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...
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?
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.
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?
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.
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.
I think that's probably true for you and it's definitely true for some, but I've tutored many students in both algebra and discrete and am very familiar with the failure modes of students' thinking. The "Door 53" example makes it very easy for people to understand that in the case of 100 doors they should switch. It doesn't, however, always get them to see that the issue is fundamentally choosing between odds of 1:X and (X-1):X (or even that they should also switch with only three doors).
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.
It's actually sufficient that he simply hasn't revealed the prize in the other 98 picks. Of course, it would make for poor drama if he just happened to open the prize in the process, but you can imagine a construction where he's picking 98/99 of the other boxes out of lotto machine and offers you the chance to switch if, at the end of the process, none of the doors he opened contained the prize. It'll just be a pretty silly show the 98% of the time that the host just opens the prize in the middle of the drawing.
Imagine you were given the choice before he opens any boxes. “Stick with your pick“ or “Switch to picking all 99 of the others”. Opening non-prize boxes before showing the contents of the last box is just part of the reveal drama - whether you’ve chosen 1 or 99 boxes.
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".
> In reality both doors are equally likely to be correct, no?
No, your original pick will always have a 1/3 chance of being correct. The door that gets opened will always have a 0/3 chance of being correct. The last door has to get all of the rest of the chance.
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edit: I haven't tried explaining it this way before, but instead of one winning door and two bad doors, there's one door with a brand new car, one door with a giant mousetrap, and one door with a stuffed toy dinosaur. These are the ways it can go.
A) You pick the door with the car. Monty either picks the dinosaur or the mousetrap, he doesn't really care. If you switch, you get the dinosaur or the mousetrap.
B) You pick the door with the dinosaur. Monty is forced to pick the mousetrap. If you switch, you win the car.
C) You pick the door with the mousetrap. Monty is forced to pick the dinosaur. If you switch, you win the car.
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edit 2: your intuition is actually comparing this problem to other problems and not seeing the difference. Take the above scenario for example, but change it so Monty always reveals the dinosaur, no matter what.
A) You pick the door with the car. Monty picks the dinosaur. If you switch, you get the mousetrap.
B) You pick the door with the dinosaur. Monty picks the same door, and shows you that it is the dinosaur. Now you will definitely switch, and have an even chance to win the mousetrap or the car.
C) You pick the door with the mousetrap. Monty picks the door with the dinosaur. If you switch, you will win the car.
This only occurs because the host knows the locations of the contents and cannot reveal the player's door and neither which has the car. So the other that he leaves closed is always the best possible one that could be found among the rest, because if any of them has the prize, that is precisely which he will avoid to discard.
If you notice, the switching door is equivalent to the selection that a second player that was cheating would make, a cheater that checked inside all the doors that the first did not pick, and took which preferred from them. In this way, it is obvious that the second player will win as long as the first starts failing, which occurs 99 out of 100 times in the 100-doors version. And the revealed doors are like which neither of them picked.
So, when there are two doors remaining, you know that one is the first player's selection and the other is the cheater's selection, where the cheater was more likely to get the correct.
It doesn't really matter if the host knows, as long as you are allowed to switch to an open door.
If the host opens other door randomly, then 1/3 times the host accidentally shows the prize, and you still have 2/3 chance of winning by switching (measured from before the door was opened), switching either to the unknown closed door or the known winning open door.
But, measured from after the doors open, with host having opened other doors uniform randomly, in the case where you didn't already win by directly being shown the prize (free win happens with frequency (doors-2)/doors, since you have to pick a loser and the host has to pick a loser to hide), then your chance of winning by switching is 1/2, making the naive answer correct (and for the correct reason)!
Yes, you could still win 2/3 of the time by switching in all games, only that with those rules the probability for each remaining door once a goat is revealed would be 1/2, as you stated. So, you could also win 2/3 of the time by staying with your selection everytime a goat is shown and only switching to the car when it is shown.
But what I think that confuses most people is why one option has more probabilities than the other when there are two remaining doors and the car must be behind one of them, which does not occur with those conditions. It is needed that the host did it deliberately avoiding to reveal the car.
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.
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?
> 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.
The probability that you have the winning door does not change when the 98 doors are opened. The simple reason for this is that it is always possible to open 98 doors without the prize regardless of which door you initially chose.
So it provides no extra information about your chosen door. But it does provide information about the single remaining door because in most situations, that specific door would need to have been opened.
> 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?
If PersonB saw everything happen, they have all the information PersonA has and therefore can make the same choice (99% correctly).
If PersonB came in after the door was chosen and the 98 doors were opened, (assuming they can't tell which door PersonA chose) their chance of picking the winning door is indeed 50%.
When thinking about probabilities, it is very important to think about the information that each person knows. The game master knows everything, so if they were playing their own game, they could win 100% of the time. PersonA can only win 99%. And PersonB can only win 50%. It all depends on what information you are given.
Not enough information. You need to know that he will always open an empty door that isn't the one you picked.
If that is the case. You can easily diagram the 9 different possibilities (you pick one of 3 doors x the car is behind one of 3 doors), and what happens if you switch.
Unless the host decides whether to open an unopened door based on whether you already picked a prize door, it's sufficient to know that the door he opened is a non-prize door that you didn't pick.
To be clear, the "trick" is that the inference is invalid if the host is differently to open a door if you picked a non-prize door, with crossover where the host is more than twice as likely to open a door if you pick a non-prize door.
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?
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.
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]
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
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.
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).
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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.
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.
> 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).
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.
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)
Well, no, it's not. We're talking in the context of decimal representations of numbers. Outside of such a context the number of digits of a number is meaningless. In any case and in any base, a leading zero would not be considered a 'digit' of a given number, unless the character '0' was used to represent a non-zero digit.
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.
