13 ms·

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

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 4y ago
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 https://en.m.wikipedia.org/wiki/Birthday_problem
- criddell 4y ago
  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?
  - andi999 4y ago
    No. The chance would be much smaller. If you choose 22 random days though...
    - criddell 4y ago
      Your comment and erehweb’s finally made it click for me. Thanks.
  - erehweb 4y ago
    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 4y ago
      [deleted]
      - deleted 4y ago
        [deleted]
  - whatever1 4y ago
    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 4y ago
      Covid cases are asymptomatically approaching zero. However, your duplicate entries will reach their target asymptotically :) https://en.wikipedia.org/wiki/Asymptote https://en.wikipedia.org/wiki/Asymptote
      - whatever1 4y ago
        I know, Apple ios keyboard does not.
  - dwighttk 4y ago
    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.
  - bena 4y ago
    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.
  - dhosek 4y ago
    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 (363364)/365² ≈ 99.2% For the fourth person it becomes (362363364)/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 θ over the range [0,π] 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.
  - munificent 4y ago
    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.
    - moffkalast 4y ago
      https://c.tenor.com/u1V_-f-jXmcAAAAC/malreynolds-firefly.gif https://c.tenor.com/u1V_-f-jXmcAAAAC/malreynolds-firefly.gif
    - flafla2 4y ago
      Complete yet succinct explanation — bravo and i might steal this way of explaining it!
- WhitneyLand 4y ago
  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 https://en.wikipedia.org/wiki/Monty_Hall_problem
  - pchristensen 4y ago
    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.
    - vitus 4y ago
      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 4y ago
        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.
        lupire 4y ago
        I disagree. This version more vividly explains what's happening, making it easier to visualize, similar to illustrating an algebra problem by using easy small numbers instead of juggling variables.
sporkland 4y ago
What's the seance/deceased siblings trick? I can't work it out from your description.
- nestorD 4y ago
  This is the one: > Put down the number of your living brothers; multiply by two; add three; multiply the result by five. Now add the number of your living sisters. Multiply the total by ten. Add the number of your dead brothers and sisters, subtract 150 from the total. The right-hand figure will be the number of deaths; the middle figure will be the number of living sisters, and the left-hand figure will be the number of living brothers. I found it in a book published in 1946 and describing the work of someone operating around 1900 (I will let you try and find the book yourself if you want further information, while I believe this item to be outdated other ideas from it are still in use).

Swizec 4y ago

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 https://en.m.wikipedia.org/wiki/Birthday_problem

criddell 4y ago
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?
- andi999 4y ago
  No. The chance would be much smaller. If you choose 22 random days though...
  - criddell 4y ago
    Your comment and erehweb’s finally made it click for me. Thanks.
- erehweb 4y ago
  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 4y ago
    [deleted]
    - deleted 4y ago
      [deleted]
- whatever1 4y ago
  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 4y ago
    Covid cases are asymptomatically approaching zero. However, your duplicate entries will reach their target asymptotically :) https://en.wikipedia.org/wiki/Asymptote https://en.wikipedia.org/wiki/Asymptote
    - whatever1 4y ago
      I know, Apple ios keyboard does not.
- dwighttk 4y ago
  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.
- bena 4y ago
  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.
- dhosek 4y ago
  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 (363364)/365² ≈ 99.2% For the fourth person it becomes (362363364)/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 θ over the range [0,π] 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.
- munificent 4y ago
  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.
  - moffkalast 4y ago
    https://c.tenor.com/u1V_-f-jXmcAAAAC/malreynolds-firefly.gif https://c.tenor.com/u1V_-f-jXmcAAAAC/malreynolds-firefly.gif
  - flafla2 4y ago
    Complete yet succinct explanation — bravo and i might steal this way of explaining it!
WhitneyLand 4y ago
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 https://en.wikipedia.org/wiki/Monty_Hall_problem
- pchristensen 4y ago
  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.
  - vitus 4y ago
    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 4y ago
      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.
      - lupire 4y ago
        I disagree. This version more vividly explains what's happening, making it easier to visualize, similar to illustrating an algebra problem by using easy small numbers instead of juggling variables.

sporkland 4y ago

What's the seance/deceased siblings trick? I can't work it out from your description.

nestorD 4y ago
This is the one: > Put down the number of your living brothers; multiply by two; add three; multiply the result by five. Now add the number of your living sisters. Multiply the total by ten. Add the number of your dead brothers and sisters, subtract 150 from the total. The right-hand figure will be the number of deaths; the middle figure will be the number of living sisters, and the left-hand figure will be the number of living brothers. I found it in a book published in 1946 and describing the work of someone operating around 1900 (I will let you try and find the book yourself if you want further information, while I believe this item to be outdated other ideas from it are still in use).