P51?? Can you outsmart this logical fallacy

Meet Lucy. She was a math major in college, and aced all her courses in probability and statistics. Which do you think is more likely: that Lucy is a portrait artist, or that Lucy is a portrait artist who also plays poker? In studies of similar questions, up to 80 percent of participants chose the equivalent of the second statement: that Lucy is a portrait artist who also plays poker. After all, nothing we know about Lucy suggests an affinity for art, but statistics and probability are useful in poker. And yet, this is the wrong answer. Look at the options again. How do we know the first statement is more likely to be true? Because it’s a less specific version of the second statement. Saying that Lucy is a portrait artist doesn’t make any claims about what else she might or might not do. But even though it’s far easier to imagine her playing poker than making art based on the background information, the second statement is only true if she does both of these things. However counterintuitive it seems to imagine Lucy as an artist, the second scenario adds another condition on top of that, making it less likely. For any possible set of events, the likelihood of A occurring will always be greater than the likelihood of A and B both occurring. If we took a random sample of a million people who majored in math, the subset who are portrait artists might be relatively small. But it will necessarily be bigger than the subset who are portrait artists and play poker. Anyone who belongs to the second group will also belong to the first– but not vice versa. The more conditions there are, the less likely an event becomes. So why do statements with more conditions sometimes seem more believable? This is a phenomenon known as the conjunction fallacy. When we’re asked to make quick decisions, we tend to look for shortcuts. In this case, we look for what seems plausible rather than what is statistically most probable. On its own, Lucy being an artist doesn’t match the expectations formed by the preceding information. The additional detail about her playing poker gives us a narrative that resonates with our intuitions— it makes it seem more plausible. And we choose the option that seems more representative of the overall picture, regardless of its actual probability. This effect has been observed across multiple studies, including ones with participants who understood statistics well– from students betting on sequences of dice rolls, to foreign policy experts predicting the likelihood of a diplomatic crisis. The conjunction fallacy isn’t just a problem in hypothetical situations. Conspiracy theories and false news stories often rely on a version of the conjunction fallacy to seem credible– the more resonant details are added to an outlandish story, the more plausible it begins to seem. But ultimately, the likelihood a story is true can never be greater than the probability that its least likely component is true.

P52? ?Can you solve Einstein's Riddle

Before he turned physics upside down, a young Albert Einstein supposedly showed off his genius by devising a complex riddle involving this list of clues.?Can you resist tackling a brain teaser written by one of the smartest people in history??Let’s give it a shot.The world’s rarest fish has been stolen from the city aquarium.?The police have followed the scent to a street with five identical looking houses.?But they can’t search all the houses at once and if they pick the wrong one, the thief will know they’re on his trail.It’s up to you, the city’s best detective to solve the case.?When you arrive on the scene, the police tell you what they know.?One: each house’s owner is of a different nationality, drinks a different beverage, and smokes a different type of cigar.Two: each house’s interior walls are painted a different color.Three: each house contains a different animal, one of which is the fish.After a few hours of expert sleuthing, you gather some clues.?It may look like a lot of information, but there’s a clear logical path to the solution.?Solving the puzzle will be a lot like Sudoku, so you may find it helpful to organize your information in a grid, like this.(Pause the video on the following screen to examine your clues and solve the riddle.)Answer in:3 2 1To start, you fill in the information from clues eight and nine.?Immediately, you also realize that since the Norwegian is at the end of the street, there’s only one house next to him, which must be the one with the blue walls in clue fourteen.?Clue five says the green-walled house’s owner drinks coffee.?It can’t be the center house since you already know its owner drinks milk, but it also can’t be the second house, which you know has blue walls.?And since clue four says the green-walled house must be directly to the left of the white-walled one, it?can’t be the first or fifth house either.?The only place left for the green-walled house with the coffee drinker is the fourth spot, meaning the white-walled house is the fifth.Clue one gives you a nationality and a color.?Since the only column missing both these values is the center one, this must be the Brit’s red-walled home.?Now that the only unassigned wall color is yellow, this must be applied to the first house, where clue seven says the Dunhill smoker lives.?And clue eleven tells you that the owner of the horse is next door, which can only be the second house.?The next step is to figure out what the Norwegian in the first house drinks.?It can’t be tea, clue three tells you that’s the Dane.?As per clue twelve, it can’t be root beer since that person smokes Bluemaster, and since you already assigned milk and coffee, it must be water.?From clue fifteen, you know that the Norwegian’s neighbor, who can only be in the second house, smokes Blends.?Now that the only spot in the grid without a cigar and a drink is in the fifth column that must be the home of the person in clue twelve.?And since this leaves only the second house without a drink, the tea-drinking Dane must live there.?The?fourth house is now the only one missing a nationality and a cigar brand, so the Prince-smoking German from clue thirteen must live there.?Through elimination, you can conclude that the Brit smokes Pall Mall and the Swede lives in the fifth house, while clue six and clue two tell you that these two have a bird and a dog, respectively.?Clue ten tells you that the cat owner lives next to the Blend-smoking Dane, putting him in the first house.Now with only one spot left on the grid, you know that the German in the green-walled house must be the culprit.You and the police burst into the house, catching the thief fish-handed.?While that explanation was straightforward, solving puzzled like this often involves false starts and dead ends.?Part of the trick is to use the process of elimination and lots of trial and error to hone in on the right pieces, and the more logical puzzles you solve, the better your intuition will be for when and where there’s enough information to make your deductions.And did young Einstein really write this puzzle??Probably not.?There’s no evidence he did, and some of the brands mentioned are too mentioned.?But the logic here is not so different from what you’d use to solve equations with multiple variables, even those describing the nature of the universe.

