P71??Can you solve the jail break riddle

Your timing made you and your partner the most infamous bank robbers in the west. Now, you’ll need to use that timing to help you break out of jail. At the appointed time, you’ll be walking in the yard near the electric fence. Your partner will flash you the signal, and exactly 45 seconds later, short out the fence circuit. It’ll automatically restart after a second or two, but as long as you move fast, you’ll be home free. And then you notice, to your horror, that your watch is broken, and there’s no time to fix it. The signal is coming, and if you make even a small mistake in counting off 45 seconds, you’ll get fried. Searching your pockets, you find something that might help: a lighter and two fuses you made earlier in the prison work program. Each fuse is a length of flammable twine, built to be lit on either end and burn for precisely one minute. The problem is that even though the fuses look uniform, they don’t burn evenly, so if you cut one in half, for example, one side might burn longer than the other. Your partner is going to give the signal any minute, and you’ll have to make your move. How can you use the fuses and lighter to time exactly 45 seconds? Pause the video to figure it out yourself. Answer in 3 Answer in 2 Answer in 1 The length of the fuse may not tell you anything, but you do know the fuses take exactly 60 seconds to burn from end to end. Here’s the key insight: If you start a fuse on one side and it burns for 30 seconds, there’ll still be 30 seconds of fuse left. If you had started it from the other end, it would’ve reached the exact same spot in thirty seconds. That means that if you lit it from both ends simultaneously, it would burn out in precisely 30 seconds. But how will you time the last fifteen? That’ll have to come from the second fuse. If it were a 30 second fuse, you’d be able to use that same trick again to double the burning speed and make it last exactly 15 seconds. And, you realize, you can shorten the second fuse by lighting one end of it at the same time as you light the first. At the moment the first burns out, you’ll be left with 30 seconds on the second fuse. Just when you’ve got this all figured out, you see the signal from your partner, and spring into action. You gather the four ends of the two fuses and light three of them. The moment the first burns out, you light the other end of the second fuse. When it flickers and dies, you know that exactly 45 seconds have passed, and the electric fence is dead. By the time it hiccups back to life, you’re over the fence and home free.

P72???Can you solve the killer robo-ants riddle

The good news is that your experimental robo-ants are a success! The bad news is that you accidentally just gave them the ability to shoot deadly lasers …and you can’t turn it off. You have five minutes to stop them before the lasers go online. Until then, all of your robo-ants will walk inside their habitat at a speed of exactly 1 meter per minute. If they bump into each other or hit a dead end, they’ll instantly turn around and walk back the way they came. When five minutes are up, they’ll turn on their lasers, break free, and stream out into the world, carving a path of destruction as they go. Your one chance to stop them is to insert the two emergency vacuum nozzles into the habitat and suck the ants up before they break free. The nozzles can press into any one location in the habitat through a membrane covering its front side, and any ants that walk past will be sucked up and deactivated. You can’t move the nozzles once they’re placed without leaving a hole that the robo-ants would pour out of, so choosing the right spots will be key. The habitat is made out of meter-long tubes. When the robots reach an intersection, they will pick randomly whether to go left, right, or forward. They only go backward if they hit hit another robo-ant or a dead end. Unfortunately, there are hundreds of them inside the habitat, and if even one escapes, it’ll do a lot of damage. With just less than five minutes remaining, where should you place the 2 vacuum nozzles to suck up all the robo-ants? Pause the video now if you want to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 With robo-ants ricocheting all over the habitat, it might seem impossible to stop them before they break free. But this situation is simpler than it seems. Here's why. Imagine just two robo-ants crawling toward each other. When they collide, they immediately reverse directions. And what would that sequence of events look like if they crawled past each other instead? It would look exactly the same before and after their collision, but with their positions swapped. This is true every time a pair of robo-ants meet. Because the identities of individual ants don’t matter, you just need to figure out where you should put the nozzles to capture any single ant walking without interruption for less than 5 minutes, starting from any point in the habitat. That’s much easier to conceptualize and solve. Placing the nozzles at intersections where three or four tubes meet seems like your best bet since that’s where the robo-ants might otherwise change directions and miss your nozzles. There are only four intersections… which two should you pick? The top right intersection has to be one of them. If it isn’t, an ant crawling down from this intersection toward the dead end would crawl for four minutes to get back to the intersection, and then go in any of three directions, walking for at least another minute. Once you’ve placed a nozzle in the top right, the only other choice that has a chance to work is the bottom left. To see that this works, imagine an ant anywhere else in the habitat. Worst case scenario, the ant would start right next to the vacuum nozzle, marching away from it. But in all those worst cases, the ant would march for at most 4 meters before being sucked up into the vacuum. No other choice of two intersection points is guaranteed to get all the robo-ants within five minutes. Having vacuumed them all up, you’ve averted a major crisis. Before you mess with robo-ants again, you’ll want to have a robo-anteater ready. And wouldn’t it be cool if it could fly and breathe fire? There’s no way that could go wrong!

