tag:blogger.com,1999:blog-6894866515532737257.comments2016-09-27T17:32:47.682-07:00Probably Overthinking ItAllen Downeyhttps://plus.google.com/111942648516576371054noreply@blogger.comBlogger664125tag:blogger.com,1999:blog-6894866515532737257.post-71351507574032455592016-09-27T17:32:47.682-07:002016-09-27T17:32:47.682-07:00And I didn't mean to sound snarky towards you....And I didn't mean to sound snarky towards you. Thanks for your comments!Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-19752113461864377832016-09-27T17:18:23.267-07:002016-09-27T17:18:23.267-07:00[Allen, I didn't mean for my comment to sound ...[Allen, I didn't mean for my comment to sound snarky towards you! My cheekiness was directed at the hypothetical interviewer asking this question :) ]<br /><br />Indeed, the question ends by asking for a probability. I totally agree with your solution to treating this as a toy probability puzzle. I agree there's no Bayes-vs-frequentist difference there. And I agree it's very important to distinguish Bayes' theorem from Bayesian inference.<br /><br />On the other hand, the question *starts* with "You want to know if you should bring an umbrella."<br />Let's take this seriously as an interview question, meant to help the interviewer decide which candidates would bring the most value to the company.<br /><br />What are they really getting at?<br />If they were asking:<br />"We want to know if our company should take action U. It only makes sense to invest in performing U if there's at least 50% chance that R is true. We can (just barely) afford to run 3 expensive tests. Each test independently has 2/3 chance of correctly identifying whether R is true. If a sensible prior on U is 10%, and all 3 tests come back True, what will be the probability that R is indeed true?"<br /><br />...then who would you rather hire?<br />Candidate A, who is content to stop after getting an answer of 47%?<br />Or Candidate B, who goes on to say:<br />"Look, even if all 3 tests agree in claiming that R is true, our estimated probability that R is true will *still* be under 50%. In all other cases, it'll be even lower. You're wasting money by running these 3 tests. If we can't afford more tests, let's just skip them and spend that money somewhere useful."<br /><br />The spirit of frequentism is to step back from the given data and understand the operating characteristics of your statistical procedures---not just the analyses but the data-collection (design) too. It's a useful habit of mind, whether your final inferences/analyses end up Frequentist or Bayesian. And yep, sometimes that means refusing to answer a particular question, when that question isn't what's actually needed.Jerzy Wieczorekhttp://www.blogger.com/profile/03611849167717252118noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-14353809901579712992016-09-27T12:57:40.183-07:002016-09-27T12:57:40.183-07:00Thanks, Dimiter. You make some great points and I...Thanks, Dimiter. You make some great points and I won't address them all, but I want to clarify one. The goal of my article is not to demonstrate the "added value of a Bayesian vs frequentist answer". The point I am trying to make is that the probability question, as stated, does not have two answers, one Bayesian and one frequentist answer. It has one answer that can be computed without any commitment to a Bayesian or frequentist interpretation of probability, and without any commitment to Bayesian or frequentist inference.<br /><br />It does, as you point out, require the choice of a reference class, but that is a general difficulty with many probability problems; it is not a special difficulty for Bayesianism or frequentism.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-11569655580038207642016-09-27T12:48:10.263-07:002016-09-27T12:48:10.263-07:00two long related comments: 1) Why not use the prio...two long related comments: 1) Why not use the priors for the particular days you will be visiting (I guess the chance of rain varies significantly over the year)? But then why not use as a prior the conditional probability of rain in Seattle given the atmospheric conditions at the moment? Or even, why not commission specific research to inform your prior: why stop at consulting the Western Regional Climate Center or the weather forecast? You can say - not worth it for the problem at hand, but this requires a separate analysis of how much effort it is worth spending on establishing a good prior for this problem. In our case, the answer is probably 'close to zero', as the costs of taking an umbrella are negligible. But then for somebody with zero knowledge about the weather in Seattle a flat prior of rain/no rain, or equivalently a frequentest analysis, would seem as justified as any other. 