P53? ?Can you solve the airplane riddle

Professor Fukanō, the famous eccentric scientist and adventurer, has embarked on a new challenge: flying around the world nonstop in a plane of his own design. Able to travel consistently at the incredible speed of one degree longitude around the equator per minute, the plane would take six hours to circle the world. There's just one problem: the plane can only hold 180 kiloliters of fuel, only enough for exactly half the journey. Let's be honest. The professor probably could have designed the plane to hold more fuel, but where's the fun in that? Instead, he's devised a slightly more elaborate solution: building three identical planes for the mission. In addition to their speed, the professor's equipped them with a few other incredible features. Each of the planes can turn on a dime and instantly transfer any amount of its fuel to any of the others in midair without slowing down, provided they're next to each other. The professor will pilot the first plane, while his two assistants Fugōri and Orokana will pilot each of the others. However, only one airport, located on the equator, has granted permission for the experiment, making it the starting point, the finish line, and the only spot where the planes can land, takeoff, or refuel on the ground. How should the three planes coordinate so the professor can fly continuously for the whole trip and achieve his dream without anyone running out of fuel and crashing? Pause here if you want to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 According to the professor's calculations, they should be able to pull it off by a hair. The key is to maximize the support each assistant provides, not wasting a single kiloliter of fuel. It also helps us to think symmetrically so they can make shorter trips in either direction while setting the professor up for a long unsupported stretch in the middle. Here's his solution. All three planes take off at noon flying west, each fully loaded with 180 kiloliters. After 45 minutes, or one-eighth of the way around, each plane has 135 kiloliters left. Orokana gives 45 to the professor and 45 to Fugōri, fully refueling them both. With her remaining 45, Orokana returns to the airport and heads to the lounge for a well-deserved break. 45 minutes later, with one-quarter of the trip complete, the professor and Fugōri are both at 135 kiloliters again. Fugōri transfers 45 into the professor's tank, leaving himself with the 90 he needs to return. Professor Fukanō stretches and puts on his favorite album. He'll be alone for a while. In the meantime, Orokana has been anxiously awaiting Fugōri's return, her plane fully refueled and ready to go. As soon as his plane touches the ground, she takes off, this time flying east. At this point, exactly 180 minutes have passed and the professor is at the halfway point of his journey with 90 kiloliters of fuel left. For the next 90 minutes, the professor and Orokana's planes fly towards each other, meeting at the three-quarter mark. Just as the professor's fuel is about the run out, he sees Orokana's plane. She gives him 45 kiloliters of her remaining 90, leaving them with 45 each. But that's just half of what they need to make it to the airport. Fortunately, this is exactly when Fugōri, having refueled, takes off. 45 minutes later, just as the other two planes are about to run empty, he meets them at the 315 degree point and transfers 45 kiloliters of fuel to each, leaving 45 for himself. All three planes land at the airport just as their fuel gauges reach zero. As the reporters and photographers cheer, the professor promises his planes will soon be available for commercial flights, just as soon as they figure out how to keep their inflight meals from spilling everywhere.

