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Paradoxes of Probability and Other Statistical Strangeness
- daxfohl 9y ago"Paradox" is a pretty strong term. The items presented are more in the category of common errors and counter-intuitiveness.
- Houshalter 9y agohttps://en.wikipedia.org/wiki/Veridical_paradox https://en.wikipedia.org/wiki/Veridical_paradox
- daxfohl 9y agoFine, but I was left feeling blah by the "paradoxes" presented.
- georgewsinger 9y agoAnother "paradox": even though it's possible to randomly pick a rational number from the reals, the probability of this happening is 0.
- Sinergy2 9y agoPlease describe how it is possible to pick such a number. For example, I can readily imagine how to pick a random 32b float, but that it is an entirely problem with a nonzero probability.
- Sinergy2 9y ago*entirely different
- georgewsinger 9y agoUsually in math we assume the axiom of choice :) https://en.wikipedia.org/wiki/Axiom_of_choice https://en.wikipedia.org/wiki/Axiom_of_choice I'm assuming this could somehow lead to such a "random" pick in the technical sense. In terms of implementation, I'm not aware of an algorithm that can randomly pick a real number on an actual computer. Perhaps a mathematician could show how to pick one on some abstract machine with infinite resources, and not constrained by finite bit representations of numbers.
- red75prime 9y agoRun a random number generator (0-9) for each decimal position after the dot in parallel. This should do a trick.
- dragonwriter 9y ago> In terms of implementation, I'm not aware of an algorithm that can randomly pick a real number on an actual computer An actual (finite in time and space) computer can't even represent arbitrary real numbers, much less randomly choose them.
- SomeStupidPoint 9y agoA Turing machine can't pick random numbers of any kind. Once you accept that you have an entropy source in the physical world, you can easily be injecting random real numbers (from some range) and in fact, usually are, which are then being binned into integers by ADCs.
- dorgo 9y agoWhat about PI? We can represent it in terms of "we know what we are talking about" and we can distinguish it from other numbers.
- leni536 9y agoYou can only have countable number of first or second order logic statements each defining a specific real number.
- stiff 9y agoIn probability theory, when dealing with continuous sample spaces / random variables, events with probability 0 still have a chance of occurring, and events with probability 1 stil l have a chance of NOT occurring, see: https://en.wikipedia.org/wiki/Almost_surely https://en.wikipedia.org/wiki/Almost_surely This strange property comes from strange properties of the real numbers (and uncountably infinite sets) that give rise to things like: https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox Measure theory deals with resolving this: https://en.wikipedia.org/wiki/Measure_(mathematics) https://en.wikipedia.org/wiki/Measure_(mathematics)
- ouid 9y agoWhat's an anagram for Banach Tarski? Banach Tarski Banach Tarski. Seriously though, you can do math without invoking the axiom of choice. The formulation of probability doesn't strictly depend on it.
- btilly 9y agoThere is no uniform probability distribution on the reals. Perhaps you meant the interval from 0 to 1?
- Mattasher 9y agoThere are many ways to pick randomly from the rationals with a non-uniform distribution.
- ramanan 9y agothere are a number of similar paradoxes that arise when considering infinities of different sizes! Infinity Paradoxes - Numberphile - https://www.youtube.com/watch?v=dDl7g_2x74Q https://www.youtube.com/watch?v=dDl7g_2x74Q
- philipov 9y agoI think this only sounds like a paradox if it is phrased poorly. The accurate way to state it is "The probability of randomly picking a specific number is 0" and that sounds reasonable. The probability of successfully picking any number is 1.
- cdavid 9y agoThat's a different statement: OP is alluding to the fact the measure of Q is 0 when using the "standard" sigma algebra on the real line, while you are saying that the measure of a number of 0. [edit] strictly speaking, you would restrict yourself to a bounded interval, e.g. if you pick a random number from a uniform distribution on [0, 1], the probability that this number is rational is 0.
- philipov 9y agooh, yeah, but that's because although Q is dense, it is not a dense subset of R and locally that's equivalent to saying a single point is not dense in R
- cdavid 9y agono, it is not. The OP statement is about the probability of the event "{X in Q}" (equal to 0), with X a random variable uniformly distributed on a bounded interval. That even contains many points (infinitely many actually), but has a probability 0. You are talking about the probability of a single point event, which is also always 0 on that same sigma algebra. The OP point is not completely trivial because the event contains an infinite (but countably) number of elements. It is fairly easy to understand though since by its very definition, the P[{X in Q}] = sum P[{x}] taken over every rational number (since Q is countable), and each P[{x}] is 0. A deeper statement is that there exists uncountable sets of probability 0.
