Can a Chess Piece Explain Markov Chains? | Infinite Series 

im surprised how this series is touching on concepts that aren't as popular as topics for math videos on youtube. i was expecting it to just be another take on the same kind of topics seen on numberphile and the like, but it seems like it's really kinda doing it's own thing. really fascinating topics ^^

I love how you guys actually get into some real mathematics in this series. It's not your usual RU-vid surface level stuff.

Thank you for doing this without dumbing things down

Can a chess piece explain Markov chains? No, but PBS Infinite series can explain Markov chains with chess and radio stations.

Starting on their original squares: King = 84 Queen = 69 Bishop = 40 Knight = 112 Rook = 64

Re: The challenges Knight in the corner: 1/(2/336) = 168, by the same logic presented in the video. Rook: First, we note that a rook has the same number of moves anywhere on the board (14), so the stationary distribution is 1/64 everywhere. From there, just take the reciprocal and find the answer to be 1/(1/64) = 64.

168 for the knight. As the rook can move to the same number of squares from every square (14), the stationary distribution is 1/64, so the average number of moves is 64.

I'm studying industrial engineering and am taking a stochastic process class so this was a nice surprise as we covered markov chains last week. However, we also used linear algebra to make the math easier and delved a lot deeper.

Love your videos. Keep up the good work.

now if only this vid was uploaded yesterday before that horrible markov chain question on the exam i did this morning

Omg, these videos are amazing!!! They explore actually interesting topics, rather than boring old basic math. Please keep up the good work!!

Wow, this makes so much sense! I think this is the best way to explain Markov Chains. But 6:30 confused me as the derivation wasn't clear.

The important question is, what is Kelsey's FIDE rating?

Came back to this channel after 5 years, from some chess content that I'm really into nowadays. One of the best channels on RU-vid

Math has always been a frustrating thing for me; Algebra for example, has for some reason always been a challenge, and beyond that, I've always struggled with formulas. This channel offers intuitive explanations to things that would otherwise be like trying to read in a language I don't speak, but sort of recognize the characters and symbols... I have a deep interest in the Sciences, especially Physics, of which you can only learn so much before you need math. Thank you for making videos like these, and please continue to do so; I believe this channel and others like it provide valuable tools for understanding things that people like me (The mathematically challenged) can truly benefit from. Also loved the crossover videos with PBS Space Time.

The knight in the corner is not much of a challenge considering you've already done most of the work. The rook will return in an average of 64 moves, an answer both boring and interesting because it's just the number of squares on the board and it doesn't matter where the rook starts. A bishop starting in any edge square will take 40 moves on average to return to that square. Everything was pretty integers, then the queen had to come along with an ugly 69.3... average if she starts in her usual square.

I feel alive each time I watch an episode from this awesome channel !!!

Kpop and ska are quite specific genres just for an example. She must be a fan of them or something.

Where does the 2/3 and 1/3 come from when you changed the probabilities of the radios?

I'm so happy I found this channel. Only 8 uploads so far, yet they are already of such high quality! Your explainations are stellar, and the visuals make this higher mathematics stuff really accessible, even for someone who doesn't study it. To me this is absolutely fascinating, and I love it. It's gonna be exciting watching this channel grow!

I was just looking at Markov chains as part of a project for my psychology undergraduate! One popular model of neurons has them act as two-state Markov chains. However, I'm looking at that modelling on my own and wasn't really taught stochastic maths, so this is impeccably timed.

These videos are so great! I wonder what the next topic will be! Maybe something with integrals?

I have never heard of Markov chains. All I knew was basic probability and little bit of state space in control systems. but this makes a lot of sense and I felt like I actually understood this. this is truly amazing.

First question: Since stationary distribution of knight at state (corner) is 1/168, 168 seconds (2.8 minutes) or hops is the average time of a knight to return to state (corner). Second question: For the rook, at all states including the corner, there is a 1/14 possibility of landing at a given state. Using this, let's assign 14 rooks per square. Then we get total population of 14x64 = 896, using that to get stationary distribution of 14/896 or 1/64. With the stationary distribution of rook at state (corner) of 1/64, 64 seconds or hops is the average time of a rook to return to state (corner). Yay math! :D

We were pretty psyched to see Markov Chains on this channel. We use them for loss prediction for loans at Banks (using them to quantify and then predict risk migrations). They're a powerful and convenient tool. They're also the first practical use of the most useless of Algebra II lessons from high school... Matrix Math ;)

this mathematician is a deadly combination of beauty and brains!!!!! thanks for making it interesting and easier.😂😂😂😂

Wow! I think it's the best video explaining the usage of Markov Chain in a very playful way I've ever seen. What a amazing job you are doing here, guys! Greetings from Brazil!