P73??Can you solve the Leonardo da Vinci riddle

You’ve found Leonardo Da Vinci’s secret vault, secured by a series of combination locks. Fortunately, your treasure map has three codes: 1210, 3211000, and… hmm. The last one appears to be missing. Looks like you’re gonna have to figure it out on your own. There’s something those first two numbers have in common: they’re what’s called autobiographical numbers. This is a special type of number whose structure describes itself. Each of an autobiographical number’s digits indicates how many times the digit corresponding to that position occurs within the number. The first digit indicates the quantity of zeroes, the second digit indicates the number of ones, the third digit the number of twos, and so on until the end. The last lock takes a 10 digit number, and it just so happens that there’s exactly one ten-digit autobiographical number. What is it? Pause here if you want to figure it out for yourself! Answer in: 3 Answer in: 2 Answer in: 1 Blindly trying different combinations would take forever. So let’s analyze the autobiographical numbers we already have to see what kinds of patterns we can find. By adding all the digits in 1210 together, we get 4 – the total number of digits. This makes sense since each individual digit tells us the number of times a specific digit occurs within the total. So the digits in our ten-digit autobiographical number must add up to ten. This tells us another important thing – the number can’t have too many large digits. For example, if it included a 6 and a 7, then some digit would have to appear 6 times, and another digit 7 times– making more than 10 digits. We can conclude that there can be no more than one digit greater than 5 in the entire sequence. So out of the four digits 6, 7, 8, and 9, only one – if any-- will make the cut. And there will be zeroes in the positions corresponding to the numbers that aren’t used. So now we know that our number must contain at least three zeroes – which also means that the leading digit must be 3 or greater. Now, while this first digit counts the number of zeroes, every digit after it counts how many times a particular non-zero digit occurs. If we add together all the digits besides the first one – and remember, zeroes don’t increase the sum – we get a count of how many non-zero digits appear in the sequence, including that leading digit. For example, if we try this with the first code, we get 2 plus 1 equals 3 digits. Now, if we subtract one, we have a count of how many non-zero digits there are after the first digit – two, in our example. Why go through all that? Well, we now know something important: the total quantity of non-zero digits that occur after the first digit is equal to the sum of these digits, minus one. And how can you get a distribution where the sum is exactly 1 greater than the number of non-zero positive integers being added together? The only way is for one of the addends to be a 2, and the rest 1s. How many 1s? Turns out there can only be two – any more would require additional digits like 3 or 4 to count them. So now we have the leading digit of 3 or greater counting the zeroes, a 2 counting the 1s, and two 1s – one to count the 2s and another to count the leading digit. And speaking of that, it’s time to find out what the leading digit is. Since we know that the 2 and the double 1s have a sum of 4, we can subtract that from 10 to get 6. Now it’s just a matter of putting them all in place: 6 zeroes, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights, and 0 nines. The safe swings open, and inside you find... Da Vinci’s long-lost autobiography.

