2a:null 2b:["$","title",null,{"children":"Probability Workshop | Chris Wallace | Observable"}] 30:T595,md`# Probability Workshop Hey, thanks for showing up! Today we're going to learn some things about probability. This is the first in a series of (mostly unwritten, so far) workshops that should take you from understanding coin flips up to solving novel problems in state of the art probabilistic programming languages. You may already know some or all of this stuff. If you do, that's great. Hopefully it serves as a useful reminder. If it's all new to you, don't be concerned if you don't get it on the first pass. I promise everything is absolutely understandable, but sometimes the language will be confusing, there may be typos and there may be some flat out errors, so feel free to question anything in the sessions (if you're here via _THE INTERNET_ the searchbar is your friend, or you can ask _me_ if you're taking this in person). The goal for this session is basically "get to Bayes' theorem" in such a way that each step is understandable and the result is not mysterious. Bayes' theorem is absolutely central to modern statistics, and a thorough understanding of it will serve you well. We'll get there with four smaller goals: 1. to understand the notion of probabilities 2. to understand what is meant by _joint_ and _conditional_ probabilities 3. to understand what _Bayes' theorem_ is, and know when and how to apply it 4. to see an example of a common statistical fallacy, and to know how to reason about it `31:T5e4,md`## What is probability? Probability is the mathematical framework we use to represent uncertainty. Uncertainty can be fundamental to the world: - will it rain tomorrow? - what will be the outcome of a coin toss? - what is the quantum state of an atom? Or there may be uncertainty arising from our imperfect knowledge of the world: - does the other poker player have a straight flush? - will a particular bus arrive on time? The distinction between the two is not always clear (and indeed, in many cases is arbitrary, or is a choice we make when modelling). You may be thinking that if you knew the the exact size and weight of a coin, and exactly how it would be tossed and how far it would fall, and controlled for air resistance and such, you might be able to predict the outcome of a toss. You might even be right. One useful delineation to make is between situations in which an event could reasonably be predicted from data that is available to you, and situations where even if you knew everything, you're unsure you could predict anything. In many cases, we don't actually know ahead of time whether a phenomenon is inherently stochastic. In such circumstances, we can apply probabilistic modelling techniques to the system to characterise its uncertainty. In statistical or scientific circles, you might hear these two kinds of uncertainty referred to as _random_ (for example, background noise on the radio) and _systematic_ (for example, a scale that is only accurate to one decimal place). `32:T4cf,md` ## Representing probability We can represent probability in numerous ways. For this workshop, we'll only really be concerned with simple numerical chances. We could refer to these as "point estimates" of full probability distributions. A probability distribution is a more sophisticated way to represent uncertainty. In a probability distribution, we allow some parameter about which we're uncertain to take a range of values, and ascribe degree of confidence to each possible value. The kind of simple probabilities we're dealing with here can be considered as just a single point in the distribution that represents the whole thing. For instance, if we were talking about a "normal distribution" (a.k.a. _Gaussian_ and bell-curve), we might approximate the whole distribution with its mean in some calculations. We'll cover probability distributions in detail in the next workshop. To take a familiar example of a point-like probability, we'd represent the probability of tossing heads on a fair coin with the notation: > ${tex `P(H) = \frac{1}{2}`} Here the ${tex `H`} represents the event that the result of a single toss is a head. The ${tex `P(\dots)`} is denoting "the probability of the events inside these parens".