tag:blogger.com,1999:blog-6894866515532737257.comments2017-01-19T05:33:39.152-08:00Probably Overthinking ItAllen Downeyhttps://plus.google.com/111942648516576371054noreply@blogger.comBlogger695125tag:blogger.com,1999:blog-6894866515532737257.post-56785656768293897892017-01-18T21:13:56.190-08:002017-01-18T21:13:56.190-08:00Why is it that when calculating the variance in Ka...Why is it that when calculating the variance in Kaplan Meier survival curves, the underlying distribution of the population is not taken into account--only the number at risk is? This is counter-intuitive. The only references are to a report by Greenwood in 1926 which doesn't really seem to answer the question.Mark Phillipshttp://www.blogger.com/profile/03120283972507328308noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-47616136231365182042017-01-18T10:46:50.416-08:002017-01-18T10:46:50.416-08:00There was disagreement among statisticians about h...There was disagreement among statisticians about how to do hypothesis testing; what we ended up with is actually a weird hybrid that makes less sense than the alternatives. See https://en.wikipedia.org/wiki/Statistical_hypothesis_testing#Origins_and_early_controversy<br /><br />Regarding precision and recall, those are the terms most often used in the context of information retrieval. In the context of stats, it's usually false positives and false negatives (or type I and type II errors). And in the context of machine learning, it's usually a confusion matrix. But they are all representations of the same information.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-45725377773039094592017-01-17T15:02:17.176-08:002017-01-17T15:02:17.176-08:00I think there was some other place where I found a...I think there was some other place where I found an explanation that there are two approaches to hypothesis validation. One used p-values. The other one was something like precision and recall (false positives and negatives) or something similar. I think it mentioned some disagreement between statisticians before p-values became popular. May this be possible?trylks yeahhttp://www.blogger.com/profile/14821108955806430210noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-84018391878133304622017-01-05T09:34:34.881-08:002017-01-05T09:34:34.881-08:00Great examples. Congratulations! Great examples. Congratulations! Hugo Pireshttp://www.blogger.com/profile/00400200472463525909noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-73832185952477179532017-01-02T09:20:23.283-08:002017-01-02T09:20:23.283-08:00Have you seen my four-volunteer version? Each will...Have you seen my four-volunteer version? Each will be wakened at least once, and maybe twice, based on the same coin flip. Each will be left asleep in only one set of conditions, different for each, as defined by the cross product {Monday,Tuesday}x{Heads,Tails}. Each will be asked for her credence that the coin landed on the side that would let her sleep through one day.<br /><br />One of these volunteers is undergoing the identical problem as in the original problem. Three are undergoing a functionally equivalent one, that must have the same answer.<br /><br />On any day, exactly three volunteers will be wakened. On any day, exactly one of the three awake volunteers is in the set of conditions she is asked about. On any day, each of the three awake volunteers has the same information upon which to base her credence.<br /><br />Yet that credence is found by a "bunch of probability calculations." It is 1/3.JeffJohttp://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-14019635407645045782016-12-20T14:23:45.564-08:002016-12-20T14:23:45.564-08:00Christopher> Am I missing something here?