P54? ?Can you solve the alien probe riddle

The discovery of an alien monolith on planet RH-1729 has scientists across the world racing to unlock its mysteries. Your engineering team has developed an elegant probe to study it. The probe is a collection of 27 cube modules capable of running all the scientific tests necessary to analyze the monolith. The modules can self-assemble into a large 3x3x3 cube, with each individual module placed anywhere in the cube, and at any orientation. It can also break itself apart and reassemble into any other orientation. Now comes your job. The probe will need a special protective coating for each of the extreme environments it passes through. The red coating will seal it against the cold of deep space, the purple coating will protect it from the intense heat as it enters the atmosphere of RH-1729, and the green coating will shield it from the alien planet’s electric storms. You can apply the coatings to each of the faces of all 27 of the cubic modules in any way you like, but each face can only take a single color coating. You need to figure out how you can apply the colors so the cubes can re-assemble themselves to show only red, then purple, then green. How can you apply the colored coatings to the 27 cubes so the probe will be able to make the trip? Pause here if you want to figure it out yourself. You can start by painting the outside of the complete cube red, since you’ll need that regardless. Then you can break it into 27 pieces, and look at what you have. There are 8 corner cubes, which each have three red faces, 12 edge cubes, which have two red faces, 6 face cubes, which have 1 red face, and a single center cube, which has no red faces. You’ve painted a total of 54 faces red at this point, so you’ll need the same number of faces for the green and purple cubes, too. When you’re done, you’ll have painted 54 faces red, 54 faces green, and 54 faces purple. That’s 162 faces, which is precisely how many the cubes have in total. So there’s no margin for waste. If there’s any way to do this, it’ll probably be highly symmetrical. Maybe you can use that to help you. You look at the center cube. You’d better paint it half green and half purple, so you can use it as a corner for each of those cubes, and not waste a single face. There’ll need to be center cubes with no green and no purple too. So you take 2 corner cubes from the red cube and paint the 3 blank faces of 1 purple, and the 3 blank faces of the other green. Now you’ve got the 6 face cubes that each have 1 face painted red. That leaves 5 empty faces on each. You can split them in half. In the first group, you paint 3 faces green and 2 faces purple; In the second group, paint 3 faces purple and 2 green. Counting on symmetry, you replicate these piles again with the colors rearranged. That gives you 6 with 1 green face, 6 with 1 red face, and 6 with 1 purple face. Counting up what you’ve completely painted, you see 8 corner cubes in each color, 6 edge cubes in each color, 6 face cubes in each color, and 1 center cube. That means you just need 6 more edge cubes in green and purple. And there are exactly 6 cubes left, each with 4 empty faces. You paint 2 faces of each green and 2 faces of each purple. And now you have a cube that’s perfectly painted to make an incredible trip. It rearranges itself to be red in deep space, purple as it enters RH-1729’s atmosphere, and green when it flies through the electric storms. As it reaches the monolith, you realize you’ve achieved something humans have dreamt of for eons: alien contact.