`33:T55f,md`## Joint probability of events and independence How did you arrive at your answer? If you got it right, then consciously or otherwise, you probably recognised that the two events are _independent_: the result of one does not impact the result of the other. Assuming the coin is fair, the probability of two heads is just the probability of one heads times the probability of one heads again. The probability of two events each occurring is known as the _joint probability_, and we usually denote it as ${tex `P(A \cap B)`} or simply ${tex `P(A,B)`}. One subtlety to this is that there is no ordering of events in the joint probability. If you were to toss a single coin twice, of course you have to do it first once and then again, but the joint probability doesn't reflect that. Consider tossing two separate identical coins instead; the result does not change. Since no order is implied, ${tex `P(A,B) = P(B,A)`}. For independent events, the joint probability is just the probability of each multiplied together (which we can reorder freely, because we're just multiplying real numbers): > ${tex `P(A, B) = P(A)P(B)`} For something simple like tossing a coin, we can easily compute the full joint probability table and lookup the probability of any given combination of events. In the more general case, a coin can have a bias, which you can set with the slider. `34:T59b,md`The joint probability can be expressed in terms of the conditional probability like so: > ${tex `P(A,B) = P(A|B)P(B)`} This is known as the _chain rule_ of probability. Let's just check that it makes sense. The joint probability is the probability of ${tex `A`} _and_ ${tex `B`} both being true. Thinking of it as a series of events, one way for both ${tex `A`} and ${tex `B`} to happen is for ${tex `B`} to happen first, which gives us a factor of ${tex `P(B)`}. Then, we need ${tex `A`} to also happen _in the context that ${tex `B`} has already happened_, which brings in a factor of ${tex `P(A|B)`}. This looks a lot like our formula for the joint probability of independent events, except instead of ${tex `P(A)`}, we have ${tex `P(A|B)`}. There's a reason for that, which is that, since ${tex `P(A|B)`} already takes ${tex `B`} to be true, it actually _is_ independent of ${tex `B`}. One note of caution for that example. When we stated that event ${tex `B`} happened _before_ event ${tex `A`}, that was a strictly didactic tool. Neither the conditional nor joint probability assumes or requires any time ordering of events, though the math all still works if events happen to be time ordered. Extending our example from above, to get to the joint probability ${tex `P(\rm{square}, \rm{red}) = P(\rm{square} | \rm{red}) P(\rm{red})`}, we first compute just the probability of the glyph being red, ${tex `P(\rm{red})`}: `35:T40c,md`## Bayes' theorem Now that we understand conditional probabilities, we can get to the heart of probabilistic reasoning: Bayes' theorem (sometimes also called Bayes' _rule_ or Bayes' law). The rule is named for the 18th century statistician Reverend Thomas Bayes, though the relevant work on probability theory was post-humously published by a fellow 18th century statistician. Given the previous work we've looked at today, it's relatively straightforward to demonstrate Bayes' theorem mathematically. Consider the joint probability ${tex `P(A,B)`}. We know from the chain rule that > ${tex `P(A,B) = P(A|B)P(B)`} We also know that in the joint probability the order of events doesn't matter, so also > ${tex `P(A,B) = P(B|A)P(A)`} These are just two expressions for the same thing, so we can equate them. > ${tex `P(A|B)P(B) = P(B|A)P(A)`} Finally giving us Bayes' Theorem: > ${tex `P(A|B) = \frac{P(B|A)P(A)}{P(B)}`} I _know_ it just looks like some equation, so I'll do what I can to convince you how important it is. `36:Tba8,md`The accuracy of the test is ${Number(100*accuracyOfTest).toFixed(0)}%. The question of interest to us is "What is the probability that someone actually has the disease given that they have tested positive." That sounds like a job for Bayes' theorem. Writing it out: > ${tex`P(\rm{disease} | \rm{test\ positive}) = \frac{P(\rm{test\ positive} | \rm{disease}) P(\rm{disease})}{P(\rm{test \ positive})}`} We already know _P_(_test positive_|_disease_), the probability that someone will test positive given that have the disease. It's just the accuracy of the test, ${Number(100*accuracyOfTest).toFixed(0)}% (or ${Number(accuracyOfTest).toFixed(2)}). We also know the probability that someone has the disease, _P_(_disease_) = ${probInfected}, since we set that above. We can calculate the probability of testing positive using the sum rule of probabilities (that the probability of an event is the sum of all the ways it can happen). There are only two ways to test positive. Either one has the disease and tests positive, or one does not have the disease and tests positive. The total probability of testing positive is just the sum of those two possibilities: > ${tex`P(\rm{test \ positive}) = P(\rm{test \ positive} | \rm{disease})P( \rm{disease}) + P(\rm{test \ positive}| \rm{no \ disease})P(\rm{no \ disease})`} The probability of not