2) Which brings me to the second point. The problem is introduced as a decision-theoretic one (bring an umbrella or not) but then it asks for a probability that, however defined and computed, is not sufficient to answer the decision-theoretic motivating problem. And it seems to me that the added value of a Bayesian vs. a frequentist answer to the probability question cannot be demonstrated outside of a decision-theoretic setup in which the costs of establishing a prior are compared to the benefits of increased precision of the answer. (And you cannot just say, oh but everybody knows the prior chance of rain in Seattle is 0.5 or 0.1 or whatever, as this info is not provided in the set-up of the problem). Dimiter Toshkovhttp://www.blogger.com/profile/09098718685262606935noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-28998916468175189542016-09-27T11:48:02.294-07:002016-09-27T11:48:02.294-07:00Thanks, Jerzy. The question asks for a probabilit...Thanks, Jerzy. The question asks for a probability; I think your analysis answers a different question.<br /><br />But refusing to answer the question is certainly in the spirit of frequentism.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-79765744841293372322016-09-27T11:19:18.817-07:002016-09-27T11:19:18.817-07:00Perhaps a more Frequentist-spirited answer would b...Perhaps a more Frequentist-spirited answer would be to discuss the study design:<br /><br />Your prior prob. of rain is under 50%, and in fact it's so low (at 10%) that *nothing* your 3 friends say could convince you the (posterior) prob. of rain is over 50%, even when they all agree it is raining.<br /><br />In other words, your study has no power to change your mind! (from the prior decision that it's probably not raining.)<br /><br />So why did you hassle your friends by asking them in the first place? Maybe that's why they lie to you 2/3 of the time :)Jerzy Wieczorekhttp://www.blogger.com/profile/03611849167717252118noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-36817919654265621822016-09-27T09:14:51.665-07:002016-09-27T09:14:51.665-07:00Hi Russ. A helpful reader submitted the following...Hi Russ. A helpful reader submitted the following explanation, which I accidentally rejected instead of publishing. So, with apologies to the helpful reader:<br /><br />my test blog has left a new comment on your post "Bayes's Theorem is not optional": <br /><br />If you look at the answer linked to in the post: https://www.glassdoor.com/Interview/You-re-about-to-get-on-a-plane-to-Seattle-You-want-to-know-if-you-should-bring-an-umbrella-You-call-3-random-friends-of-y-QTN_519262.htm<br /><br />and subsitute 10% chance of rain for 25%, you should get the answer listed here:<br />0.1*(8/27) / ( 0.1*8/27 + 0.9*1/27 )<br />8/270 / 8/270 + 9/270<br />8/17<br />47.06%<br /><br />In Downey's version, he uses both odds and probabilities, which makes the calculations in this case easier but, maybe, harder to follow. <br /><br />Here's the top answer from the linked post:<br /><br />Bayesian stats: you should estimate the prior probability that it's raining on any given day in Seattle. If you mention this or ask the interviewer will tell you to use 25%. Then it's straight-forward:<br /><br />P(raining | Yes,Yes,Yes) = Prior(raining) * P(Yes,Yes,Yes | raining) / P(Yes, Yes, Yes)<br /><br />P(Yes,Yes,Yes) = P(raining) * P(Yes,Yes,Yes | raining) + P(not-raining) * P(Yes,Yes,Yes | not-raining) = 0.25*(2/3)^3 + 0.75*(1/3)^3 = 0.25*(8/27) + 0.75*(1/27)<br /><br />P(raining | Yes,Yes,Yes) = 0.25*(8/27) / ( 0.25*8/27 + 0.75*1/27 )<br /><br />**Bonus points if you notice that you don't need a calculator since all the 27's cancel out and you can multiply top and bottom by 4.<br /><br />P(training | Yes,Yes,Yes) = 8 / ( 8 + 3 ) = 8/11<br /><br />But honestly, you're going to Seattle, so the answer should always be: "YES, I'm bringing an umbrella!"