P74? ?Can you solve the locker riddle

Your rich, eccentric uncle ?just passed away,and you and your 99 nasty relatives have been invited to the reading of his will.He wanted to leave ?all of his money to you,but he knew that if he did, your relatives would pester you forever.So he is banking on the fact?that he taught you everything ?you need to know about riddles.Your uncle left the following note in his will:"I have created a puzzle.If all 100?of you answer it together, you will share the money evenly.However, if you are the first to find the pattern and solve the problem?without going through all of the leg work,you will get the entire inheritance ?all to yourself.Good luck."The lawyer takes you and your 99?relatives to a secret room in the mansion?that contains?100?lockers,each hiding a single word.He explains:Every relative is assigned a number from 1?to?100.Heir 1?will open every locker.Heir 2?will then?close every second locker.Heir 3?will change the status of every third locker,specifically if it's open, ?she'll close it,but if it's closed, she'll open it.This pattern will continue until all 100?of you have gone.The words in the lockers that remain open at the end?will help you crack the code for the safe.Before cousin Thaddeus can even start down the line,you step forward and tell the lawyer you know which lockers will remain open.But how?Pause the video now if you want to figure it out for yourself!Answer in: 1Answer in: 2Answer in:? 3The key is realizing that the number of times a locker is touched?is the same as the number of factors in the locker number.For example, in locker #6,Person 1?will open it,Person 2?will close it,Person 3?will open it,and Person 6?will close it.The numbers 1, 2, 3, and? 6?are the factors of .So when a locker has an even number of factors?it will remain closed,and when it has an odd number of factors,it will remain open.Most of the lockers ?have an even number of factors,which makes sense because factors naturally pair up.In fact, the only lockers that have an odd number of factors?are perfect squares?because those have one factor that when multiplied by itself equals the number.For Locker?9, 1?will open it,3?will close,and 9?will open it.3?x 3?= 9,but the 3?can only be counted once.Therefore, every locker that is ?a perfect square will remain open.You know that these ten lockers are the solution,so you open them immediately and read the words inside:"The code is the first five lockers touched only twice."You realize that the only lockers ?touched twice have to be prime numbers?since each only has two factors:1?and itself.So the code is?2-3-5-7-11.The lawyer brings you to the safe,and you claim your inheritance.Too bad your relatives were always too busy being nasty to each other?to pay attention to your eccentric uncle's riddles.

P75? ?Can you solve the Mondrian squares riddle

Dutch artist Piet Mondrian’s abstract, rectangular paintings inspired mathematicians to create a two-fold challenge. First, we must completely cover a square canvas with non-overlapping rectangles. All must be unique, so if we use a 1x4, we can’t use a 4x1 in another spot, but a 2x2 rectangle would be fine. Let’s try that. Say we have a canvas measuring 4x4. We can’t chop it directly in half, since that would give us identical rectangles of 2x4. But the next closest option - 3x4 and 1x4 - works. That was easy, but we’re not done yet. Now take the area of the largest rectangle, and subtract the area of the smallest. The result is our score, and the goal is to get as low a score as possible. Here, the largest area is 12 and the smallest is 4, giving us a score of 8. Since we didn’t try to go for a low score that time, we can probably do better. Let’s keep our 1x4 while breaking the 3x4 into a 3x3 and a 3x1. Now our score is 9 minus 3, or 6. Still not optimal, but better. With such a small canvas, there are only a few options. But let’s see what happens when the canvas gets bigger. Try out an 8x8; what’s the lowest score you can get? Pause here if you want to figure it out yourself. Answer in: 3 Answer in: 2 Answer in: 1 To get our bearings, we can start as before: dividing the canvas roughly in two. That gives us a 5x8 rectangle with area 40 and a 3x8 with area 24, for a score of 16. That’s pretty bad. Dividing that 5x8 into a 5x5 and a 5x3 leaves us with a score of 10. Better, but still not great. We could just keep dividing the biggest rectangle. But that would leave us with increasingly tiny rectangles, which would increase the range between the largest and smallest. What we really want is for all our rectangles to fall within a small range of area values. And since the total area of the canvas is 64, the areas need to add up to that. Let’s make a list of possible rectangles and areas. To improve on our previous score, we can try to pick a range of values spanning 9 or less and adding up to 64. You’ll notice that some values are left out because rectangles like 1x13 or 2x9 won’t fit on the canvas. You might also realize that if you use one of the rectangles with an odd area like 5, 9, or 15, you need to use another odd-value rectangle to get an even sum. With all that in mind, let’s see what works. Starting with area 20 or more puts us over the limit too quickly. But we can get to 64 using rectangles in the 14-18 range, leaving out 15. Unfortunately, there’s no way to make them fit. Using the 2x7 leaves a gap that can only be filled by a rectangle with a width of 1. Going lower, the next range that works is 8 to 14, leaving out the 3x3 square. This time, the pieces fit. That’s a score of 6. Can we do even better? No. We can get the same score by throwing out the 2x7 and 1x8 and replacing them with a 3x3, 1x7, and 1x6. But if we go any lower down the list, the numbers become so small that we’d need a wider range of sizes to cover the canvas, which would increase the score. There’s no trick or formula here – just a bit of intuition. It's more art than science. And for larger grids, expert mathematicians aren’t sure whether they’ve found the lowest possible scores. So how would you divide a 4x4, 10x10, or 32x32 canvas? Give it a try and post your results in the comments.