having the disease is just one minus the probability of having the disease, we can substitute in those numbers. > ${tex `P(\rm{test \ positive}) = ${Number(accuracyOfTest).toFixed(3)}*${Number(probInfected).toFixed(3)}+${Number(1-accuracyOfTest).toFixed(3)}*${Number(1-probInfected).toFixed(3)}`} So before we even know whether the person has the disease or not, we know the probability of them testing positive is about ${Number(probInfected*accuracyOfTest + (1-probInfected)*(1-accuracyOfTest)).toFixed(3)}. Substituting everything into Bayes' rule, we find that the probability of someone having the disease given that they've tested positive is ${Number( accuracyOfTest * probInfected/(probInfected*accuracyOfTest + (1-probInfected)*(1-accuracyOfTest)) ).toFixed(2)}. In the case of a rare disease (try a probability of infection of, say, 0.001) and an effective test (an accuracy of, say, 99%), do you find the result surprising? Below we show a population of 400 people who are all potentially infected with a disease. We're assuming the disease is independently and identically distributed among everyone. That means that one person having the disease does not make another more likely to have it. It's probably not a good model of infectious disease - maybe you could try dreaming up something it might be a good model for. Adjust the sliders here or above to see new randomly generated sample outcomes.`37:T773,md`## A note on Frequentism and Bayesianism You may have heard something about there being two camps of statisticians: the Bayesians and the frequentists. The Bayesians are supposedly an airy-fairy bunch, putting their _subjective_ (meant derogatorily) priors all over the place and talking about beliefs, like idiots. The hard-nosed frequentists run only repeatable experiments, and as such will believe only things that they can run in a lab. This is nonsense. There _are_ camps of people who identify as being frequentist or Bayesian. Reading about their differences can be entertaining: [relevant XKCD](https://xkcd.com/1132/). I would strongly suggest that you consider frequentist and Bayesian approaches as tools you can apply in appropriate situations. In a frequentist appoach, probabilities are thought of as frequencies in long running experiments (hence the name). An obvious place to apply this is working out the bias of a coin - simply flip it a million times. The field is more subtle than that though, and events that may occur only once can be reasoned about as a single sample from an imagined population. The Bayesian approach considers probabilities as degrees of belief in an outcome. While this does introduce subjectivity, since this is expressed as a prior distribution, the assumption is part of the model, and should always be made explicit. Indeed, the result of Bayesian inference is conditional on the priors selected, so it's part of the definition. The good news is that with enough observations, the two approaches usually converge to the same answer. In the wild, you'll often find frequentist approaches applied when there is a large amount of data, and Bayesian approaches applied when there is only a small amount of data. I like to note that [if you have big data, you should ask better questions.](http://andrewgelman.com/2018/01/22/big-data-needs-big-model-2/) `38:Tcfd,md`## Prosecutor's fallacy Probability and statistics are useful, and give us a principled means of reasoning about the world, especially under conditions of uncertainty. I think statistical literacy is important for those reasons, but moreover, statistical illiteracy is dangerous. We've already seen that applying Bayesian reasoning to medical diagnosis reveals that it's much less likely that you have a rare disease _even when you test positive for it_ than you might think. One even more pernicious fallacy is known as the "prosecutor's fallacy" and involves mistaking conditional probabilities. The heart of the fallacy is a fundamental misunderstanding of conditional probabilities, which can be neatly encapsulated by the incorrect equality > ${tex`P(A|B) = P(B|A)`} Now that we're familiar with Bayes' theorem, we can instantly recognise this as false - there are missing factors of ${tex `P(A)`} and ${tex `P(B)`}. ### An example in court Imagine a criminal case, where there is some evidence against the defendent. Further, imagine that the evidence is extremely unlikely to occur if the defendent was innocent. As statisticians, we'd describe this with a conditional probability and say "the probability of the evidence, given innocence is low". We'll encode a low probability graphically as a small dot. > ${tex`P(\rm{evidence} | \rm{innocence}) \sim`} Thinking now as a member of the jury, is