<br />(yeah yeah, unless your friends mess with you ALL the time ;)<br /><br />Interview Candidate on Sep 12, 2013 Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-36036828069056310512016-09-27T02:24:01.914-07:002016-09-27T02:24:01.914-07:00Thanks a lot for sharing this!!Thanks a lot for sharing this!!Biagio Chiricohttp://www.blogger.com/profile/18044407043088889045noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-32960971274446582392016-09-26T22:15:26.763-07:002016-09-26T22:15:26.763-07:00I was hoping to understand how to use Bayesian rea...I was hoping to understand how to use Bayesian reasoning, but I was completely lost by the Bayesian argument. Would you mind elaborating the reasoning behind the segment of the post that starts at "A base rate of 10 ... " and ends at "Probability(8 Odds(p))". (I don't even understand how to read that final bit of notation!) Thanks.Russ Abbotthttp://www.blogger.com/profile/15431389045571531450noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-61692278484356774412016-09-19T18:04:57.020-07:002016-09-19T18:04:57.020-07:00I thought this is how you get tenure...I thought this is how you get tenure...Jason Moorehttp://www.blogger.com/profile/15362357639624306439noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-22300994389967245382016-09-19T06:11:03.899-07:002016-09-19T06:11:03.899-07:00I agree with Creosote. Halfers and thirders can no...I agree with Creosote. Halfers and thirders can not be separated by any decision problem - if they could, the puzzle would indeed be trivial. <br /><br />Both halfers and thirders would bet on 1/3 if asked to make one bet per wake-up, and on 1/2 if asked to make one bet per experiment (although their argumentation might be slightly different.)<br /><br />Further, if the experiment is modified so that we have different people (or Sleeping Beauty-clones) waking up at the possible wake-up events, everyone agrees on the 1/3 answer. <br /><br />The halfer position is an ordinary proposition about an ordinary probability. It is consistent with the facts and leads to correct decisions.Mikkel Schmidthttp://www.blogger.com/profile/06840892229326863085noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-87872123183485905252016-09-19T05:54:20.567-07:002016-09-19T05:54:20.567-07:00Regarding the red dice problem, a halfer would cla...Regarding the red dice problem, a halfer would claim (correctly I think) that once the experimental procedure has been explained, one should believe there is a 2/3 chance the last die is mostly red.<br /><br />When you then say that the actual outcome is red (an event which happens with probability one) that provides no new information, and the belief remains the same.Mikkel Schmidthttp://www.blogger.com/profile/06840892229326863085noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-73964928618752496382016-09-18T16:11:20.168-07:002016-09-18T16:11:20.168-07:00Hi Jason, Good to hear from you, and thanks for t...Hi Jason, Good to hear from you, and thanks for this comment. But maybe you should wait until you have tenure before you say things like this.<br /><br />Just kidding! (I hope.)Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-25817518731030557672016-09-17T10:44:53.499-07:002016-09-17T10:44:53.499-07:00Great post Allen. I so get this now after my first...Great post Allen. I so get this now after my first year at UCD where I'm supposedly hired to be an innovator in engineering education. The walls became all too clear as soon as I started asking more provocative questions and trying to do things just a little differently. I certainly don't feel that I'm empowered to make waves.<br /><br />I experimented in my classes right from the get go, but as soon as some of my superiors got wind, I never received "that's awesome keep it up". It was more like, "you better talk to the old guard and see if they approve".<br /><br />Just this week we visited a demo of a "service" class. The Circuits 101 course for non-electrical engineering majors. That course is quite disconnected from what the students in my mechanical engineering department actually need to be able to do when implementing their labs and projects that require some electrical engineering.<br /><br />And the casual warnings of "don't do anything crazy till after you get tenure" are so annoying.Jason Moorehttp://www.blogger.com/profile/15362357639624306439noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-91060316911878996462016-09-16T08:22:25.195-07:002016-09-16T08:22:25.195-07:00That's very helpful. Thanks!That's very helpful. Thanks!Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-67205422141194663592016-09-15T12:08:21.616-07:002016-09-15T12:08:21.616-07:00Thanks for that link. The response from Adriano s...Thanks for that link. The response from Adriano shows another way to answer the question, but it doesn't use Bayes's theorem. I think the method using Bayes's theorem is easier, but maybe not everyone thinks so.