[HN Gopher] The math behind mind-reading tricks ___________________________________________________________________ The math behind mind-reading tricks Author : theafh Score : 84 points Date : 2022-05-27 14:12 UTC (8 hours ago) (HTM) web link (www.quantamagazine.org) (TXT) w3m dump (www.quantamagazine.org) | bern4444 wrote: | My own fun math mind reading trick: | https://sambernheim.com/blog/the-party-math-trick | kick_in_the_dor wrote: | I came here for magic tricks and was disappointed. | curiousgal wrote: | What is your partner's age? Add two zeros to the end. Minus | your birth year from that number. Add the current year to this | equation. The result should be yours and your partner's current | ages. | Someone wrote: | Not if the one asked to do that calculation hasn't had their | birthday yet this year (in which case "current year - birth | year" doesn't yield age in years) or their birth year was | over 99 years ago (in which case adding "current year - birth | year" overflows into the partner's age) | neogodless wrote: | Close but not quite. If it helps, both birthdays have not | occurred yet this year. If I assume BOTH have, the trick | works. | aaaaaaaaaaab wrote: | Here's a trick I came up with: 1. think of any number! 2. Add | your age to the number! 3. Subtract the original number! | | _magical hand motions_ | | The result is your age! | nestorD wrote: | Fun fact, about a hundred years ago, some pseudo-mediums would | use a similar kind of math trick to "guess" the number of | _deceased_ brother and sisters of someone coming to see them in | order to do a Seance. | | Morality aside, discovering that highly surprised me because I | always thought that those techniques were transparent to all | people. But I guess math literacy in the US wasn't always | great... | Swizec wrote: | Statistics and probabilities are incredibly unintuitive to | people. That's why it's called The Birthday Paradox and not the | birthday obvious result | | > The birthday paradox is that, counterintuitively, the | probability of a shared birthday exceeds 50% in a group of only | 23 people. | | https://en.m.wikipedia.org/wiki/Birthday_problem | WhitneyLand wrote: | They are deceptively unintuitive also. I went back to the | Monty Hall problem more than once after I previously thought | I had it figured out: | | Suppose you're on a game show, and given the choice of three | doors: Behind one door is a car; behind the others, goats. | You pick a door, say No. 1, and the host, who knows what's | behind the doors, opens another door, say No. 3, which has a | goat. He then says to you, "Do you want to pick door No. 2?" | Is it to your advantage to switch your choice? | | https://en.wikipedia.org/wiki/Monty_Hall_problem | pchristensen wrote: | Here's how I was able to understand it. There are 100 | doors, you choose 1, then host opens 98 doors with goats. | Do you keep your door or take the 1 remaining? | | In the original, the 3 doors muddles the odds. But with | 100, you're choosing between the 1 you chose on the first | try, or the 99 you didn't choose. | jwilk wrote: | This doesn't help my intuition at all. | derriz wrote: | You're actually being given the choice between "my | initial pick contains the prize" or "the prize is | contained in any of the other 99 boxes". Okay - the host | does a routine to add to the "drama" by revealing 98 non- | prize boxes but since he will never reveal the prize | during this process, he isn't giving you any more | information. | quesera wrote: | This is how I think about it: | | * Your first choice had a natural chance of 1% to be | correct. | | * After eliminating the 98 other options, a random choice | of the remaining two is 50% likely to be correct. | | * BUT, you know that the _reason_ the door you picked was | not opened is because you picked it! Its inclusion in the | set-of-two is _not_ random. | | * So your first choice is isn't any more likely to be | correct that it was when you first you chose it. | patrickdavey wrote: | The way I think about it (which I hope is correct): when | you first chose your probability of guessing correctly | was 1 in 100. It doesn't magically become 1 in 2 when the | other 98 doors are opened. | lijogdfljk wrote: | Funny, i think (but am not arguing i'm correct) it _does_ | change probability when the doors change. As i | explained[1] in the sister post. | | Perhaps this is some mathematical concept? To me i'm | viewing every choice as a dice roll. I have no question | that the first dice roll was worse than the 2nd. However | to me the 2nd is a die with two faces. Two choices, of | which both are equally possible. | | To look at it differently. The doors changing probability | to me sounds like.. imagine two people, the PersonA has | this primary scenario. They had 100 doors, chose one, and | are then asked if they'd like to switch. PersonB then is | asked the second question, of the 2 doors, which do they | want? PersonA and B are standing next to each other. The | doors are the same for both of them. Why would PersonA | have different odds than PersonB? And how could PersonB's | odds be any different than 50/50 with no prior history of | the doors? | | Is there some fundamental difference between the grand | idea of probability and "reality" as i'm trying to | describe it here? | | [1]: https://news.ycombinator.com/item?id=31535557 | lijogdfljk wrote: | Man, i still don't get it. Well, maybe i do? :shrug: | | The way i see it, the other door has a higher chance of | being correct than the door you _originally_ chose. | However the two are fundamentally different, are they | not? | | In reality both doors are equally likely to be correct, | no? So yes, changing would be better compared to your | original answer - but "to change or not" evaluates both | doors in my mind, and each is a 50% chance, no? | | Which i _think_ is what you 're saying here: | | > So your first choice is isn't any more likely to be | correct that it was when you first you chose it. | | But nevertheless the problem has me a bit confused in | what is expected to be "right". It seems to me there is | no right, it is a 50/50, unless you are strictly | answering the moot point of "Which is better? The door | you first chose, or the potential other door?". | | Ie i guess i view the original choice as irrelevant the | moment the variables change. I feel i misunderstand the | goal of the "problem". | vitus wrote: | Or, an even more telling variation: | | You pick door #1. | | The host goes and, in