it likely that this person is guilty? It's extremely unlikely that the evidence would be found were they innocent, after all. We might claim that it's likely they are guilty, and that, considering the evidence, the probability of innocence is low. > ${tex`P(\rm{innocence} | \rm{evidence}) \sim`} Hopefully, when phrased in terms of conditional probabilities, you have spotted our error. We just equated two different conditional probabilities > ${tex`P(\rm{innocence} | \rm{evidence}) = P(\rm{evidence} | \rm{innocence})`} committing the prosecutor's fallacy. We know this is wrong, because Bayes' theorem tells us the correct way to relate conditional probabilities: > ${tex`P(\rm{innocence} | \rm{evidence}) = \frac{P(\rm{evidence} | \rm{innocence})P(\rm{innocence})}{P(\rm{evidence})}`} Now, this doesn't mean that the defendent is innocent. It means we need to properly consider the prior probabilities of innocence and evidence. In the most egregious case, the prior probability of the evidence is low regardless of innocence, and the prior probability of innocence is high. In this case, the low probability of the evidence in general cancels out the low probability of the evidence given innocence and we find that actually, the defendent is likely to be innocent, > ${tex`P(\rm{innocence} | \rm{evidence}) \sim`}${tex`\sim`} ` 39:[] 2e:{"id":"7f744c4faff359c6","slug":"probability-workshop","title":"Probability Workshop","thumbnail":"628988b5f44427a02c6c0e08c34301211069d24d6762afbc7eb28de5b8751a6d","access_level":"public","create_time":"$D2018-07-05T12:03:34.598Z","update_time":"$D2018-07-10T11:08:19.123Z","view_time":null,"first_public_time":"$D2018-07-10T08:43:09.933Z","has_unpublished_edits":null,"fork":false,"stars":17,"starred":false,"owner":{"type":"team","id":"0ac9abcf38e64f7b","login":"cjwallace","name":"Chris Wallace","avatar_url":"https://avatars-next.observableusercontent.com/ddff0ab0c8f4670e38fe8871e77257cde4558310a5516931b79aa56e27b59005","tier":"starter_2024"},"version":748,"publish_version":748,"role":null,"authored":false,"authors":[{"type":"individual","id":"b6d36fa8c41d8f92","login":"cjwallace","name":"Chris Wallace","avatar_url":"https://avatars-next.observableusercontent.com/ddff0ab0c8f4670e38fe8871e77257cde4558310a5516931b79aa56e27b59005","tier":null}],"members":[{"type":"individual","id":"b6d36fa8c41d8f92","login":"cjwallace","name":"Chris Wallace","avatar_url":"https://avatars-next.observableusercontent.com/ddff0ab0c8f4670e38fe8871e77257cde4558310a5516931b79aa56e27b59005","tier":null,"role":"editor"}],"comments":[],"custom_thumbnail":false,"trashed":false,"lapsed":null,"subscribed":null,"forks":14,"collection":null,"type":"notebook","body":{"titling":"auto","title":"Probability Workshop","cells":[{"id":0,"value":"$30","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":16,"value":"$31","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":257,"value":"exercise(`\n#### Exercise: Uncertainty\n_Take a moment to think of an example of one thing that you might consider a fundamental uncertainty and also one thing that is uncertain as a result of your imperfect knowledge of the system, and write them down._\n`)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":261,"value":"$32","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":217,"value":"exercise(`\n#### Exercise: Are you smarter than an MP?\n_Back in 2012, as part of the Royal Statistical Society's \"Getstats\" campaign, 97 British Members of Parliament were asked the question:_\n\n> _\"if you spin a coin twice, what is the probability of getting two heads?\"_\n\n_47% of them got it wrong. Assume the coin is fair. What is the answer?_\n[BBC Article](https://www.bbc.co.uk/news/uk-19801666)\n`)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":78,"value":"$33","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":119,"value":"viewof headProb = html``","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":132,"value":"md`If the probability of tossing a head is ${headProb}, the joint probability table is:\n\n| | H | T |\n|------- | ---------------------------------- | ------------------------------------- |\n| **H** | ${decimal(headProb*headProb)} | ${decimal(headProb*(1-headProb))} |\n| **T** | ${decimal((1-headProb)*headProb)} | ${decimal((1-headProb)*(1-headProb))} |\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":267,"value":"exercise(`\n#### Exercise: Joint probability table\n_If the probability of getting a head on a single toss is 70% and you spin the coin twice, what is the probability of getting one heads and one tails?_\n\n_Do you notice anything about the numbers in the table. What do they sum to? Do they always sum to the same thing? Why?