Math Girl

link to desmos graph. I did not include an explanation for how any of it works. good luck.

In a matter of seconds Gauss had an answer for him: 5050. See, he noticed a pattern:

and 50 times 101 is 5050. Done and done.

This puzzle is impossible and has no answer. Mathematically.

The super short and no-math lingo explanation is this: If you are in a room, one of three things must be true

1) This is your starting position (you will exit the room, using one door)

2) This is the end of your path (you have entered the room, but you will not exit it, using one door)

3) You are in the middle of your path (You have entered the room, you’re about to exit it, using two doors)

So since option (1) and (2) only happen once apiece, every time you’re in a room, you’re going to be entering it and exiting it, using two doors. So if you want to be able to go through every door exactly one time, then you can only have a maximum of two rooms with an odd number of doors (one will be your starting point, the other will be your endpoint). This puzzle has four rooms with an odd number of doors (if you count the outside as a room)

Each dot (vertex) represents a room, with an extra one labeled zero for the outside, and each line (edge) represents a direct opening between those two rooms. This is a mathematical “graph,” defined by its vertices and edges, that represents any way you can move from one room to another. For example, room 1 is connected to rooms 2, 3, and 4, and has two openings to 0, the outside.

What we’re looking for in this graph is something called an Eulerian path (or an Eulerian trail)-- a continuous line drawn along the edges of this graph that visits each edge exactly once. We need to know whether it’s even possible to find an Eulerian path in this graph. From Wikipedia, here’s the relevant bit we need to know this:

An undirected graph has an Eulerian trail if and only if exactly zero or two vertices have odd degree, and all of its vertices with nonzero degree belong to a single connected component.

Some vocabulary:

Undirected graph: a graph where the edges don’t come with a specified direction. Our graph is undirected, since you can move between rooms (or along the lines) in any direction you like.

Degree: the number of edges connected to a vertex. For example, vertex 1 has a degree of 5 since that’s how many lines are coming out of it, while vertex 0 has a degree of 9. A vertex with odd degree would have an odd number as its degree.

Connected component: A piece of a graph where you can travel between any two vertices by some path, directly or indirectly. Since our graph is one connected shape and there are no separate pieces floating around, we can ignore this part.

Count up the degree of each vertex on our graph, and you’ll see that they have degrees of 5, 5, 4, 5, 4, and 9 for the outside. There are four vertices with an odd degree. This isn’t exactly zero or two, and therefore, our theorem tells us that there is no Eulerian path in this graph. Therefore, there is no way to travel through each opening exactly once.

This is exactly what @chongoblog​ just explained very clearly and concisely! But well, if you like lingo, there you go. Next time you see a problem like this, try drawing its corresponding graph! It may make the problem easier to process.

I need to credit our old friend Euler for doing the heavy lifting on this problem. As it happens, he famously tackled a very similar problem long ago, known as the Seven Bridges of Königsberg, which actually laid the foundation for modern graph theory! It was a question about whether you could plan a path around the city that crosses each of its seven bridges exactly once.

I’m gonna take a wild guess and say the average number of new worlds generated in any given Minecraft game is 10. With 200 million copies sold, this estimates that 2 billion worlds have been generated and played in.

The Minecraft fan wiki tells me that in Java edition, there are 2^48 possible seeds that can generate meaningful, unique worlds. (To keep things simple, I’m ignoring differences between bedrock and java editions here.)

The question I’m asking is: what are the chances that two players have ever generated the exact same seed by chance? Or inversely: what are the chances that every seed generated so far has been unique? This is equivalent to the Birthday Paradox, which I’d love to discuss in detail, but I’ll try to keep it brief.

The birthday paradox asks the question: in a room full of people, what are the chances that two people share the same birthday? You can think of a person’s birthday like some random value from 1 to 365 that is assigned to them (excluding February 29 for the sake of simplicity; sorry leap year babies). Any two people have a pretty low chance of having the same birthday, but this problem is called a paradox because the answer isn’t intuitive, and the odds are higher than you’d expect. In a room of 23 people, there’s a more than 50% chance that two individuals share a birthday.