P55? ?Can you solve the bridge riddle

Taking that internship in a remote mountain lab might not have been the best idea. Pulling that lever with the skull symbol just to see what it did probably wasn't so smart, either, but now is not the time for regrets because you need to get away from these mutant zombies fast. With you are the janitor, the lab assistant, and the old professor. You've gotten a headstart, but there's only one way to safety: across an old rope bridge spanning a massive gorge. You can dash across in a minute, while the lab assistant takes two minutes. The janitor is a bit slower and needs five minutes, and the professor takes a whole ten minutes, holding onto the ropes every step of the way. By the professor's calculations, the zombies will catch up to you in just over 17 minutes, so you only have that much time to get everyone across and cut the ropes. Unfortunately, the bridge can only hold two people at a time. To make matters worse, it's so dark out that you can barely see, and the old lantern you grabbed on your way only illuminates a tiny area. Can you figure out a way to have everyone escape in time? Remember: no more than two people can cross the bridge together, anyone crossing must either hold the lantern or stay right next to it, and any of you can safely wait in the dark on either side of the gorge. Most importantly, everyone must be safely across before the zombies arrive. Otherwise, the first zombie could step on the bridge while people are still on it. Finally, there are no tricks to use here. You can't swing across, use the bridge as a raft, or befriend the zombies. Pause the video now if you want to figure it out for yourself! Answer in: 3 Answer in: 2 Answer in: 1 At first it might seem like no matter what you do, you're just a minute or two short of time, but there is a way. The key is to minimize the time wasted by the two slowest people by having them cross together. And because you'll need to make a couple of return trips with the lantern, you'll want to have the fastest people available to do so. So, you and the lab assistant quickly run across with the lantern, though you have to slow down a bit to match her pace. After two minutes, both of you are across, and you, as the quickest, run back with the lantern. Only three minutes have passed. So far, so good. Now comes the hard part. The professor and the janitor take the lantern and cross together. This takes them ten minutes since the janitor has to slow down for the old professor who keeps muttering that he probably shouldn't have given the zombies night vision. By the time they're across, there are only four minutes left, and you're still stuck on the wrong side of the bridge. But remember, the lab assistant has been waiting on the other side, and she's the second fastest of the group. So she grabs the lantern from the professor and runs back across to you. Now with only two minutes left, the two of you make the final crossing. As you step on the far side of the gorge, you cut the ropes and collapse the bridge behind you, just in the nick of time. Maybe next summer, you'll just stick to the library.

P56? ?Can you solve the buried treasure riddle

After a massive storm tears through the Hex Archipelago, you find five grizzled survivors in the water. Shivering their timbers, they explain that they’re the former crew of the great pirate Greenbeard, who marooned them after they tried to mutiny. Each was bound up in a different spot on a small island, until the storm washed them out to sea. In gratitude for saving them, they reveal a secret: the island they were on is also where Greenbeard has buried his treasure hoard. But when the sailors try to describe the island, something seems off. All agree it was flat and barren with no prominent features except for some trees. Yet each pirate claims they saw a different number of trees, ranging from two to six. The pirate who saw two trees says the treasure was buried right at his feet. When you fly your hot air balloon over the area to investigate, you see hundreds of small islands, each with exactly six trees. The next storm will be here soon, so you’ll have to hurry and narrow your search. What does the island with Greenbeard’s treasure look like from the sky? And where will the treasure be on that island? Pause here if you want to figure it out for yourself! Answer in 3 Answer in 2 Answer in 1 It might seem like the pirates are delirious from dehydration. But that’s not what’s going on. Remember, each was confined to a separate point on the island, and no two of them could see the same number of trees. That means that for all but one pirate, something was blocking their view. And since there are no other features on the island, that something could only have been other trees. A pirate would see fewer trees when two or more fell along a straight line from their vantage point. So we need to find the island where five different pirates standing in different spots would each see a different number of trees. Virtually every island has a position from which you can see six trees. And on most islands there’s a position where 5 trees can be seen by standing in line with two of them. It turns out that the hardest locations to find are those with fewer visible trees precisely because they require more trees to line up with the viewer’s position. So how can we see just two trees? One way would be if all the trees were lined up in single file, such as on this island. Then, you could stand at the end of the line and see one, stand in the middle and see two, or stand anywhere else and see all six. But there’s no place from which you can see only three, four, or five, so one straight line of trees is out. So what about two lines of trees? So long as the lines aren’t parallel and they intersect over land, there’ll always be a position where the two lines converge from which you could see exactly two trees. And if they’re grouped two and four, or three and three, there are many arrangements in which you could also see three, four, five, and six trees. Fortunately for us, there’s only one island in the archipelago with two non-parallel lines of trees, and it’ll be buried at the intersection of the two lines. You land on this island and dig up a chest containing a massive pile of tree seeds, ready for planting. Was this treasure really worth all that trouble? That’s a matter of perspective.