_\n`)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":263,"value":"md`\nAnother quantity that's sometimes of interest is the so-called _union_ of probabilities - the probability of either of two (or more!) events happening. It's denoted by:\n\n> ${tex `P(A \\cup B) = P(A) + P(B) - P(A \\cap B)`}\n\nThis is known as the _sum rule_ of probabilities.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":271,"value":"exercise(md`\n#### Exercise: Union\n_Why do we subtract the joint probability ${tex `P(A \\cap B)`} in the above formula? Thinking about the probabilities in a Venn diagram where the area of each circle represents the probability of the event happening might help._\n\n\n\n`)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":571,"value":"html`

PAUSE HERE

`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":84,"value":"md`## Conditional probability\n\nIn the above, we assumed that events ${tex `A`} and ${tex `B`} were independent. What if they aren't? When we want to compute the probability of one thing _given_ that another thing is true, we need the _conditional_ probability. The conditional probability of ${tex `A`} given ${tex `B`} is written as ${tex `P(A|B)`}.\n\nIf we wanted to write the conditional probability given several other events, we could write ${tex `P(A | B, C, D, \\dots)`}. We could also ask questions about more complicated objects like ${tex `P(A, B, C | D, E)`}, which you should read as \"what is the joint probability of ${tex `A`}, ${tex `B`} and ${tex `C`}, _conditioned on_ ${tex `D`} and ${tex `E`}?\"\n\nTo illustrate conditional probability, let's look at the following dotty chart.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":498,"value":"glyphs = {\n const viz = d3.select(DOM.svg(600, 180));\n viz.append('g')\n\t .selectAll('circle')\n .data(circles)\n .enter()\n .append('circle')\n \t.attr('cx', d => grid(d.x))\n \t.attr('cy', d => grid(d.y))\n \t.attr('r' , '4')\n \t.attr('fill', d => d.colour)\n \n viz.append('g')\n\t .selectAll('square')\n .data(squares)\n .enter()\n .append('rect')\n \t.attr('x', d => grid(d.x) - 4)\n \t.attr('y', d => grid(d.y) - 4)\n \t.attr('width' , '8')\n .attr('height', '8')\n \t.attr('fill', d => d.colour)\n \n return viz.node()\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":545,"value":"md`To find the unconditional probability of a square, ${tex `P(\\rm{square})`} we would just add up the squares and divide that number by the total number of glyphs (squares or circles) on the chart. We can see that two rows out of ten are squares, so ${tex `P(\\rm{square}) = \\frac{2}{10} = \\frac{1}{5}`}. This gives us a nice visual intuition for probability - what we're asking is \"if I pick a place on the grid at random, what is the chance it's a square?\"\n\nNow, to find the _conditional_ probability of a square, we have to put a condition on the grid. What is the conditional probability of a square, given that the glyph is red, ${tex `P(\\rm{square}|\\rm{red})`}? We first put our condition on the grid, and ignore all the non-red marks.`\n","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":559,"value":"redGlyphs = {\n const viz = d3.select(DOM.svg(600, 180));\n viz.append('g')\n\t .selectAll('circle')\n .data(circles)\n .enter()\n .append('circle')\n .filter(d => d.colour == red)\n \t.attr('cx', d => grid(d.x))\n \t.attr('cy', d => grid(d.y))\n \t.attr('r' , '4')\n \t.attr('fill', d => d.colour)\n \n viz.append('g')\n\t .selectAll('square')\n .data(squares)\n .enter()\n .append('rect')\n .filter(d => d.colour == red)\n \t.attr('x', d => grid(d.x) - 4)\n \t.attr('y', d => grid(d.y) - 4)\n \t.attr('width' , '8')\n .attr('height', '8')\n \t.attr('fill', d => d.colour)\n \n return viz.node()\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":563,"value":"md`Now that we've conditioned our grid, we've taken care of the \"${tex `|\\rm{red}`}\" in the conditional probability formula, so we just ask for the ${tex `P(\\rm{square})`} on this restricted set. We can do just as we did before and count up the marks to find that ${tex `P(\\rm{square}|\\rm{red}) = \\frac{8}{24} = \\frac{1}{3}`}.