- Houshalter 9y agoThe paradox is that, after picking a random number, you have just done a thing which has probability zero. Doing a thing that has zero probability shouldn't be possible. Ever.
- cdavid 9y agoI find that result fairly intuitive, when you understand how measure theory came up to be. A much more surprising result is that most irrational are normal numbers, but we know almost no normal number (morally speaking, a normal number is an irrational number where each digit is equiprobable in any base).
- BeetleB 9y agoThis is virtually an axiom for continuous distributions. One of the axioms of probability is that if you have an event (i.e. a set), then the probability of a countable union of disjoint sets is the sum of the probability of each set (event) occurring. Assume a uniform distribution between 0 and 1. Now consider point sets of the rationals (i.e. the number 0.5 is represented by a set with just 0.5 in it). Since the distribution is uniform, each set has the same probability (i.e. the likelihood of picking a random rational). Now consider this question: What is the probability of picking any rational between 0 and 1? Well, that's just the sum of the probabilities over all rationals (because it is a countable sum of disjoint sets). If the probability of picking any particular rational was non-zero, this sum would be infinite, which violates the laws of probability. Thus, by convention, it's just simpler to define it to be 0. There's no magic here. These properties were picked merely to make analysis with measure theory clean. Don't try to ascribe any real world meaning to picking a point.
- pella 9y agohttps://en.wikipedia.org/wiki/Category:Statistical_paradoxes https://en.wikipedia.org/wiki/Category:Statistical_paradoxes
- fitchjo 9y agoMy favorite statistical/probability paradox has always been the birthday paradox.
- curiousgal 9y agoThere is a 98.75% chance of some match of birthdays of the users who upvoted this post (57 at this moment)
- beefield 9y agoI don't know if Monty hall problem counts as a paradox, but that is quite high on my favourite list of counterintuitive probability results.
- thousandautumns 9y agoIn my experience the only reason the Monty Hall problem comes off as paradoxical is because it is usually poorly explained.
- noam87 9y agoFor me it's Simpson's paradox: it throws everyone off -- it's caused (and will continue to cause) real-world damage, it's everywhere once you see it -- it's in how newspapers report science, it's in our social policy and how we talk about social issues, it's in court cases --, and finally, it's really hard to explain to a non-math person; so even when it's happening, you sound like the irrational one for pointing it out. ... and don't get cocky once you know about it, because it's so pernicious it'll get you too if you're not careful!
- yiyus 9y agoI do not think this is a particularly difficult concept to explain to anyone. A typical example (not difficult to find) and a simple graph are usually good enough for most people.
- Houshalter 9y agoBy far the most unintuitive paradox for me personally is the one presented here: https://youtu.be/go3xtDdsNQM?t=3m27s https://youtu.be/go3xtDdsNQM?t=3m27s "Mr. Jones has 2 children. What is the probability he has a girl if he has a boy born on Tuesday?" Somehow knowing the day of the week the boy was born changes the result. It's completely bizarre.
- Jenya_ 9y agoThe comments to this video actually say (with proof) that this was an error in the video.
- Houshalter 9y agoThere seems to be quite a bit of debate about it in the comments and I'm not sure who to believe. At one point someone coded a simulation to test it and the results were as predicted by the video. Even if the video is incorrect, the fact it's so confusing still makes it an interesting paradox.
- Retric 9y agoIt's fairly simple. If you flip 2 coins then say whatever the first coin was the the odds if you said H was HH, or HT and if you said T it would be TH, TT. However, if you flip two coins and then say if you got at least one head independently from whatever you flipped then the odds you have 3 options HT, HH, TH with equal odds. So, the question is if the full statement was based on the data or only the truth value of the statement is based on the data. PS: Now assuming it's truth value is based on data. if you look at all options there are 14 gender day combinations per kid and 14 * 14 = 196 gender day combinations in totoal. Only 14 of of those 196 start BT which is then split evenly 7 BTB_, 7 BTG_. However that leaves 196 - 14 other options to consider. 7 * 14 of them Start G, and 6 * 14 of them start with B not on a Tuesday, but out of those you only keep 1/14 as you need BT on the second roll. Now add them up 13B and 14G out of (13 + 14) = 27. Or 13/27 B, and 14/27G.