P57? ?Can you solve the control room riddle

As your country's top spy, you must infiltrate the headquarters of the evil syndicate, find the secret control panel, and deactivate their death ray. But all you have to go on is the following information picked up by your surveillance team. The headquarters is a massive pyramid with a single room at the top level, two rooms on the next, and so on. The control panel is hidden behind a painting on the highest floor that can satisfy the following conditions: Each room has exactly three doors to other rooms on that floor, except the control panel room, which connects to only one, there are no hallways, and you can ignore stairs. Unfortunately, you don't have a floor plan, and you'll only have enough time to search a single floor before the alarm system reactivates. Can you figure out which floor the control room is on? Pause now to solve the riddle yourself. Answer in: 3 Answer in: 2 Answer in: 1 To solve this problem, we need to visualize it. For starters, we know that on the correct floor there's one room, let's call it room A, with one door to the control panel room, plus one door to room B, and one to C. So there must be at least four rooms, which we can represent as circles, drawing lines between them for the doorways. But once we connect rooms B and C, there are no other connections possible, so the fourth floor down from the top is out. We know the control panel has to be as high up as possible, so let's make our way down the pyramid. The fifth highest floor doesn't work either. We can figure that out by drawing it, but to be sure we haven't missed any possibilities, here's another way. Every door corresponds to a line in our graph that makes two rooms into neighbors. So in the end, there have to be an even number of neighbors no matter how many connections we make. On the fifth highest floor, to fulfill our starting conditions, we'd need four rooms with three neighbors each, plus the control panel room with one neighbor, which makes 13 total neighbors. Since that's an odd number, it's not possible, and, in fact, this also rules out every floor that has an odd number of rooms. So let's go one more floor down. When we draw out the rooms, low and behold, we can find an arrangement that works like this. Incidentally, the study of such visual models that show the connections and relationships between different objects is known as graph theory. In a basic graph, the circles representing the objects are known as nodes, while the connecting lines are called edges. Researchers studying such graphs ask questions like, "How far is this node from that one?" "How many edges does the most popular node have?" "Is there a route between these two nodes, and if so, how long is it?" Graphs like this are often used to map communication networks, but they can represent almost any kind of network, from transport connections within a city and social relationships among people, to chemical interactions between proteins or the spread of an epidemic through different locations. So, armed with these techniques, back to the pyramid. You avoid the guards and security cameras, infiltrate the sixth floor from the top, find the hidden panel, pull some conspicuous levers, and send the death ray crashing into the ocean. Now, time to solve the mystery of why your surveillance team always gives you cryptic information. Hi everybody. If you liked this riddle, try solving these two.

P58? ?Can you solve the counterfeit coin riddle?