Rather than going over the math here, I’ll let you check it out on wikipedia, but it basically boils down to this: add people to the group one-by-one, and check the likelihood of everyone in the group having a different birthday. The first person is alone, and therefore has a 100% chance of having a unique birthday. Add a second person. In order for their birthday to be unique, it has to fall on one of the 364 days that aren’t the first person’s birthday, giving them a very high 364/365 chance of having a unique birthday. The pool of possible unique birthdays shrinks again for the third person, but they still have a 363/365 chance of being unique. And so on.

I won’t get into the pigeonhole principle, fascinating as it is, since it isn’t relevant to our Minecraft problem. Hopefully you can see that if you replace the days of the year with possible random seeds, and replace people in a room with randomly generated seeds, it’s the same exact question.

Wikipedia helpfully shows us how to calculate the probability that a group of n people all have unique birthdays, and gives us this formula:

Just replace 365 with our total potential number of seeds (2^48) and interpret n as the number of randomly generated worlds, and we’re well on the way to finding the chance that every generated world has been unique.

Er… the only problem is, these are quite big numbers to be calculating factorials and binomial coefficients for. I’m kind of scared to ask a machine to do that for me. Luckily, Wolfram Alpha has a built in birthday paradox calculator, but… it’s not quite up to the task either.

Um… help?

You can use Stirling’s approximation (like you alluded to), and if you substitute 2^48 with a variable (here it’s alpha) it’s easy to simplify. I think you get this:

(note that this is the probability of getting two same birthdays, not the probability of all unique birthdays, so it’s the opposite of the result you guys found. Plugging in 2^48 for d and 2 billion for n gives a result that rounds to 1.)

So what does this mean for Minecraft? I think it’s safe to say that the chance of every randomly generated seed being unique is negligibly small, meaning it’s certain that two people somewhere have rolled the same seed. In fact, there are probably lots of seed soulmates out there. If you believe, then there’s a chance your favorite speedrunner may have stumbled across your own world seed while they were going for the world record.

I think the reason the result is so clear hearkens back to the birthday problem itself. Out of 365 possible birthdays you can have, the chance of two people sharing a birthday ramps up really fast for every person you introduce to the group.

This makes sense if you think about the number of pairs of people in a group, since that’s what you’re really checking when looking for a match. For every person you add, the web of pairs grows more complex. In the case of Minecraft, the number of pairs you can draw between 2 billion worlds is absolutely enormous. I just guessed those numbers, but I’m sure you’d get a similar result no matter what reasonable estimate you make.

This was a fun exercise! Thanks for the help. And good luck finding your Minecraft soulmate.

There are far more ways to shuffle a deck of cards (52! ~ 8*10^67) than there are possible minecraft seeds, but the number of times a deck of cards has been shuffled should vastly exceed the number of number of minecraft worlds generated.

Using the same approximation for the generalised birthday problem we find that, for the chance of at least one pair of decks to have been identically shuffled in all of history to be 50%, we need to have shuffled…. ~10^34 decks of playing cards.

Wikipedia says that standard 52 card decks have existed since around 1500, so ~500 years. To shuffle 10^34 decks of cards in this time, you’d need to shuffle ~6.7*10^23 decks per second, on average. Even if every living human spent their entire life shuffling cards at an incredible rate, you wouldn’t approach an appreciable fraction of that.

So, it’s more likely than not that no two (fairly shuffled) decks have ever been shuffled in the same order.

Now let’s put a upper bound on this.

It’s reasonable to assume the overwhelming majority of card game activity has occurred in the last 100 years, given population increase and the emergence of casinos and mass production. Let’s very generously approximate that this entire time the world has been using cards at the present day rate, and that the world population is a constant 10 billion.

