaHScd8teKCwuLa+sFtfWNza3Sts7DR2nikOdxzJWLZ9pkCKCOgqU0EoUsNCX0PQfrob95iMoLeLoFgcJdEJ2F4lAcIaG6pbuPYQnVGFWU7F/2M+ff/aQH9Fz6gWK8WxKA7+avtF4KELQdPqYozyfMRmmWyo7FWdUdB64E1Amk6p1S69eL+ZpCBFyybRuu06CnYwpFFxCXvRSDQnjD+wO2gZGzAzSyUaZ5PTAMD0axMqsCOmI/evIWKj1IPSNMmTY17O9Iflfr51icNbJRJSkCBEfXxSkkmJMhwHTnlDAUQ4MYFwJMyvlfWZiRPMNRROCO/vkedA4rrjVSvXmpHxxOYljheyRfXJIXHJKLsg1qZE64eSFfJAvi1hv1qdt24Wx1LYmnl0yVfbGN7YntsA= probability of Prob(h|e) = hypothesis h, given Prob(e) evidence e. Expectedness: the probability of evidence e, whether or not the hypothesis h is true.  Bayes’ Theorem The more likely e is given h, the stronger The higher the prior e is as evidence for h. probability of h, the higher the posterior probability of h. Posterior probability: the Prob(e|h) → Prob(h) AAACU3icbVHLSgMxFM2M9VVfVZdugkXQTZkRqW4E0Y3LCvYBnVIy6R0bzTxI7ohlnH8UwYU/4saFpg/Rtl4InJx7TnJz4idSaHScd8teKCwuLa+sFtfWNza3Sts7DR2nikOdxzJWLZ9pkCKCOgqU0EoUsNCX0PQfrob95iMoLeLoFgcJdEJ2F4lAcIaG6pbuPYQnVGFWU7F/2M+ff/aQH9Fz6gWK8WxKA7+avtF4KELQdPqYozyfMRmmWyo7FWdUdB64E1Amk6p1S69eL+ZpCBFyybRuu06CnYwpFFxCXvRSDQnjD+wO2gZGzAzSyUaZ5PTAMD0axMqsCOmI/evIWKj1IPSNMmTY17O9Iflfr51icNbJRJSkCBEfXxSkkmJMhwHTnlDAUQ4MYFwJMyvlfWZiRPMNRROCO/vkedA4rrjVSvXmpHxxOYljheyRfXJIXHJKLsg1qZE64eSFfJAvi1hv1qdt24Wx1LYmnl0yVfbGN7YntsA= probability of Prob(h|e) = hypothesis h, given Prob(e) evidence e. The more we can expect e, whether or not h is true, the weaker e is as evidence for h. Bayes’ Theorem: Evidence Bayesian thinking is very useful, even when we don’t have exact numerical probabilities to plug into Bayes’ theorem. This is because it shows us what factors are required to determine how strongly a piece of evidence supports a hypothesis and how reasonable a hypothesis is once we factor in that evidence. likelihood → prior 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 Posterior = !"#$!"%&&' tells us how strongly a piece of evidence supports expectedness $()$*+$',$-- a hypothesis In addition, Bayes’ theorem reminds us that we must also take into account how reasonable a hypothesis was before gaining the evidence, in order to know how reasonable a hypothesis is once we factor in that evidence. 56 Bayes’ Theorem: Evidence Suppose a palm reader says that you will pass THINK1010, and you do! Is this good evidence that palm reading works? Even though we can’t plug in exact probabilities in this case, Bayes’ theorem shows us it isn’t. It is very likely that you will pass THINK1010 anyway. I.e. likelihood → prior 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 expectedness is high, so the palm reader’s correct prediction Posterior = gives very little support (at most) to the hypothesis that palm expectedness reading works. (If the palm reader correctly predicted very unexpected events, that would support the hypothesis that palm reading works.) Bayes’ Theorem: Priors Recall the previous example of presenting with the symptoms (having blue spots on your tongue) of disease X. Does this evidence e make the hypothesis h that you have disease X probable? Even if we don’t have data on exact probabilities, Bayes’ theorem shows us that it is impossible to know whether the likelihood → prior 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 hypothesis is likely without knowing the prior probability of Posterior = having disease X. expectedness So unless we have a rough idea of a prior probability, we should suspend judgement on how probable some evidence makes some hypothesis. Activity Turn the argument below into a Bayesian argument. That is, formulate the argument in terms of claims about: 1. Posterior: the probability of the hypothesis h given the evidence e 2. Likelihood: the probability of the evidence e given the hypothesis h 3. Prior: the prior probability of the hypothesis h before considering the evidence e 4. Expectedness: the probability of the evidence e whether or not the hypothesis h is true You do not need to use precise numbers/values. You can use values like ‘high’, ‘low’ etc. Argument: Given the evidence, there is strong reason to believe COVID19 was developed in a laboratory. The virus has features that are extremely unlikely to develop naturally. Furthermore, many laboratories develop viruses, so it’s not an unreasonable hypothesis in the first place. Bayes’ Theorem & Frequency Trees Bayes’ Theorem & Frequency Trees Let’s use an example from before to see how frequency trees correspond to Bayes’ theorem. This frequency tree shows that the base rate of HIV for men in low risk groups is 1 in every 10,000. This corresponds to a prior probability of the hypothesis (Prob(p)) of 0.0001. Bayes’ Theorem & Frequency Trees For anyone with HIV, there is a 99.9% chance of testing positive. This corresponds to a likelihood of the evidence (a positive test result) given the the hypothesis (having HIV) of 0.999. Bayes’ Theorem & Frequency Trees For anyone without HIV, there is a 99.99% chance of testing negative, and so a 0.01% chance of testing positive. This corresponds to a probability of the evidence (a positive test result) given the negation of the hypothesis (not having HIV) of 0.0001. Bayes’ Theorem & Frequency Trees The frequency tree allows us to calculate expectedness Prob(e). Out of 10,000 men, ~2 men have the evidence of a positive test. ! So Prob(e) ≈ = 0.0002. "#,### Bayes’ Theorem & Frequency Trees The frequency tree also allows us to calculate the posterior probability Prob(h|e). Out of ~2 with the evidence, ~1 is HIV positive. " So Prob(h|e) = ~ = ~0.5. !