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title: 'Square visualization of probabilities on two events: (example) Diseasitis',
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text: '$$\n\\newcommand{\\bP}{\\mathbb{P}}\n$$\n\n[summary: $$ \\newcommand{\\bP}{\\mathbb{P}} $$\n\nWe can visualize the [22s]:\n\n<img src="http://i.imgur.com/ssokgJ2.png" width="601" height="322">\n\n\nThen we can see at a glance that the probability $\\bP(D \\mid B)$ of the patient having Diseasitis given that the tongue depressor is black, despite some intuitions, isn't all that big: \n\n$$\\bP(D \\mid B) = \\frac{\\bP(B, D)}{ \\bP(B, D) + \\bP(B, \\neg D)} = \\frac{3}{7} $$\n<img src="http://i.imgur.com/NJOydOD.png" width="541" height="324">\n\n]\n\nFrom the [22s]: \n\n> You are screening a set of patients for a disease, which we'll call Diseasitis. Based on prior epidemiology, you expect that around 20% of the patients in the screening population will in fact have Diseasitis. You are testing for the presence of the disease using a tongue depressor containing a chemical strip. Among patients with Diseasitis, 90% turn the tongue depressor black. However, 30% of the patients without Diseasitis will also turn the tongue depressor black. Among all the patients with black tongue depressors, how many have Diseasitis? \n\nIt seems like, since Diseasitis very strongly predicts that the patient has a black tongue depressor, it should be the case that the [-1rj] $\\bP( \\text{Diseasitis} \\mid \\text{black tongue depressor})$ is big. But actually, it turns out that a patient with a black tongue depressor is more likely than not to be completely Diseasitis-free.\n\nCan we see this fact at a glance? Below, we'll use the [-496] to draw pictures and use our visual intuition.\n\n\nTo introduce some notation: our [1rm prior probability] $\\bP(D)$ that the patient has Diseasitis is $0.2$. We think that if the patient is sick $(D)$, then it's 90% likely that the tongue depressor will turn black $(B)$: we assign conditional probability $\\bP(B \\mid D) = 0.9$. We assign conditional probability $\\bP(B \\mid \\neg D) = 0.3$ that the tongue depressor will be black even if the patient isn't sick. We want to know $\\bP(D \\mid B)$, the [1rp posterior probability] that the patient has Diseasitis given that we've seen a black tongue depressor.\n\nIf we wanted to, we could solve this problem precisely using [1lz]: \n\n$$\n\\begin{align}\n\\bP(D \\mid B) &= \\frac{\\bP(B \\mid D) \\bP(D)}{\\bP(B)}\\\\\n&= \\frac{0.9 \\times 0.2}{ \\bP(B, D) + \\bP(B, \\neg D)}\\\\\n&= \\frac{0.18}{ \\bP(D)\\bP(B \\mid D) + \\bP(\\neg D)\\bP(B \\mid \\neg D)}\\\\\n&= \\frac{0.18}{ 0.18 + 0.24}\\\\\n&= \\frac{0.18}{ 0.42} = \\frac{3}{7} \\approx 0.43\\ .\n\\end{align}\n$$\n\nSo even if we've seen a black tongue depressor, the patient is more likely to be healthy than not: $\\bP(D \\mid B) < \\bP(\\neg D \\mid B) \\approx 0.57$. \n\nNow, this calculation might be enlightening if you are a real expert at [1lz]. A better calculation would probably be the [1x9 odds ratio form of Bayes's rule].\n\nBut either way, maybe there's still an intuition saying that, come on, if the tongue depressor is such a strong indicator of Diseasitis that $\\bP(B \\mid D) = 0.9$, it must be that $\\bP(D \\mid B) =big$.\n\nLet's use the [496 square visualization of probabilities] to make it really visibly obvious that $\\bP(D \\mid B) < \\bP(\\neg D \\mid B)$, and to figure out why $\\bP(B \\mid D) = big$ doesn't imply $\\bP(D \\mid B) =big$. \n\nWe start with the probability of $\\bP(D)$ (so we're [factoring_probability factoring] our probabilities by $D$ first): \n\n<img src="http://i.imgur.com/oFctjqn.png" width="330" height="363">\n\nNow let's break up the red column, where $D$ is true and the patient has Diseasitis, into a block for the probability $\\bP(B \\mid D)$ that $B$ is also true, and a block for the probability $\\bP(\\neg B \\mid D)$ that $B$ is false.\n\n>Among patients with Diseasitis, 90% turn the tongue depressor black.\n\nThat is, in 90% of the outcomes where $D$ happens, $B$ also happens. So $0.9$ of the red column will be dark ($B$), and $0.1$ will be light:\n\n<img src="http://i.imgur.com/ra6maoF.png" width="478" height="358">\n\n\n>However, 30% of the patients without Diseasitis will also turn the tongue depressor black.\n\nSo we break up the blue $\\neg D$ column by $\\bP(B \\mid \\neg D) = 0.3$ and $\\bP(\\neg B \\mid \\neg D) = 0.7$: \n\n\n<img src="http://i.imgur.com/ssokgJ2.png" width="668" height="358">\n\nNow we would like to know the probability $\\bP(D \\mid B)$ of Diseasitis once we've observed that the tongue depressor is black. Let's break up our diagram by whether or not $B$ happens: \n\n<img src="http://i.imgur.com/yKv723w.png" width="413" height="402">\n\nConditioning on $B$ is like only looking at the part of our distribution where $B$ happens. So the probability $\\bP(D \\mid B)$ of $D$ conditioned on $B$ is the proportion of that area where $D$ also happens: \n\n<img src="http://i.imgur.com/imiJdSH.png" width="541" height="324">\n\nHere we can see why $\\bP(D \\mid B)$ isn't all that big. It's true that $\\bP(B,D)$ is big relative to $\\bP(\\neg B,D)$, since we know that $\\bP(B \\mid D)$ is big (patients with Diseasitis almost always have black tongue depressors): \n\n<img src="http://i.imgur.com/kJvtM1X.png" width="471" height="340">\n\nBut this ratio doesn't really matter if we want to know $\\bP(D \\mid B)$, the probability that a patient with a black tongue depressor has Diseasitis. What matters is that we also assign a reasonably high probability $\\bP(B, \\neg D)$ to the patient having a black tongue depressor but nevertheless *not* suffering from Diseasitis:\n\n<img src="http://i.imgur.com/NJOydOD.png" width="541" height="324">\n\nSo even when we see a black tongue depressor, there's still a pretty high chance the patient is healthy anyway, and our [-1rp] $\\bP(D\\mid B)$ is not that high. Recall our square of probabilities: \n\n<img src="http://i.imgur.com/ssokgJ2.png" width="601" height="322">\n\nWhen asked about $\\bP(D\\mid B)$, we think of the really high probability $\\bP(B\\mid D) = 0.9$:\n\n<img src="http://i.imgur.com/Hy2FyYP.png" width="601" height="322">\n\nReally, we should look at the part of our probability mass where $B$ happens, and see that a sizeable portion goes to places where $\\neg D$ happens, and the patient is healthy:\n\n<img src="http://i.imgur.com/Kk4IgfK.png" width="601" height="322">\n\nSide note\n---\n\nThe square visualization is very similar to [1x1 frequency diagrams], except we can just think in terms of probability mass rather than specifically frequency. Also, see that page for [1x1 waterfall diagrams], another way to visualize updating probabilities.\n',
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