order, one by one, opens doors | 2-52, conspicuously skips door 53, and then opens 54-100. | Would you like to stick with your original choice? | cgriswald wrote: | I think this variation makes it more intuitively clear | one should switch in the case with 100 doors, but I think | it makes the explanation of _why_ less clear because of | the extra, unnecessary information. | criddell wrote: | I've never been able to wrap my head around this one. | | If in a group 23 people at least two share a birthday with a | ~50% chance, would that mean if I have a group of 22 people | and pick a random day of the year, there's also a ~50% chance | of somebody having a birthday on that day? | erehweb wrote: | No. There's about a 22/365 chance that somebody has a | birthday on that day. But this is getting close to the | intuition. Go through the 23 people. For each one of them, | there's about a 22/365 somebody else in the group has the | same birthday as they do. So adding these probabilities up | over all 23, you get 23 * 22/365. Of course, adding up the | probabilities is obviously wrong since the events are not | independent (and 23 * 22/365 > 1) but it gives the | intuition that the probability of a shared birthday grows | like N^2. [and actually, you can add up expectations] | [Edits for correctness] | [deleted] | [deleted] | whatever1 wrote: | For all these problems I find it easier to just simulate it | and plot the results. Do a loop with 100k iterations. In | each iteration generate 23 numbers between 1,365 from a | uniform distribution. Then check in how many iterations you | had duplicate entries. | | You will see that asymptomatically you will reach 50k cases | conanite wrote: | Covid cases are asymptomatically approaching zero. | However, your duplicate entries will reach their target | _asymptotically_ :) | | https://en.wikipedia.org/wiki/Asymptote | whatever1 wrote: | I know, Apple keyboard does not. | dhosek wrote: | I think the easiest way to think about it is to invert the | problem. What is the probability that no one shares a | birthday? | | One person, nobody shares a birthday because they're alone. | | Two people, it's going to be a 364/365 [?] 99.7% that they | don't share a birthday.1 | | Now add a third person. We have the initial probability for | the first pair and then the third person can't share a | birthday with either of them, so the probability is going | to be 363/365 that they don't share a birthday with either, | but since we have two events that both must occur, we | multiply the probabilities. That gives us (363 _364) /3652 | [?] 99.2% | | For the fourth person it becomes (362_363 _364) /3653 [?] | 98.4% | | You'll notice that the probability is decreasing more and | more with each step.2 | | In general, for _n* people we would have3 _P_ = 365! | /((365- _n_!)365n). At _n_ = 23, we drop below 50%. | | This technique is useful for a lot of statistical | calculations, e.g., for calculating the probabilities of | poker hands, some of them are easier to calculate the | probability of not getting the hand then the probability of | getting the hand (a pair being the most obvious case). | | [?] | | 1. We're mostly programmers here so we're going to pretend | leap years don't exist. | | 2. This will not continue indefinitely, obviously. The | first derivative hits a minimum4 around _n_ = 20 for this | formula and at _n=365_. The graph kind of looks like cos th | over the range [0,p] but not exactly, it's just the first | well-known graph that comes to mind of this shape. | | 3. If you're eagle eyed, you might notice that my examples | before had a denominator of 365 raised to the _n_ -1 power. | Then you'll also notice that my factorials have _n_ terms | and end with 365 and not 364. Remember our lonely birthday | dude? His probability of not sharing a birthday (1) can be | thought of as 365 /365 and should also be part of the | product, but when we're doing the math for small numbers, | it's easier to just ignore that but for a general formula, | keeping him in the product makes the formula clearer. | | 4. Minimum because we're _decreasing_ , remember. | bena wrote: | Statistics aren't reflective. They're aggregate. | | It's not a 50% chance of being on the date you chose, but | that any two dates are the same. | | If you pick a random date, Alice has a 364/365 chance of | _not_ having a birthday that day. Bob has a 363 /365 chance | of not having a birthday on that day or on Alice's birthday | independently of whether or not Alice's birthday is on the | random date. So we take Alice's chances and Bob's chances | and we can multiply them together to get our odds that none | of these people share a date. And of course, if we know the | odds of something not happening, the odds of it happening | is the opposite. | munificent wrote: | The trick is to realize that "same birthday" isn't a | property of _people_ , it's a question about _pairs_ of | people. | | A group of 23 people has 253 unique pairs. So if I ask, "In | a room with 253 pairs of people, what are the odds that in | one of those pairs, both have the same birthday?" Now | around 50% seems a lot more intuitive. | | It's unintuitive because the number of pairs increases | quadratically with the number of people. | dwighttk wrote: | that's just 1 date that has to be matched, so you get a | lower probability of matching, comparing 22 dates to 1... | in the 23 person scenario, you have 23 dates that can match | with any of the 23 dates. | andi999 wrote: | No. The chance would be much smaller. If you choose 22 | random days though... | criddell wrote: | Your comment and erehweb's finally made it click for me. | Thanks. | charcircuit wrote: | I was expecting an article talking about probability instead of | algebra. People are bad at coming up with random numbers | especially if you put some restrictions on what number they are | allowed to generate. | vincnetas wrote: | show me a kid that can multiply by 13 tree digit number in theyr | head and that would be enough magic trick for me :) | cgriswald wrote: | Define 'kid.' By middle school, at least, I could have done | that even without the (10x + 3x) simplification. | | The real trick would be finding a three digit number that can | be multiplied by 77 and still be a three digit number. | slingnow wrote: | I don't think any one of these could be classified as a "mind- | reading magic trick". ___________________________________________________________________ (page generated 2022-05-27 23:00 UTC)