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-27775106159205120922016-09-15T12:04:53.165-07:002016-09-15T12:04:53.165-07:00If Oliver left blood at the scene, he accounts for...If Oliver left blood at the scene, he accounts for the O sample, leaving only one sample to be explained. The chance that one random person has AB blood is 1%.<br /><br />If Oliver did not leave blood at the scene, there are two samples to be explained. If you draw random pairs from the population, you might get [AB, O] or [O, AB] with equal probability, so there are two ways to generate the evidence.<br /><br />If that's not clear, the best way to prove it to yourself is to enumerate all possible pairs.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-80902387206358174252016-09-15T02:35:42.551-07:002016-09-15T02:35:42.551-07:00I enjoyed this blog post very much. I didn't t...I enjoyed this blog post very much. I didn't think about scale free networks for some years. However, I see some waek points:<br />- the only parameter that alows you to controll the degree sequence and the clustering coefficient independently of the average degree is the rewiring probability. this means that the clustering coefficient and the distribution do strongly dependend on each other. <br />-every edge is part of a triangle. <br /><br />There is a the model of hyperbolic randomg raphs which became very popular in the last few years. Its possible to control its clustering, average degree and degree sequence. The degree sequence can be power law distributed, the clustering can be strong and the average degree can be small. The model was introduced here:<br />http://arxiv.org/pdf/1006.5169.pdf<br />Later we analyzed it rigorously here: http://arxiv.org/pdf/1205.1470.pdf<br /><br />Since then many other usefull properties like small average path lengh, efficient greedy routing etc. have been proven. Moreover, on a phylosophical level, it give (in my opinion) a very nice explanation for the properties of social networks, the internet graphs and other small world network. <br />Ueli Peterhttp://www.blogger.com/profile/07399700772199193833noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-53371462267221401922016-09-14T07:09:19.827-07:002016-09-14T07:09:19.827-07:00Thanks for the positive response.
IMO, this contr...Thanks for the positive response.<br /><br />IMO, this controversy stresses how the term "new information" gets tossed about without a definition. Specifically, halfers use it as the cornerstone of their solution, without defining it.<br /><br />Here's what I think it means: First, define a non-trivial partition X to be a set of (1) disjoint events that (2) span the sample space and (3) all have non-zero probability. If an information event I occurs such that the union of I with each member of X is no longer a non-trivial partition, then I is "new information" that requires probabilities to be updated by dividing by the new sum.<br /><br />Usually, this situation comes about when the information tells us that certain events are no longer possible. If tell you that a die-roll was even, then Pr(1∩I)=Pr(3∩I)=Pr(5∩I)=0, and these events can't be in a non-trivial partition. The updated probabilities are found by dividing Pr(2∩I)=Pr(4∩I)=Pr(6∩I)=1/6 by their sum, 1/2.<br /><br />In the Sleeping Beauty Problem on Sunday, {Coin=H∩Wake=Mon, Coin=T∩Wake=Mon} suffices to make a non-trivial partition. Each event has probability 1/2, and they are disjoint events. Coin=T∩Wake=Mon and Coin=T∩Wake=Tue are not disjoint because they represent the same future/world/whatever. But when SB is wakened, Coin=T∩Wake=Mon and Coin=T∩Wake=Tue become disjoint, so an update is required.JeffJohttp://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-47571357797154577452016-09-13T17:03:04.712-07:002016-09-13T17:03:04.712-07:00Thank you for publishing these problems. I had so...Thank you for publishing these problems. I had some trouble figuring out how P(E) was derived in the M&M problem (#2). I asked for help on math overflow and got a good answer.. May be helpful to others who got similarly stuck -> http://math.stackexchange.com/questions/1924904/urn-type-problem-with-bayes-theorem-and-mms-dont-understand-how-probability<br />Chris Bedfordhttp://www.blogger.com/profile/15601997591378792422noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-62871977694719436302016-09-13T15:11:33.791-07:002016-09-13T15:11:33.791-07:00I'm having troubles going through the fourth p...I'm having troubles going through the fourth problem.