No-o...Christopher> Am I missing something here?<br /><br />No-one's disputing the maths:<br /> * [A] One half of experiments in which Beauty wakes are those in which the coin is Heads.<br /> * [B] One third of Beauty's experimental wake-ups are those in which the coin is Heads.<br /><br />Beauty knows the facts, but she is obliged to pick just one of them as the basis of her "credence for Heads". Halfers insist she pick [A], Thirders insist on [B]. She's not allowed to state the reference class (w.r.t. experiments/wakeups) in her answer: her "credence" must be unqualified and absolute.<br /><br />So there's an interesting philosophical question lurking at the heart of this: when Beauty can see both perspectives, [A] and [B], what then do we mean, fundamentally, by her "credence"? If there is a correct answer to the Sleeping Beauty problem then it's a philosophical one, not bunch of probability calculations.Creosotehttp://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-72168284068907806242016-12-17T21:28:06.322-08:002016-12-17T21:28:06.322-08:00Alright, let's assume Beauty believes the thir...Alright, let's assume Beauty believes the thirder perspective and decides that she will answer "I think it landed on Tails, the odds are 2 in 3" every time she is questioned because she thinks she'll be correct more often. She gets asked twice if it comes up Tails after all, and all these mathematicians can't be wrong.<br /><br />As soon as she enters sleep/stasis and the coin is flipped there emerge two timelines, T1 where the fair coin came up Heads and T2 where the fair coin came up Tails. Since it's a fair coin T1 happens 50% of the time and T2 happens 50% of the time. This means that in T1 Beauty says Tails and is incorrect, whereas in T2 Beauty says Tails and is correct.<br /><br />Doesn't this mean that she's correct in 50% of timelines and incorrect in 50% of timelines? There are only two possible timelines and they happen 50% of the time each since it's a fair coin. In fact she could be asked a thousand times in T2 and since her memories are wiped and reset she'll always answer Tails. Furthermore, doesn't this mean that if you repeat the experiment 1,000 times, you get 500 Heads and 500 Tails, and she answers Tails every time... that she's correct in 500 experiments and incorrect in 500 experiments?<br /><br />The fact that she answers incorrectly once in T1 and correctly twice in T2 doesn't make her answer twice as correct. You're only tracking whether she answered correctly, not how many times she gets it right. I mean, you're not giving her 100$ every time she gets the answer right. If you did then she'd always say Tails because she has a 50% of getting 200$ and a 50% of getting 0$, whereas if she said Heads she'd have a 50% chance of getting 100$ and a 50% chance of getting 0$. What you're looking at is her certainty that the coin came up Heads or Tails. And if my previous logic is correct, that means that guessing Tails makes her correct in 50% of timelines... so she'd break even instead of being ultimately correct approximately 666-667 times out of 1,000 experiments.<br /><br />I can't believe that the process of being interviewed would lead her to believe that the odds of it being Heads or Tails had changed. This isn't a Monty Haul problem, she's not getting any new information. She knows that she's going to be woken up at least once and has no capacity to distinguish between Heads(Monday), Tails(Monday), and Tails(Tuesday)... or the 997th Tails day for that matter.<br /><br />Am I missing something here?Christopherhttp://www.blogger.com/profile/16259364826167235453noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-31672788098146932342016-11-30T13:28:12.526-08:002016-11-30T13:28:12.526-08:00Yes, this "natural frequency" way of sol...Yes, this "natural frequency" way of solving problems like this is excellent.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-43918002372392650512016-11-30T13:16:22.615-08:002016-11-30T13:16:22.615-08:00Simplest way to answer this question is to assume ...Simplest way to answer this question is to assume a number for total population.<br />Assume that 400 is the combined population of all three states. Then each of state1,state2,state3 will be with population 160,100,140 respectively. From the data provided we can derive that number of people supporting party1 in state1,state2,state3 are 0.5*160,0.6*10,0.35*140 (= 80,60,49). <br />From the above derived information we can evaluate the probability that a party1 supporter will be from state2 is 60/(80+60+49) = 0.317 Approx.Unknownhttp://www.blogger.com/profile/04089618677994568066noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-46134659329321860422016-11-29T18:25:50.126-08:002016-11-29T18:25:50.126-08:00^_^^_^张亮http://www.blogger.com/profile/11113984241425650034noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-63620806772327350092016-11-29T13:26:31.343-08:002016-11-29T13:26:31.343-08:00At least some of the forecasters had the uncertain...At least some of the forecasters had the uncertainty of the future included in their models, so they generated two predictions: one if the election were held today and another that took into account the remaining time until the election. The second was generally closer to 1/2 because of the greater uncertainty.<br /><br />But I don't think it's right to say that if the level of uncertainty is high, the probability is necessarily 1/2.<br /><br />To put that differently: a year ago I would not have bet on Trump with even odds.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-50178042642302586312016-11-29T08:24:53.292-08:002016-11-29T08:24:53.292-08:00I think your 41% differs from my 30% because you a...I think your 41% differs from my 30% because you actually computed your answer, while I just made mine up as an example. I'm actually mildly surprised that I was as close as I was. Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-38944453231781143462016-11-29T03:46:24.607-08:002016-11-29T03:46:24.607-08:00What do you think about the criticism which insist...What do you think about the criticism which insists that their probability must be 1/2 until the election approaches, since the poll is so stochastic that we can never predict the future in long term.Stanhttp://www.blogger.com/profile/16725096203497460549noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-35498289081639575272016-11-26T06:21:48.920-08:002016-11-26T06:21:48.920-08:00This comment has been removed by the author.Unknownhttp://www.blogger.com/profile/02468670897396780927noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-5114770962749636732016-11-23T03:10:59.223-08:002016-11-23T03:10:59.223-08:00I think the biggest problem was the bogus precisio...I think the biggest problem was the bogus precision: "71.4%" - this implicates that the model could give a precision of one per thousand - which is of course bs.vonjdhttp://www.blogger.com/profile/12488764399725481497noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-75230078812694086762016-11-16T18:06:17.255-08:002016-11-16T18:06:17.255-08:00Always enjoy your writing Allen, thank you.