P76? ?Can you solve the multiplying rabbits riddle

After years of experiments, you’ve finally created the pets of the future– nano-rabbits! They’re tiny, they’re fuzzy… and they multiply faster than the eye can see. In your lab there are 36 habitat cells, arranged in an inverted pyramid, with 8 cells in the top row. The first has one rabbit, the second has two, and so on, with eight rabbits in the last one. The other rows of cells are empty… for now. The rabbits are hermaphroditic, and each rabbit in a given cell will breed once with every rabbit in the horizontally adjacent cells, producing exactly one offspring each time. The newborn rabbits will drop into the cell directly below the two cells of its parents, and within minutes will mature and reproduce in turn. Each cell can hold 10^80 nano-rabbits – that’s a 1 followed by 80 zeros – before they break free and overrun the world. Your calculations have given you a 46-digit number for the count of rabbits in the bottom cell– plenty of room to spare. But just as you pull the lever to start the experiment, your assistant runs in with terrible news. A rival lab has sabotaged your code so that all the zeros at the end of your results got cut off. That means you don’t actually know if the bottom cell will be able to hold all the rabbits – and the reproduction is already underway! To make matters worse, your devices and calculators are all malfunctioning, so you only have a few minutes to work it out by hand. How many trailing zeros should there be at the end of the count of rabbits in the bottom habitat? And do you need to pull the emergency shut-down lever? Pause the video now if you want to figure it out for yourself. Answer in 3 Answer in 2 Answer in 1 There isn’t enough time to calculate the exact number of rabbits in the final cell. The good news is we don’t need to. All we need to figure out is how many trailing zeros it has. But how can we know how many trailing zeros a number has without calculating the number itself? What we do know is that we arrive at the number of rabbits in the bottom cell through a process of multiplication – literally. The number of rabbits in each cell is the product of the number of rabbits in each of the two cells above it. And there are only two ways to get numbers with trailing zeros through multiplication: either multiplying a number ending in 5 by any even number, or by multiplying numbers that have trailing zeroes themselves. Let’s calculate the number of rabbits in the second row and see what patterns emerge. Two of the numbers have trailing zeros – 20 rabbits in the fourth cell and 30 in the fifth cell. But there are no numbers ending in 5. And since the only way to get a number ending in 5 through multiplication is by starting with a number ending in 5, there won’t be any more down the line either. That means we only need to worry about the numbers that have trailing zeros themselves. And a neat trick to figure out the amount of trailing zeros in a product is to count and add the trailing zeros in each of the factors – for example, 10 x 100 = 1,000. So let’s take the numbers in the fourth and fifth cells and multiply down from there. 20 and 30 each have one zero, so the product of both cells will have two trailing zeros, while the product of either cell and an adjacent non-zero-ending cell will have only one. When we continue all the way down, we end up with 35 zeros in the bottom cell. And if you’re not too stressed about the potential nano-rabbit apocalypse, you might notice that counting the zeros this way forms part of Pascal’s triangle. Adding those 35 zeros to the 46 digit number we had before yields an 81 digit number – too big for the habitat to contain! You rush over and pull the emergency switch just as the seventh generation of rabbits was about to mature – hare-raisingly close to disaster.