<br /><br />Why would p(E|X) be equal to .01 <br /><br />I think you get that result from multiplying .01 (The probability of finding AB type blood) by 1, given that it makes sense to find O type blood if Oliver was actually involved in the crime.<br /><br />But why didn't you multiply that by a factor of 2? Why wouldn't the permutation factor matter in one case (When testing the hypothesis of Oliver being guilty), but it does in the other? <br /><br />Thank you! :D Unknownhttp://www.blogger.com/profile/07599875152333798994noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-43661707978311479082016-09-13T08:50:33.423-07:002016-09-13T08:50:33.423-07:00I agree.I agree.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-36060764206107818952016-09-13T05:56:05.905-07:002016-09-13T05:56:05.905-07:00What creates controversy in this problem, is how t...What creates controversy in this problem, is how to treat the fact the SB may be awake on a different day. Specifically, is that a different event (Elga), or the same event (Lewis). But there is a simple way to eliminate that issue by addressing a different, but equivalent, question.<br /><br />Four women volunteer to participate in an experiment that uses all but one of the procedures in the original Sleeping Beauty Problem.. One coin is flipped on Sunday night after all four are put to sleep. All four will be wakened (in separate rooms) once, or twice, depending on the result of that coin flip:<br /><br />SB1 will be wakened on Monday, and also on Tuesday if the result is Tails.<br /><br />SB2 will be wakened on Monday, and also on Tuesday if the result is Heads.<br /><br />SB3 will be wakened on Tuesday, and also on Monday if the result is Tails.<br /><br />SB4 will be wakened on Tuesday, and also on Monday if the result is Heads.<br /><br />This way, exactly three of the women will be awakened on each day of the experiment. Each will be asked the question "What is your credence, now, for the proposition that today is the only day you will be wakened?"<br /><br />It is trivial to see that SB1's experiment is identical to the original Sleeping Beauty Problem. It is also trivial to see that SB2, SB3, and SB4 are in an experiment that is functionally equivalent, varying only it the specifics of the schedule.<br /><br />When SB1 finds herself awake, she knows that exactly two other volunteers are awake. She also knows that the proposition "today is the only day you will be wakened" is true for exactly one of these three, and that the Principle of Indifference applies to which it is.<br /><br />The only possible answer to this version of the problem is 1/3. The answer to the original Sleeping Beauty Problem has to be 1/3, because SB1's is problem is the same problem.JeffJohttp://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-50357071917685529842016-09-04T09:34:14.859-07:002016-09-04T09:34:14.859-07:00Allen, the Thirder position can be equally inconse...Allen, the Thirder position can be equally inconsequential: just ask a Thirder how her credence for Heads leads Beauty to pick the correct jar of beans (http://allendowney.blogspot.co.uk/2015/06/the-sleeping-beauty-problem.html?showComment=1446826222182#c466456992572929693).<br /><br />Imagine we used slightly different language for the problem. The interviewer asks Beauty, "what do you believe about the state of the coin?". Halfer-Beauty answers, "with respect to experiments, a half-chance of Heads". Thirder-Beauty answers, "with respect to wake-ups, a third-chance of Heads". Both are correct, and both lead to exactly the same decisions.<br /><br />That's what I reckon is really happening here: a mostly-unconscious tendency to express the coin flip w.r.t. experiments or w.r.t. wakeups. And the fact that contributors to threads like this cannot be separated in decision problems is some evidence in favour of that interpretation.Creosotehttp://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-27193407979861196832016-09-04T06:16:26.342-07:002016-09-04T06:16:26.342-07:00I think you and I have a different understanding o...I think you and I have a different understanding of the Halfer position. I think it's an ordinary proposition about an ordinary probability, and because it's wrong, someone who uses it to make decisions will make bad ones.<br /><br />It sounds like you interpret the Halfer position as a belief that has no consequences. If so, it's not very interesting.<br /><br />I don't think that's what Halfers mean, but maybe it's not worth debating what Halfers mean.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.com