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":322,"value":"exercise(md`\n#### Exercise: Conditional probability\n_Think of two \"real world\" things that are conditionally dependent._\n`)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":554,"value":"$34","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":579,"value":"redGlyphs2 = {\n const viz = d3.select(DOM.svg(600, 180));\n viz.append('g')\n\t .selectAll('circle')\n .data(circles)\n .enter()\n .append('circle')\n \t.attr('cx', d => grid(d.x))\n \t.attr('cy', d => grid(d.y))\n \t.attr('r' , '4')\n \t.attr('fill', d => d.colour == red ? red : '#ccc')\n \n viz.append('g')\n\t .selectAll('square')\n .data(squares)\n .enter()\n .append('rect')\n \t.attr('x', d => grid(d.x) - 4)\n \t.attr('y', d => grid(d.y) - 4)\n \t.attr('width' , '8')\n .attr('height', '8')\n \t.attr('fill', d => d.colour == red ? red : '#ccc')\n \n return viz.node()\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":599,"value":"md`Then bring in the factor of ${tex `P(\\rm{square}| \\rm{red})`}:`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":582,"value":"squareReds = {\n const viz = d3.select(DOM.svg(600, 180)); \n viz.append('g')\n\t .selectAll('circle')\n .data(circles)\n .enter()\n .append('circle')\n .filter(d => d.colour == red)\n \t.attr('cx', d => grid(d.x))\n \t.attr('cy', d => grid(d.y))\n \t.attr('r' , '4')\n \t.attr('fill', '#ccc')\n \n viz.append('g')\n\t .selectAll('square')\n .data(squares)\n .enter()\n .append('rect')\n .filter(d => d.colour == red)\n \t.attr('x', d => grid(d.x) - 4)\n \t.attr('y', d => grid(d.y) - 4)\n \t.attr('width' , '8')\n .attr('height', '8')\n \t.attr('fill', d => d.colour)\n \n return viz.node()\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":606,"value":"md`Counting the dots in each case, we get ${tex `P(\\rm{square}, \\rm{red}) = \\frac{24}{100} \\cdot \\frac{8}{24} = \\frac{8}{100}`}, which we can easily verify in this simple situation by counting just the red squares and the total number of glyphs.\n\nNotice that since ${tex `P(A,B) = P(B,A)`} as we mentioned earlier, we could have arrived at the same answer the other way around, ${tex `P(\\rm{square}, \\rm{red}) = P(\\rm{red} | \\rm{square}) P(\\rm{square})`}, by first calculating the probability of a square:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":608,"value":"squareGlyphs = {\n const viz = d3.select(DOM.svg(600, 180));\n viz.append('g')\n\t .selectAll('circle')\n .data(circles)\n .enter()\n .append('circle')\n \t.attr('cx', d => grid(d.x))\n \t.attr('cy', d => grid(d.y))\n \t.attr('r' , '4')\n \t.attr('fill', '#ccc')\n \n viz.append('g')\n\t .selectAll('square')\n .data(squares)\n .enter()\n .append('rect')\n \t.attr('x', d => grid(d.x) - 4)\n \t.attr('y', d => grid(d.y) - 4)\n \t.attr('width' , '8')\n .attr('height', '8')\n \t.attr('fill', d => d.colour)\n \n return viz.node()\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":613,"value":"md`And then multiplying by the probability of the glyph being red, given that it's a square:`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":617,"value":"redSquares = {\n const viz = d3.select(DOM.svg(600, 180));\n viz.append('g')\n\t .selectAll('square')\n .data(squares)\n .enter()\n .append('rect')\n \t.attr('x', d => grid(d.x) - 4)\n \t.attr('y', d => grid(d.y) - 4)\n \t.attr('width' , '8')\n .attr('height', '8')\n \t.attr('fill', d => d.colour == red ? red : '#ccc')\n \n return viz.node()\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":469,"value":"exercise(md`#### Exercise: Two child families\n\n_Suppose you know a family to have two children. One day, walking past their house, you see one of their children playing in the garden and it is a boy. What is the probability that the other child is a girl?_\n\n_You may assume that in general in the population, the split of girls and boys is even. It might help to draw out a probability table. Try to phrase the question as a conditional probability._\n\n_Now suppose that the parent sees you passing by and says hello. They introduce the child as their youngest and say that their sibling is playing inside. What is the probability that the elder sibling is a girl? Again try to phrase this as a conditional probability._\n`)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":88,"value":"$35","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":373,"value":"{\n const bayes = md`> ${tex `P(A|B) = \\frac{P(B|A)P(A)}{P(B)}`}`;\n bayes.style.color = d3.interpolateRainbow(now / 1000);\n return bayes;\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":417,"value":"md`OK, great, math. What has that ever done for me? Let's consider what this buys us. We just proved that if we know the total probabilities of ${tex `A`} and ${tex `B`}, we can convert between the conditional probabilities of the two. The ramifications of this won't have sunk in yet, but it's _HUGE_. We can now reason backwards. We'll spend more time on this in future workshops.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":441,"value":"exercise(md`\n#### Exercise: Medical Diagnostics\n\n_Imagine that you visit the doctor because you're worried you have a disease. The disease is quite rare, affecting only one in one thousand people. Fortunately, there's a test, which is 99% accurate. If you have the disease, 99% of the time you will test positive. 1% of the time you'll test negative. If you don't have the disease, 99% of the time you'll test negative, and 1% of the time you'll test positive._\n\n_The doctor tests you, and your test comes back positive. Oh no! What's the probability that you have the disease?_\n\n_If this is the first time you've seen a problem like this, you're **expected** to struggle and maybe get it wrong._\n`)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":569,"value":"html`