- Chinjut 9y agoYour problem is that you are thinking there's a "the boy". But there's not a "the boy". Mr. Jones could have two boys. He could have two boys both born on Tuesday, even. The term "the boy" does not denote any particular boy, in that case, and causes you to think about the situation erroneously. If the question were "There's Kid 1 and Kid 2, each independently selected with random gender and birth-day-of-the-week. Out of those cases where Kid 1 is a boy born on Tuesday, what proportion are cases where Kid 2 is a girl?", then the answer would indeed be a straightforward 50%; the status of Kid 1 is entirely independent of the status of Kid 2. But that's not the question. The question is "There's Kid 1 and Kid 2, each independently selected with random gender and birth-day-of-the-week. Out of those cases where at least one (either one, and possibly both) of Kid 1 and Kid 2 is a boy born on Tuesday, what proportion are cases where at least one of Kid 1 and Kid 2 is a girl?". This is very different, and of course just drawing out the possibilities (all 2 * 7 * 2 * 7 equiprobable-by-stipulation choices of gender and birth-day-of-the-week for Kid 1 and Kid 2) and circling which pairs of subsets are the relevant ones for the two questions reveals the difference, the probabilities for either question elementarily calculable in this way by basic counting.
- boreas 9y agoFor those who might be interested, and in a slightly different vein than the examples in the article, there's the "sleeping beauty" paradox: https://en.wikipedia.org/wiki/Sleeping_Beauty_problem https://en.wikipedia.org/wiki/Sleeping_Beauty_problem Basically, an agent is put to sleep and told they will be woken up once or twice, depending on the results of a fair coin flip, without the ability to remember other awakenings. What probability does the agent assign to the event that the coin landed heads? The intuitive response is 1/3, but this poses obvious epistemological problems. The agent has, ostensibly, no new information at all, and their prior is surely 1/2. Hope someone else finds this as interesting as I do!
- colonelxc 9y agoI mostly find it interesting in that people could think that the chance is 1/3 (and that it may even be obvious!). After reading the description I can understand what they are getting at, but I think the conditional probability is messed up. Instead of P(Monday | Heads) = P(Monday | Tails) = P(Tuesday | Tails) it is really P(Monday | Heads&Awake) = P(Monday | Tails&Awake) = P(Tuesday| Tails&Awake) or something like that. But the interviewer isn't asking about that, they are asking for the probability of the coin. The 3 positions are only exhaustive given that you are awake to be interviewed about them, not exhaustive of possible states (it's missing P(Tuesday | Heads&Asleep)). Since you're always awakened at least once, I find the argument that being awake has 'given you information that it is not tuesday AND heads' is pretty weak. While true, both heads and tails expect to be awoken while it is not both tuesday AND heads.
- tzs 9y agoHere is how one might decide that 1/3 is obvious. Imagine that N people simultaneous undergo the experiment, for a very large N. Half of them end up in the heads group. They wake up on Monday and are questioned. Then they sleep until Wednesday and are released. The other half end up in the tails group, and so are questioned twice (Monday and Tuesday) then released on Wednesday. Because we gain no information during the experiment, we can make our decision before the experiment. Let's count. There will be 3N/2 interviews conducted. N/2 of the will be 'heads' interviews and N will be 'tails' interviews. So going in, we can see that when someone experiences the event 'being asked about the coin', 1/3 of the time the coin will be heads and 2/3 o the time it will be tails. Hence, our credence in the coin being heads should be 1/3. Here is a counterargument. Imagine a slightly different experiment. The people are not asked what their credence in the coin being heads is. They are asked to guess if it is heads or tails. If they are right, the experiment continues and they are eventually released. If they are wrong, this is noted, and the experiment continues until Wednesday, and then they are killed and their home planet is destroyed. As before, we gain no information during the experiment, and so can decide our answer beforehand. No matter what strategy one picks for making that decision, there is a 50/50 chance that one ends up with a destroyed planet. That indicates that our credence in heads should be 1/2.