There are a few thousand casinos in the world today. Let’s say there’s 10,000. The largest casino (for our purposes) is The Venetian Macao which has 800 tables. Let’s round that up to 1000 tables and say they’re all for card games, and let’s assume every casino is the same size. That means there’s 10 million card tables in casinos worldwide. Let’s assume they each shuffle 1 deck of cards every second, for 100 years - that’s 3.2*10^16 shuffles total. Nowhere near our 10^34 from before.

Now let’s say that every person in the world plays a card game outside of a casino once per day, and that each person shuffles the deck 10 times per game. That’s 100 billion shuffles per day, or 3.7*10^15 shuffles total.

So, between casino and personal usage, we have a very generous upper limit on the total number of card decks ever shuffled of ~3.6x10^16.

Plugging this into the solution from before gives a maximum probability that any two decks ever shuffled in all of human history ended up in the exact same order of… 8x10^-36, or:

0.0000000000000000000000000000000008%

I’m gonna take a wild guess and say the average number of new worlds generated in any given Minecraft game is 10. With 200 million copies sold, this estimates that 2 billion worlds have been generated and played in.

The Minecraft fan wiki tells me that in Java edition, there are 2^48 possible seeds that can generate meaningful, unique worlds. (To keep things simple, I’m ignoring differences between bedrock and java editions here.)

The question I’m asking is: what are the chances that two players have ever generated the exact same seed by chance? Or inversely: what are the chances that every seed generated so far has been unique? This is equivalent to the Birthday Paradox, which I’d love to discuss in detail, but I’ll try to keep it brief.

The birthday paradox asks the question: in a room full of people, what are the chances that two people share the same birthday? You can think of a person’s birthday like some random value from 1 to 365 that is assigned to them (excluding February 29 for the sake of simplicity; sorry leap year babies). Any two people have a pretty low chance of having the same birthday, but this problem is called a paradox because the answer isn’t intuitive, and the odds are higher than you’d expect. In a room of 23 people, there’s a more than 50% chance that two individuals share a birthday.

Rather than going over the math here, I’ll let you check it out on wikipedia, but it basically boils down to this: add people to the group one-by-one, and check the likelihood of everyone in the group having a different birthday. The first person is alone, and therefore has a 100% chance of having a unique birthday. Add a second person. In order for their birthday to be unique, it has to fall on one of the 364 days that aren’t the first person’s birthday, giving them a very high 364/365 chance of having a unique birthday. The pool of possible unique birthdays shrinks again for the third person, but they still have a 363/365 chance of being unique. And so on.

I won’t get into the pigeonhole principle, fascinating as it is, since it isn’t relevant to our Minecraft problem. Hopefully you can see that if you replace the days of the year with possible random seeds, and replace people in a room with randomly generated seeds, it’s the same exact question.

Wikipedia helpfully shows us how to calculate the probability that a group of n people all have unique birthdays, and gives us this formula:

Just replace 365 with our total potential number of seeds (2^48) and interpret n as the number of randomly generated worlds, and we’re well on the way to finding the chance that every generated world has been unique.

Er... the only problem is, these are quite big numbers to be calculating factorials and binomial coefficients for. I’m kind of scared to ask a machine to do that for me. Luckily, Wolfram Alpha has a built in birthday paradox calculator, but... it’s not quite up to the task either.

Um... help?

You can use Stirling's approximation (like you alluded to), and if you substitute 2^48 with a variable (here it's alpha) it's easy to simplify. I think you get this:

(note that this is the probability of getting two same birthdays, not the probability of all unique birthdays, so it’s the opposite of the result you guys found. Plugging in 2^48 for d and 2 billion for n gives a result that rounds to 1.)

So what does this mean for Minecraft? I think it’s safe to say that the chance of every randomly generated seed being unique is negligibly small, meaning it’s certain that two people somewhere have rolled the same seed. In fact, there are probably lots of seed soulmates out there. If you believe, then there’s a chance your favorite speedrunner may have stumbled across your own world seed while they were going for the world record.

I think the reason the result is so clear hearkens back to the birthday problem itself. Out of 365 possible birthdays you can have, the chance of two people sharing a birthday ramps up really fast for every person you introduce to the group.