For m...Always enjoy your writing Allen, thank you.<br /><br />For me, this whole post screamed 'Probably Overthinking it'. It seems like the problem fits neatly into a use case of a discriminative statistical model like logistic regression for example. I think anyone who claims that the results of such a model are unfair or not useful is probably overthinking it.Adam Levinhttp://www.blogger.com/profile/07717303642060714074noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-90890487169420147272016-11-02T16:10:40.004-07:002016-11-02T16:10:40.004-07:00TheRingshifter> Does the "amnesia" el...TheRingshifter> Does the "amnesia" element really change it that much?<br /><br />The amnesia is necessary in order to argue about Beauty's credence within the experiment itself, i.e. to prevent Tuesday Beauty saying "oh, hang on, you woke me yesterday ... this must be a Tails time-line".<br /><br />But should Beauty feel any differently about the coin flip within the experiment than she did prior to it? That's a key sticking-point of the Sleeping Beauty problem.Creosotehttp://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-74342049259542223102016-11-01T09:04:43.904-07:002016-11-01T09:04:43.904-07:00Am I wrong saying this...
Probabilistically, the ...Am I wrong saying this...<br /><br />Probabilistically, the question is quite simple. Every time the coin is flipped, it IS 1/2 odds (it's a coin). But every time she is WOKEN UP it's 2/3 tails and 1/3 heads. Is it really that different to just saying something like, "I'm going to slap you twice on a tails, and once on a heads. How likely is it, at each point I've slapped you, that you got a heads or a tails?". Does the "amnesia" element really change it that much? It's the same here... if you are betting on the coin flip, it's still 1/2, but if you are being slapped, then it's more likely it's because you got a tails since 2/3rds of the slaps will end up being attributable to tails and only 1/3rd of them to heads. <br /><br />So to me... it seems like although the "wager" problem is interesting, you are indeed "overthinking" it. Surely the thing that makes the difference is that she is possibly being asked twice about what the coin landed on when she wakes up, but when the bet is resolved, she is only being asked once. <br /><br />Simply, if you do this, say, 6 times, and get 3 heads and 3 tails, and she bets on heads, she will earn money by being correct about the wager, but will also be MOSTLY correct about saying tails is more likely - because she has been asked about the coin NINE times (3 on heads, 6 on tails) and been correct SIX times.TheRingshifterhttp://www.blogger.com/profile/18363611127790007978noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-32594185956365346362016-10-31T19:14:02.503-07:002016-10-31T19:14:02.503-07:00Prof Downey,
Thanks for the article. How did you ...Prof Downey, <br />Thanks for the article. How did you get these values:P(H|FF) = 0.26 after the second, and P(H|FFF) = 0.07 after the third.<br /><br />Thanks<br />Anonymoushttp://www.blogger.com/profile/10202239324994703634noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-6883500172906903422016-10-14T15:04:15.174-07:002016-10-14T15:04:15.174-07:00It's hard to want to get married when I saw so...It's hard to want to get married when I saw so many people in the previous generation or two got divorced. As a man, the idea of losing half my stuff sucks. On top of that if there are kids involved I would only get to see them on Tuesdays and every other weekend, and I'd also get the pleasure of watching some other guy raise them. This makes me very cautious to anyone I'd consider marrying. Greg Bhttp://www.blogger.com/profile/08033946748326670909noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-72125112234101430192016-10-14T13:26:45.120-07:002016-10-14T13:26:45.120-07:00Excellent data viz! As a married 32yo millennial, ...Excellent data viz! As a married 32yo millennial, I can see the shift in my friend group, many of whom have never been married. My wedding was in 2014 when I was 30.Joshua Oliveirahttp://www.blogger.com/profile/04949363293575817646noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-36946818009232125082016-10-13T06:06:30.186-07:002016-10-13T06:06:30.186-07:00Hi. I just added a worksheet that shows how to so...Hi. I just added a worksheet that shows how to solve this problem using Bayes's Theorem.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-56656754707304130282016-10-13T05:50:47.838-07:002016-10-13T05:50:47.838-07:00Hi Emile, I'm glad it is making sense.