P77? ?Can you solve the multiverse rescue mission riddle

It was a normal Tuesday at the superconductor, until a bug in the system created a small situation. Now your team is trapped in eleven separate pocket dimensions. Luckily for you, there’s a half-finished experimental teleportation robot that may be able to get you all home, if you can figure out how to work through the quirks of its design. Over interdimensional radio, your engineers explain that the robot can teleport into the alternate universes you’re trapped in, but it’ll do so completely at random. The robot has two levers and one big button. When it appears, you just switch the position of one of the levers from A to B or vice versa, and then the robot will note your dimensional position and teleport to another of the eleven dimensions at random. If it shows up again, you’ll have to pull a lever before it’ll teleport away. When anyone presses the button, the robot will bring everyone who pulled a lever back home. Anyone who didn’t will be lost in the multi-verse forever. The challenge is to make sure everyone has pulled a lever before anyone hits the button. While you can talk to each other now over the interdimensional radio and agree on a plan, the robot’s teleportation technology will interfere with all attempts at communication once it arrives. You won’t be able to attach messages to the robot or scratch notes into its superstrong alloy body. Your only way to communicate information is to change the position of exactly one lever or hit the button. What plan will make sure everyone gets home? Pause the video now if you want to figure it out for yourself. Answer in 3 Answer in 2 Answer in 1 It would be nice if you could set different combinations of the levers to indicate who’s already been visited by the robot. But it has only two levers. That gives four combinations— far too few to communicate about 11 people, especially when you’re forced to flip one to send the robot onward. There must be another way. The critical insight is that not everyone has to know when every pocket dimension has been visited. If one person accepts responsibility ahead of time for hitting the button, then only they need to know who the robot has visited. In fact, they don’t even need to know exactly who’s been visited… just how many people have been. You volunteer to be the person in charge of pressing the button when the moment is right, and give the following directions to everyone else. Your plan is simple: you’ll use the left lever to count visits, and the right lever will have no meaning, so there’s no harm in moving it up or down. Each of the others will pull the left lever from position A to position B exactly once. If the robot appears with the left lever already pulled down, or if an individual has previously pulled the left lever down at any point in the past, then they should move the right lever. You, meanwhile, will be the only one who ever resets the left lever from position B to position A. This gives you a way to count how many people have been visited by the robot. Everyone needs to pull the left lever down exactly once, and you’re the only one to pull it back up. So you know that the tenth time the robot visits you with its left lever in the down position, it must have visited all ten of the others. And that means you’re safe to press the button and teleport everyone home. It may take a while– most likely the robot will need to teleport around 355 times; but better that than leave anyone behind. Your teammates phase back into your home dimension one at a time. The mission proves a great success. Well...mostly.