This makes sense if you think about the number of pairs of people in a group, since that’s what you’re really checking when looking for a match. For every person you add, the web of pairs grows more complex. In the case of Minecraft, the number of pairs you can draw between 2 billion worlds is absolutely enormous. I just guessed those numbers, but I’m sure you’d get a similar result no matter what reasonable estimate you make.

This was a fun exercise! Thanks for the help. And good luck finding your Minecraft soulmate.

I’m gonna take a wild guess and say the average number of new worlds generated in any given Minecraft game is 10. With 200 million copies sold, this estimates that 2 billion worlds have been generated and played in.

The Minecraft fan wiki tells me that in Java edition, there are 2^48 possible seeds that can generate meaningful, unique worlds. (To keep things simple, I’m ignoring differences between bedrock and java editions here.)

The question I’m asking is: what are the chances that two players have ever generated the exact same seed by chance? Or inversely: what are the chances that every seed generated so far has been unique? This is equivalent to the Birthday Paradox, which I’d love to discuss in detail, but I’ll try to keep it brief.

The birthday paradox asks the question: in a room full of people, what are the chances that two people share the same birthday? You can think of a person’s birthday like some random value from 1 to 365 that is assigned to them (excluding February 29 for the sake of simplicity; sorry leap year babies). Any two people have a pretty low chance of having the same birthday, but this problem is called a paradox because the answer isn’t intuitive, and the odds are higher than you’d expect. In a room of 23 people, there’s a more than 50% chance that two individuals share a birthday.

Rather than going over the math here, I’ll let you check it out on wikipedia, but it basically boils down to this: add people to the group one-by-one, and check the likelihood of everyone in the group having a different birthday. The first person is alone, and therefore has a 100% chance of having a unique birthday. Add a second person. In order for their birthday to be unique, it has to fall on one of the 364 days that aren’t the first person’s birthday, giving them a very high 364/365 chance of having a unique birthday. The pool of possible unique birthdays shrinks again for the third person, but they still have a 363/365 chance of being unique. And so on.

I won’t get into the pigeonhole principle, fascinating as it is, since it isn’t relevant to our Minecraft problem. Hopefully you can see that if you replace the days of the year with possible random seeds, and replace people in a room with randomly generated seeds, it’s the same exact question.

Wikipedia helpfully shows us how to calculate the probability that a group of n people all have unique birthdays, and gives us this formula:

Just replace 365 with our total potential number of seeds (2^48) and interpret n as the number of randomly generated worlds, and we’re well on the way to finding the chance that every generated world has been unique.

Er... the only problem is, these are quite big numbers to be calculating factorials and binomial coefficients for. I’m kind of scared to ask a machine to do that for me. Luckily, Wolfram Alpha has a built in birthday paradox calculator, but... it’s not quite up to the task either.

Um... help?

wohaohowhao. gears are impossibly difficult to fabricate. way, way beyond what people might expect looking at them. “oh, it is just a shape, it is simple” - imbeciles. morons

this will be VERY LONG and i am doing most of you idiots a favor by putting this here:

You can cook a chicken by slapping it at 3725.95 mph, an impossible task by any human means. If you do succeed however, you will not only cook the chicken but also decimate its entire structure, causing a violent explosion. This is a simulation of the slap using FEA. 

For example, something you hear a lot is: “why am I losing points for not showing my work when I got the correct answer?”, or even “why are we being told to use this procedure at all when the answers are so obvious?”.

There answer to both of those questions, of course, is: “Because what’s actually being taught is a problem-solving method that works for big and complicated problems as well as small and simple ones. We practice it with the simple ones first so that you can easily compare your intuitive solution with the results of applying the method and know whether you did it right. That way, when we get to the complicated ones where the intuitive approach doesn’t work, you can have confidence that you practised the method correctly.”

Not once in two decades of schooling did I hear that rationale offered – if an instructor deigned to address the objections at all, their response typically boiled down to some variation of “because this is how it’s done”.