But I...Hi Emile, I'm glad it is making sense.<br /><br />But I want to clarify the point of my article. I am not saying that there are two answers to this question, one Bayesian and one frequentist, and the Bayesian one is right.<br /><br />I am saying that there is only one answer to this question, and it is neither Bayesian nor frequentist. It is just a consequence of the laws of probability.<br /><br />The issue Dimiter raised is called the reference class problem: https://en.wikipedia.org/wiki/Reference_class_problem<br /><br />And while the reference class problem is relevant to the problem, it is a general problem for all of probability, and not specifically Bayesian or frequentist, either.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-25120124062450485922016-10-12T18:28:14.111-07:002016-10-12T18:28:14.111-07:00I deleted my initial comment because I was probabl...I deleted my initial comment because I was probably overthinking it. I grappled with your explanation and the other posts, but now understand your statement that only a Bayesian approach can lead to the right answer. <br /><br />We want Prob(raining|YYY).<br />This is equal to Prob(raining and YYY)/Prob(YYY). We don't know Prob(raining and YYY).<br /><br />However Prob(raining and YYY) also equals Prob(raining)*Prob(YYY|raining)<br /><br />So we need Prob(raining)but don't know this. So as already mentioned by Dimiter, there are infinitely many answers in the absence of information. <br />However, we do know Prob(YYY|raining)=8/27<br /><br />Furthermore, <br />Prob(YYY) = Prob(YYY|raining)*Prob(raining) + Prob(YYY|not raining)* Prob (not raining)<br /><br />We also know Prob(YYY|not raining) = 1/27. <br /><br />The frequentist solution is really a relative odds ratio: Prob(YYY|raining)/{ Prob(YYY|raining) + Prob(YYY|not raining)}. It is not really a probability at all. <br /><br />Now when Prob(raining)=1/2 ( the best guess if you have no information) then the frequentist relative odds will equal the Bayesian probability. <br /><br />If instead of Seattle the trip was to Los Angeles, and the Prob(raining) =0.01, then having three friends saying "Yes it's raining" will result in Prob(raining|YYY) =~0.075.<br /><br />However if you asked 14 of your mostly truthful friends, and all 14 said,"yes dude, it's raining", then the conditional probability increase to ~0.993 (frequentist relative odds = ~0.99994). <br /><br />This makes sense.<br /> <br />Thanks for posting this.<br /><br />Emile Elefteriadishttp://www.blogger.com/profile/06682927722971051966noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-71304155594207886632016-10-11T23:24:50.902-07:002016-10-11T23:24:50.902-07:00Hello Allen, I'm having difficulty solving thi...Hello Allen, I'm having difficulty solving this problem using Bayes Theorem but have no idea where I'm going wrong. Could you please shed me a light?<br /><br />We have:<br /><br />P(rain|YYY) = P(YYY|rain)*P(rain) / P(YYY)<br /><br />P(YYY|rain) = 8/27 ?<br />P(rain) = 0.1<br />P(YYY) = 8/27 ?<br /><br />Are the previous values correct? If they are then P(rain|YYY) = 0.1*8/27 / 8/27 = 0.1<br /><br />This would imply that P(rain|YYY) is not dependent on P(YYY) at all. What am I doing wrong?Unknownhttp://www.blogger.com/profile/15299634454831159125noreply@blogger.com