PAUSE HERE

`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":648,"value":"md`## Bayes' theorem for diagnosis\n\nThe ability to reason under uncertainty is essential to many areas of enquiry: science, business, life. Fortunately, formal mathematical tools for such have long since been developed.\n\nRecall that for two events _A_ and _B_, Bayes' theorem may be written,\n> ${tex`P(A|B) = \\frac{P(B|A)P(A)}{P(B)}`}\n\nOr, in words:\n> the probability of _A_ _given_ _B_ is the product of the probability of _B_ given _A_ and the total (marginal) probability of _A_, divided by the total probability of _B_.\n\nLet's make this more concrete with an example in medical diagnostics.\n\nSuppose there is a disease outbreak in an enclosed population. Maybe it's an airborn virus that turns people into zombies. Maybe it's hiccups. You can choose the probability of a person being infected by dragging this slider.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":654,"value":"viewof infectedSlider1 = html``","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":656,"value":"md`The probability of a person being infected is ${probInfected}.\n\nNow let's say this is a subtle disease, but you've put your best scientists on it and they've come up with a test. While we're playing Elder God, you can also choose the accuracy of a test for the disease. An accuracy of 100% means that the test is always correct and 50% means it is no better than random. (Thought experiment: according to this formulation, why can the test not be less than 50% accurate?)`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":673,"value":"viewof accuracySlider1 = html``","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":675,"value":"$36","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":677,"value":"viewof infectedSlider2 = html``","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":679,"value":"md`The probability of a person being infected is ${probInfected}.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":681,"value":"viewof accuracySlider2 = html``","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":683,"value":"md`The accuracy of the test is ${Number(100*accuracyOfTest).toFixed(0)}%`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":685,"value":"bayesChart = {\n const viz = d3.select(DOM.svg(600, 180));\n viz.append('g')\n\t .selectAll('circle')\n .data(bayesCircles)\n .enter()\n .append('circle')\n \t.attr('cx', d => grid(d.x))\n \t.attr('cy', d => grid(d.y))\n \t.attr('r' , '4')\n \t.attr('fill', d => d.colour)\n \n return viz.node()\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":687,"value":"md` are people who don't have the disease and tested negative - the general populace.\n are people who don't have the disease and tested positive - \"false alarms\".\n are people who have the disease and tested positive - correct diagnoses.\n are people who have the disease and tested negative - missed diagnoses.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":319,"value":"$37","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":95,"value":"$38","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":449,"value":"md`I'd like to tell you about the case of Sally Clark. It is among the most heartbreaking stories I've ever heard, and I need to provide you with a **trigger warning** because the case involves two instances of sudden infant death syndrome and a gross miscarriage of justice. I do not believe it to be flippant to use such an emotive example. This is a real thing that happened and I think it's important we understand it, so that we can not repeat similar mistakes, and so that we understand why statistics is not an abstract subject.\n\n[Wikipedia entry for Sally Clark](https://en.wikipedia.org/wiki/Sally_Clark)`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":700,"value":"md`To my shame as a statistician, I first heard about the case at [Open Vis Conf](http://www.openvisconf.com/) 2018, where [Heather Krause](https://twitter.com/datassist) gave an excellent talk highlighting four common and pernicious statistical fallacies. I recommend you to watch it when the videos are released.