You're the realm's greatest mathematician,but ever since you criticized ?the Emperor's tax laws,you've been locked in the dungeonwith only a marker to count the days.But one day, you're suddenly brought before the Emperorwho looks even angrier than usual.One of his twelve governors has been convicted of paying his taxeswith a counterfeit coinwhich has already made its way into the treasury.As the kingdom's greatest mathematician,you've been granted a chance to earn your freedom by identifying the fake.But the Emperor's not a patient man.You may only use the scale three timesbefore you'll be thrown back ?into the dungeon.You look around for anything else you can use,but there's nothing in the room -just the coins,the scale,and your trusty marker.How do you identify the counterfeit?Pause here if you want ?to figure it out for yourself!Answer in:3Answer in:2Answer in: 1Obviously you can't weigh each coin against all of the others,so you'll have to weigh several coins at the same timeby splitting the stack ?into multiple pilesthen narrowing down ?where the false coin is.Start by dividing the twelve coins into three equal piles of four.Placing two of these on the scale gives us two possible outcomes.If the two sides balance, all eight coins on the scale are real,and the fake must be among ?the remaining four.So how do you keep track of these results?That's where the marker comes in.Mark the eight authentic coins with a zero.Now, take three of them and weigh them against three unmarked coins.If they balance, the remaining ?unmarked coin must be the fake.If they don't, draw a plus on the three unmarked coins if they're heavieror a minus if they're lighter.Now, take two of the newly marked coins and weigh them against each other.If they balance, the third coin is fake.Otherwise, look at their marks.If they are plus coins, ?the heavier one is the imposter.If they are marked with minus, it's the lighter one.But what if the first two piles you weigh don't balance?Mark the coins on the heavier side with a plusand those on the lighter side ?with a minus.You can also mark the remaining four coins with zerossince you know the fake one is already somewhere on the scale.Now, you'll need to think strategicallyso you can remove all remaining ambiguity in just two more weighings.To do this, you'll need ?to reassemble the piles.One method is to replace ?three of the plus coinswith three of the minus coins,and replace those ?with three of the zero coins.>From here, you have three possibilities.If the previously heavier side of the scale is still heavier,that means either the remaining plus coin on that sideis actually the heavier one,or the remaining minus coin on the lighter sideis actually the lighter one.Choose either one of them, and weigh it against one of the regular coinsto see which is true.If the previously heavier side became lighter,that means one of the three minus coins you movedis actually the lighter one.Weigh two of them against each other.If they balance, the third is counterfeit.If not, the lighter one is.Similarly, if the two sides balanced after your substitution,then one of the three plus coins you removedmust be the heavier one.Weigh two of them against each other.If they balance, the third one is fake.If not, then it's the heavier one.The Emperor nods approvingly at your finding,and the counterfeiting Lord takes your place in the dungeon.

P59? ?Can you solve the cuddly duddly fuddly wuddly riddle

For your son’s sixth birthday, you’ve promised to get him the cutest creature in creation: the cuddly. It’s hard to believe that it’s a cousin of the terrifying duddly or the hideous fuddly. They’re all members of the Wuddly species, and the process of adopting them is deeply peculiar. It takes 100 eggs to make a single animal in genus Wuddly. When 100 eggs are placed together in an incubator, they undergo egg fusion, and combine in the following way. Blue and purple combine to make red eggs. Red and blue combine to make purple eggs, and red and purple combine to make blue eggs. The most plentiful eggs pair up first, and if two piles are even, an egg comes from one of them at random. They keep combining until there’s just one left. If the final egg is blue, a Cuddly hatches out of it. Purple eggs give you Duddlies, and Red eggs give you Fuddlies. The incubator currently has 99 eggs in it. 23 are blue, 33 are purple, and 43 are red. You can begin the process of egg fusion by adding an egg of any color to the room. When all the eggs have combined into a single egg, the creature that hatches will bond with you on sight, which is why getting a Cuddly is so important. After all, you made a promise to your son. Which color egg should you add to the incubator to get a cuddly? Pause the video to figure it out for yourself. Answer in 3 Answer in 2 Answer in 1 It’s easy to get mixed up with all the cuddlies, duddlies, and fuddlies coming from different colored eggs. If we ignore how many total eggs of each color there are, and just look at the process of egg fusion, we might notice something that will make this problem simpler. When two eggs fuse, the number of eggs of each of those colors decreases by one, and the number of the third color increases by one. That means they all change parity, or evenness and oddness, at the same time. Right now all three piles are odd, but you get to add an egg to one color, which means that it’ll be even and the other two will be odd. Whichever color you choose will always be the opposite parity of the other two piles: odd when they’re even and even when they’re odd, since every egg fusion flips each pile’s parity simultaneously. We want to end with 1 blue, 0 purple, and 0 red eggs, or odd, even, even. That means we want the blue egg pile to be the opposite parity of the other two piles at the start as well. So you add a blue egg into the room, and 99 egg fusions later, only a single blue egg remains. The Cuddly that hatches is sure to make your 6-year-old as happy as can be. Just be sure to follow the shopkeeper’s warning, and never feed it after midnight.

