Timeline for Do I have a misconception about probability?
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 22, 2023 at 4:01 | audit | First answers | |||
| Aug 22, 2023 at 4:02 | |||||
| Aug 20, 2023 at 19:29 | audit | First answers | |||
| Aug 20, 2023 at 19:30 | |||||
| Jul 29, 2023 at 15:17 | comment | added | plm | @AnderBiguri, ok, thank you. I just wanted to underline that the formulation is really totally unambiguous. And to be honest, when i read the question at first i thought there was a mistake in the "misconception", i really thought that we were supposed to compute the expected position; so it is a matter preconditioned or automatic thought. | |
| Jul 29, 2023 at 13:49 | comment | added | Ander Biguri | @plm I do get it, just pointing out why the misconception arises | |
| Jul 28, 2023 at 8:44 | comment | added | plm | @AnderBiguri , from your remark i think you are not understanding what is done here. If you are in your sofa then your distance from home is 0, regardless of whether you have gone "around the world". Same for the random walk, if its final position is O (to distinguish from distance $0$) then its distance from O is $0$, regardless how far it travelled. And if i ask you a distance you wont give me a negative number. When you are averaging you are not averaging $|S_i|$ over the intermediate steps: you are averaging $|S_{100}|$ over $2^{100}$ full paths, not intermediate steps. -See my answer too. | |
| Jul 28, 2023 at 7:09 | comment | added | Ander Biguri | @plm it's just a language thing. If I ask someone "how far I am from home" I don't expect them to give me a large positive number after going around the world and being back in my sofa. I don't thing it's unambiguous, otherwise there would net be as much discussion there is here hehe. | |
| Jul 27, 2023 at 12:59 | comment | added | plm | @AnderBiguri expected location/position = "where am i on average ?" = "what is my average position ?" ; expected distance = "how far am i on average?" (from where i started). This distance is just the absolute value of your walk after 100 steps, so you take the expectation of this. "Absolute value of the average [position]", which is $0$, is just not what is asked (for instance, again, the random variable "position" can be negative while the rv "distance" cannot). The original text is written unambiguously. | |
| Jul 25, 2023 at 13:01 | audit | First answers | |||
| Jul 25, 2023 at 13:11 | |||||
| Jul 25, 2023 at 9:01 | comment | added | Davis Yoshida | @AnderBiguri I think it's more that there's yet a third thing you could say which is "what is my average position". That's distinct from both "how far am I on average from where I started" and "how far I have traveled on average". | |
| Jul 24, 2023 at 21:23 | audit | First answers | |||
| Jul 24, 2023 at 21:53 | |||||
| Jul 24, 2023 at 19:37 | comment | added | Vincent Batens | Other viewpoint: variance is $E\left[(X-\mu)^2\right]$ where $E$ stands for expected value and $\mu$ for the expected value of $X$ (meaning average). Another type of measure for variance could be $E\left[|X-\mu|\right]$, the average distance between values and the expected value. The reason we don't use that is because squaring is much easier for a computer than taken abs and $x^2$ is continuous everywhere so easier to describe mathematically. (It's just another norm) The average distance expresses (almost) the same as variance: bigger means more spread out data. 0 means no variation of data | |
| Jul 24, 2023 at 14:22 | comment | added | Ennar | @Falco, you are right that the probability of position to be $0$ is the biggest, but it's not true that the probability that distance is $0$ is the biggest. The probability that the distance is $0$ is $\frac 1{2^{100}}\binom{100}{50} \approx 0.0796$, while the probability that the distance is $2$ is $2\cdot\frac 1{2^{100}}\binom{100}{49}\approx 0.1561$. Intuitively, probability that we get $50$ heads is approximately the same as the probability that we get $51$ or $49$ heads, so the probability that the distance is $2$ is approximately double the probability that the distance will be $0$. | |
| Jul 24, 2023 at 13:55 | comment | added | Falco | The whole catch lies in how different results are summed up to get the "average distance" and distance ignoring direction aka the negative sign. | |
| Jul 24, 2023 at 13:50 | comment | added | Falco | @AntiHeadshot in contrast - 50 heads and 50 tails is the most likely single scenario, if you paint a curve of the chance of finishing at each individual location, it would be a bell curve around zero. And this is the intuitive answer - if I had to bet on any single location to win, I would bet on zero. The same for distance - if I had do bet on a single most likely distance it would be zero. Zero is the most likely result for location and distance. | |
| Jul 24, 2023 at 12:36 | comment | added | Ennar | @AntiHeadshot, ok, I misunderstood the point you are trying to make. | |
| Jul 24, 2023 at 12:00 | comment | added | lonza leggiera | If someone were to ask me the question, my suspicion would be they meant it to be interpreted in the way that Denker does. But the ambiguity highlighted by Ander Biguri's comment is also sufficiently obvious that I'd certainly want to clarify the question before I attempted to answer it. | |
| Jul 24, 2023 at 11:59 | comment | added | lonza leggiera | I agree that the misconception isn't about probabilities at all, but is one which Denker himself shares with anyone else who is labouring under it—namely, that the phrase "how far away am I, on average, from where I started?" is completely unambiguous. Denker's misconception is that it must be interpreted as synonymous with "what is my expected distance from where I started?" rather than "what is my expected location, relative to where I started?", which is probably the way that those who think the answer is zero interpret it. | |
| Jul 24, 2023 at 11:44 | comment | added | AntiHeadshot | @Ennar I just want to say the question is not about the average distance while tossing the coin, it is about the average result of doing this experiment multiple times. And thus only exactly 50 heads an 50 tails will end up in 0 steps while all the other cases will increase the average. | |
| Jul 24, 2023 at 11:27 | comment | added | Ennar | @AntiHeadshot, that's not how expected value works. It's like saying that expected value of normal distribution $\mathcal N(\mu,\sigma^2)$ can't be $\mu$ since the probability that $\mu$ occurs exactly is $0$. | |
| Jul 24, 2023 at 11:17 | comment | added | AntiHeadshot | @Falco I would argue it is not. The Average result is not 0 steps, because the only way you are 0 steps from the start is, if you tossed exactly 50 heads and 50 tails. Which is not that common. | |
| S Jul 24, 2023 at 10:54 | history | suggested | psmears | CC BY-SA 4.0 |
Fix a number of typos
|
| Jul 24, 2023 at 10:18 | comment | added | Falco | I would add to the answer: This is not really a misconception of probabilities, this is mostly a misunderstanding of "distance" vs "location" in the question. It is a little like a trick question like "what is heavier a ton of feathers or a ton of lead" | |
| Jul 24, 2023 at 9:46 | comment | added | psmears | @AnderBiguri: Not necessarily - whether you're two steps north, or two steps south, of your starting location, you're precisely 2 steps away from it. (If you were two steps due east of your start point, you'd still be 2 steps away, not zero!) But you're right that it's a matter of how the English description is interpreted in mathematical terms. | |
| Jul 24, 2023 at 9:44 | review | Suggested edits | |||
| S Jul 24, 2023 at 10:54 | |||||
| Jul 24, 2023 at 9:42 | comment | added | Ander Biguri | But in the text, it says, "how far away am I, on average, from where I started" not "how far have I travelled", which I would translate to the expected location, in absolute value. I would take the absolute value of the average, not the average of the absolute value | |
| Jul 23, 2023 at 23:41 | history | answered | Vincent Batens | CC BY-SA 4.0 |