P78??Can you solve the passcode riddle

In this dystopian world, your resistance group is humanity's last hope.Unfortunately, you've all been captured by the tyrannical rulers?and brought to the ancient colosseum for their deadly entertainment.Before you're thrown into the dungeon,you see many numbered hallways leading outside.But each exit is blocked by ?an electric barrier?with a combination keypad.You learn that one of you will be allowed to try to escape by passing a challenge?while everyone else will be fed to the mutant salamanders the next morning.With her perfect logical reasoning, Zara is the obvious choice.You hand her a concealed audio transmitter so that the rest of you can listen along.As Zara is led away,you hear her footsteps echo ?through one of the hallways,then stop.A voice announces ?that she must enter a code?consisting of three positive whole numbers in ascending order,so the second number is greater than or equal to the first,and the third is greater than or equal to the second.She may ask for up to three clues,but if she makes a wrong guess,or says anything else,she'll be thrown back into the dungeon.For the first clue, the voice says the product of the three numbers is 36.When Zara asks for the second clue,it tells her the sum of the numbers?is the same as the number ?of the hallway she entered.There's a long silence.You're sure Zara remembers ?the hallway number,but there's no way for you to know it,and she can't say it out?loud.If Zara could enter the pass?code ?at this point, she would,but instead, she asks for the third clue,and the voice announces that the largest number appears only once?in the combination.Moments later, the buzz of the electric barrier stops for a few seconds,and you realize that Zara has escaped.Unfortunately, her transmitter ?is no longer in range,so that's all the information you get.Can you find the solution?Pause on the next screen to work out the solution.321You're worried about the fact that you don't know Zara's hallway number,but you decide to start ?from the beginning anyways.From the first clue, you work out all of the eight possible combinations?that come out to a product of 36.One of these must be right, but which one?Now comes the hard part.Even though you don't know which number you're looking for,you decide to work out the sum of each combination's three numbers.That's when it hits you.All but two of the sums are unique,and if the hallway number had matched any of these,Zara would have known the correct combination right then and there?without asking for the third clue.Since she did ask for the clue,the hallway number must have ?matched the only sum?that appears more than once in the list:thirteen.But which of the two combinations that add up to thirteen is correct:1,6,6,or 2,2,9?That's where the third clue comes in.Since it tells us that the largest number must be unique,2,2,9 must be the code.When night falls, you and the others escape through hallway thirteen?and rejoin Zara outside.You've freed yourselves through ?math and logic.Now it's time to free ?the rest of the world.

P79? ?Can you solve the penniless pilgrim riddle

After months of travel, you’ve arrived at Duonia, home to the famous temple that’s the destination of your pilgrimage. Entering from the northwest, you pass through the city gates and the welcome center, where you’re given a map and a brochure. The map reveals that the town consists of 16 blocks, formed by five streets that run west to east, intersecting five more that run north to south. You’re standing on the northernmost street facing east, with the two blocks containing the gate and the welcome center behind you. The temple’s only entrance lies at the very southeast corner. It’s not a long walk, but there’s a problem. As you learn from the brochure, Duonia imposes a unique tax on all visitors, which must be paid when they arrive at their destination within the city. The tax begins at zero, increases by two silver for every block you walk east, and doubles for every block you walk south. However, a recent reform to make the tax fairer halves your total bill for every block you walk north and subtracts two silver for every block you walk west. Just passing through the gate and the welcome center means you already owe four silver. As a pilgrim you carry no money and have no way of earning any. What’s more, the rules of your pilgrimage forbid you from walking over any stretch of ground more than once during your journey— though you can cross your own path. Can you figure out a way to reach the temple without owing any tax or walking the same block twice in any direction? Pause here if you want to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 You look at the map to consider your options. Walking towards the temple always increases the tax, and walking away decreases it, so it seems like you can never reach it without owing silver. But what happens when you walk around a single block? If you start out owing four silver and go clockwise starting east, your tax bill becomes six, then 12, then 10, then five. If you looped again, you’d owe seven, 14, 12, and six. It seems that each clockwise loop leaves you owing one extra silver. What about a counterclockwise loop then? Starting owing four again and going south first, your bill changes to eight, 10, five, and three. Looping again you’d owe six, eight, four, and two. Each counterclockwise loop actually earns you one silver. That’s because any tax doubled, plus two, halved, and minus two, always ends up one smaller than it started. The key here is that while the different taxes for opposite directions may seem to balance each other out, the order in which they’re applied makes a huge difference. You start off owing four silver, so four counterclockwise loops would get you down to zero. Unfortunately, it’s not that simple, since you can’t walk the same block twice. But there’s another way to reduce your bill: walking one large counterclockwise loop through the city. From your starting position, walk three blocks south. You need to leave the southernmost street clear for the final stretch, so continuing counterclockwise means going east. Walk two blocks to the eastern wall and you owe a whopping 36 silver. But now you can start reducing your bill. Three blocks north and one block west cuts it to 2.5. You can’t go west from here —that would leave you with no way out. So you go one block south, and the remaining three blocks west, leaving you with a debt of -1 silver. And since doubling a negative number still gives you a negative number, walking the three blocks to the south wall means the city owes you eight. Fortunately, that’s exactly enough to get you through the final blocks to the temple. As you enter, you realize what you’ve learned from your pilgrimage: sometimes an indirect route is the best way to reach your destination.