P60? ?Can you solve the dark coin riddle

You heard the traveler's tales, you followed the crumbling maps, and now, after a long and dangerous quest, you have some good news and some bad news. The good news is you've managed to locate the legendary dungeon containing the stash of ancient Stygian coins and the eccentric wizard who owns the castle has even generously agreed to let you have them. The bad news is that he's not quite as generous about letting you leave the dungeon, unless you solve his puzzle. The task sounds simple enough. Both faces of each coin bear the fearsome scorpion crest, one in silver, one in gold. And all you have to do is separate them into two piles so that each has the same number of coins facing silver side up. You're about to begin when all of the torches suddenly blow out and you're left in total darkness. There are hundreds of coins in front of you and each one feels the same on both sides. You try to remember where the silver-facing coins were, but it's hopeless. You've lost track. But you do know one thing for certain. When there was still light, you counted exactly 20 silver-side-up coins in the pile. What can you do? Are you doomed to remain in the dungeon with your newfound treasure forever? You're tempted to kick the pile of coins and curse the curiosity that brought you here. But at the last moment, you stop yourself. You just realized there's a surprisingly easy solution. What is it? Pause here if you want to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 You carefully move aside 20 coins one by one. It doesn't matter which ones: any coins will do, and then flip each one of them over. That's all there is to it. Why does such a simple solution work? Well, it doesn't matter how many coins there are to start with. What matters is that only 20 of the total are facing silver side up. When you take 20 coins in the darkness, you have no way of knowing how many of these silver-facing coins have ended up in your new pile. But let's suppose you got 7 of them. This means that there are 13 silver-facing coins left in the original pile. It also means that the other 13 coins in your new pile are facing gold side up. So what happens when you flip all of the coins in the new pile over? Seven gold-facing coins and 13 silver-facing coins to match the ones in the original pile. It turns out this works no matter how many of the silver-facing coins you grab, whether it's all of them, a few, or none at all. That's because of what's known as complementary events. We know that each coin only has two possible options. If it's not facing silver side up, it must be gold side up, and vice versa, and in any combination of 20 coins, the number of gold-facing and silver-facing coins must add up to 20. We can prove this mathematically using algebra. The number of silver-facing coins remaining in the original pile will always be 20 minus however many you moved to the new pile. And since your new pile also has a total of 20 coins, its number of gold-facing coins will be 20 minus the amount of silver-facing coins you moved. When all the coins in the new pile are flipped, these gold-facing coins become silver-facing coins, so now the number of silver-facing coins in both piles is the same. The gate swings open and you hurry away with your treasure before the wizard changes his mind. At the next crossroads, you flip one of your hard-earned coins to determine the way to your next adventure. But before you go, we have another quick coin riddle for you – one that comes from this video sponsor’s excellent website. Here we have 8 arrangements of coins. You can flip over adjacent pairs of coins as many times as you like. A flip always changes gold to silver, and silver to gold. Can you figure out how to tell, at a glance, which arrangements can be made all gold? You can try an interactive version of this puzzle and confirm your solution on Brilliant’s website. We love Brilliant.org because the site gives you tools to approach problem-solving in one of our favorite ways— by breaking puzzles into smaller pieces or limited cases, and working your way up from there. This way, you're building up a framework for problem solving, instead of just memorizing formulas. You can sign up for Brilliant for free, and if you like riddles a Brilliant.org premium membership will get you access to countless more interactive puzzles. Try it out today by visiting brilliant.org/TedEd and use that link so they know we sent you. The first 833 of you to visit that link will receive 20% off the annual premium subscription fee.