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":714,"value":"md`## Further reading\nThat's it for this workshop. I hope you learned something. Those who found this introduction interesting might consider reading further. Below are two resources that are freely available online and are particularly aimed at people who can code, but don't know stats.\n\n- [Think Bayes](http://greenteapress.com/wp/think-bayes/)\n- [Probabilistic Programming and Bayesian Methods for Hackers](https://github.com/CamDavidsonPilon/Probabilistic-Programming-and-Bayesian-Methods-for-Hackers)`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":110,"value":"md`#### Supporting code\nBelow here in the notebook is the code that makes the interactive parts tick.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":644,"value":"md`Utils:`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":138,"value":"decimal = function(number) {\n return (Math.round(number*100)/100).toFixed(2)\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":502,"value":"blue = \"#5DBCD2\"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":509,"value":"red = \"#D2735D\"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":214,"value":"exercise = function(text) {\n const box = md`${text}`;\n box.style.background = '#e4f6fb';\n box.style.padding = '20px 20px 5px 20px';\n box.style.border = '1px solid black';\n box.style.maxWidth = '640px';\n return box;\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":537,"value":"grid = d3.scaleLinear()\n .domain([0, 40])\n .range([20, 580])","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":637,"value":"md`Conditional probability:`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":540,"value":"circles = {\n var arr = [];\n for (let i = 1; i <= 10; ++i) {\n for (let j = 3; j <= 10; ++j) {\n if (i >= 9) {\n var col = red;\n } else {\n var col = blue;\n }\n \t arr.push({x: i, y: j, colour: col})\n \t}\n }\n return arr\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":543,"value":"squares = {\n var arr = [];\n for (let i = 1; i <= 10; ++i) {\n for (let j = 1; j <= 2; ++j) {\n if (i >= 7) {\n var col = red;\n } else {\n var col = blue;\n }\n \t arr.push({x: i, y: j, colour: col})\n \t}\n }\n return arr\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":646,"value":"md`Bayes:`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":671,"value":"bayesCircles = {\n var arr = []\n for (let i = 1; i <= 40; ++i) {\n for (let j = 1; j <= 10; ++j) {\n\t var col = '#eaeded' // light grey\n \n let infected = (Math.random() <= probInfected) ? true : false\n \n if (infected) {\n\t let test_positive = Math.random() <= accuracyOfTest\n\t col = test_positive ?\n \t \t'#f5b7b1' : // light red\n \t \t'#a93226' // dark red\n } else {\n let test_positive = Math.random() > accuracyOfTest\n\t col = test_positive ?\n \t\t '#f7dc6f' : // yellow\n \t\t\t col\t // still light grey\n }\n \n \t arr.push({x: i,\n y: j,\n colour: col\n })\n \t}\n }\n return arr\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":658,"value":"mutable accuracyOfTest = 0.99","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":660,"value":"{\n if (accuracySlider1 !== mutable accuracyOfTest) {\n mutable accuracyOfTest = accuracySlider1;\n viewof accuracySlider2.value = accuracySlider1;\n viewof accuracySlider2.dispatchEvent(new CustomEvent(\"input\"));\n }\n} ","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":662,"value":"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":663,"value":"{\n if (accuracySlider2 !== mutable accuracyOfTest) {\n mutable accuracyOfTest = accuracySlider2;\n viewof accuracySlider1.value = accuracySlider2;\n viewof accuracySlider1.dispatchEvent(new CustomEvent(\"input\"));\n }\n} ","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":665,"value":"mutable probInfected = 0.01","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":667,"value":"{\n if (infectedSlider1 !== mutable probInfected) {\n mutable probInfected = infectedSlider1;\n viewof infectedSlider2.value = infectedSlider1;\n viewof infectedSlider2.dispatchEvent(new CustomEvent(\"input\"));\n }\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":669,"value":"{\n if (infectedSlider2 !== mutable probInfected) {\n mutable probInfected = infectedSlider2;\n viewof infectedSlider1.value = infectedSlider2;\n viewof infectedSlider1.dispatchEvent(new CustomEvent(\"input\"));\n }\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":642,"value":"md`Imports:`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":235,"value":"d3 = require('d3')","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":279,"value":"import {color, button} from \"@jashkenas/inputs\"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"}],"files":"$W39","stdlib":"1"}} 2f:"$2e:body"