P80??Can you solve the pirate riddle

It's a good day to be a pirate.Amaro and his four mateys,Bart,Charlotte,Daniel,and Eliza?have struck gold:a chest with 100 coins.But now, they must divvy up the booty according to the pirate code.As captain, Amaro gets to propose how to distribute the coins.Then, each pirate, ?including Amaro himself,gets to vote either yarr or nay.If the vote passes, or if there's a tie, the coins are divided according to plan.But if the majority votes nay,Amaro must walk the plank?and Bart becomes captain.Then, Bart gets to propose a new distribution?and all remaining pirates vote again.If his plan is rejected, he walks the plank, too,and Charlotte takes his place.This process repeats,with the captain's hat moving to Daniel and then Eliza?until either a proposal is accepted or there's only one pirate left.Naturally, each pirate wants to stay alive while getting as much gold as possible.But being pirates, ?none of them trust each other,so they can't collaborate in advance.And being blood-thirsty pirates,if anyone thinks they'll end up with the same amount of gold either way,they'll vote to make the captain walk the plank just for fun.Finally, each pirate is excellent at logical deduction?and knows that the others are, too.What distribution should Amaro propose to make sure he lives?Pause here if you want to figure it out for yourself!Answer in: 3Answer in: 2Answer in: 1If we follow our intuition,it seems like Amaro should try to bribe the other pirates with most of the gold?to increase the chances of his plan being accepted.But it turns out he can do much better than that. Why?Like we said, the pirates all know each other to be top-notch logicians.So when each votes, they won't just be thinking about the current proposal,but about all possible outcomes down the line.And because the rank order is known in advance,each can accurately predict how the others would vote in any situation?and adjust their own votes accordingly.Because Eliza's last, she has the most outcomes to consider,so let's start by following ?her thought process.She'd reason this out by working backwards from the last possible scenario?with only her and Daniel remaining.Daniel would obviously propose to keep all the gold?and Eliza's one vote would not be enough to override him,so Eliza wants to avoid this situation at all costs.Now we move to the previous decision point?with three pirates left ?and Charlotte making the proposal.Everyone knows that if she's outvoted, the decision moves to Daniel,who will then get all the gold while Eliza gets nothing.So to secure Eliza's vote,Charlotte only needs to offer her slightly more than nothing, one coin.Since this ensures her support,Charlotte doesn't need to offer Daniel anything at all.What if there are four pirates?As captain, Bart would still only need one other vote for his plan to pass.He knows that Daniel wouldn't want ?the decision to pass to Charlotte,so he would offer Daniel one coin for his support?with nothing for Charlotte or Eliza.Now we're back at the initial vote with all five pirates standing.Having considered all the other scenarios,Amaro knows that if he goes overboard,the decision comes down to Bart,which would be bad news ?for Charlotte and Eliza.So he offers them one coin each, keeping 98 for himself.Bart and Daniel vote nay,but Charlotte and Eliza ?grudgingly vote yarr?knowing that the alternative ?would be worse for them.The pirate game involves some interesting concepts from game theory.One is the concept of common knowledge?where each person is aware of what the others know?and uses this to predict their reasoning.And the final distribution is an example of a Nash equilibrium?where each player knows every other players' strategy?and chooses theirs accordingly.Even though it may lead to a worse outcome for everyone?than cooperating would,no individual player can benefit by changing their strategy.So it looks like Amaro gets to keep most of the gold,and the other pirates might need to find better ways?to use those impressive logic skills,like revising this absurd pirate code.