Excellent work and content! Kudos to all involved. As an aside, this video just hit me in that this would be an excellent way to intuitively explain the "probabilistic" nature of quantum mechanics and stuff like the many-worlds interpretation.

Highly recommended video on the Markov chain, I thought I was stuck during my stochastic signals course and yet this video has just explain what my lecturer did for the past 3 hrs

Rooks take, on average, 64 moves to return to their starting position. I used the same logic as presented in the video! Great video by the way. I had heard of markov chains before, but this was still enlightening. I'm so happy that PBS made a math channel!

The production value is crazy good on these, all of the diagrams are very well done. I wish they were still making them.

Can't get over how amazing this channel is. Seriously thank you.

Ah, Markov chains! I learned about them as a computer science concept many decades ago, but I hadn't learned about them as a way of predicting expected values in games until 2 years ago (almost exactly), when I was pondering "How many moves does the average game of the infuriating preschool boardgame Hi-Ho Cherry-O! last?" (it was hard to figure out with simple E(v) calculations, because of the variable value of the "lose all your cherries" spin.) Turns out the answer is approximately 15.8 rounds.

Kelsey is really elegant. ...btw, plz, make more videos on hyper dimensional geometry :D

Well the corner knight would take on average 168 moves to return, since 1/(1/168) = 168. Now for the rook, since on each square the rook has the same number of squares it can go to for any starting square (14), it is quite easy to say that it must take on average 64 moves to return, since the stationary distribution is equal on all squares, therefore it must be 1/64. Cool video!

168 average hops for the knight in the corner. The rook, at least as I understand this, should have an equal distribution all over the board, which is 14 (according to the rules of chess as I understand them). The total number of rooks should be 896, 896/14 is 64. So I guess it should be 64 moves on average for the rook to get back to its starting position. It can be any starting position on the board.

Very comprehensive explanation, thanks! Great channel!

For the rook: on an empty board every square has the same set of moves(14). the total number for computing the probability is 14*64, or 896. 1/(14/896) = 896/14, or 64. On average it will take 64 moves. As a side note, this number feels like a measure of the mobility of a particular piece and location.

The rook is simple once you draw out there is (n - 1) vertical possibilities and (n - 1) horizontal, giving u 2 * (n - 1) possibilities or 2 * (8 - 1) = 14. So then A = { {0, 1/14, ... 1/14}, {1/14, 0, 1/14, ... 1/14}, ... {1/14, ... 0, 1/14}, {1/14, ... 1/14, 0} }. This gives u a steady state transition vector of so for every starting position, the average time to return is 64 moves by the theorem :). Survival Mode: Solve the average moves til return for every piece from every starting position and make heat maps for each piece.

To elaborate on the stationary distribution, it's not that easy to extend the logic of multiple knights on one space to the radio example. But it is easy to see that it holds. In the example where K-Pop is more popular, you can think of 6 radio channels that operate under that transition function. Four will be turned to K-Pop and two will be turned to Ska. When the song changes, 1/4 of all K-Pop channels (which is one) will switch to Ska while 1/2 of the Ska channels (one again) will switch to K-Pop. This distribution is thus stable and stationary. How to get to that distribution in the first place is generally not trivial, but pretty easy in this case. You can think of the distribution as thus: k portion of these radio channels will be turned to K-Pop. s portion of these radio channels will be turned to Ska. Obviously, s+k = 1 because the sum of all probabilities in the distribution must be 100%. Now, when the song changes, 1/4 of the time, a K-Pop station will switch to Ska, so that's k * 1/4. 1/2 of the time, a Ska station will stay with Ska, so s * 1/2. And here's the important bit: After this switcheroo, the distribution has to be the same, so: k * 1/4 + s * 1/2 = s k * 1/4 = s * 1/2 k/2 = s Now, putting that into s+k = 1 gives us: s + k = 1 k/2 + k = 1 3/2 * k = 1 k = 2/3 Additionally: s = k/2 s = (2/3)/2 s = 1/3 And this will work every time with any Markov chain. If you have more states, the system of equations becomes more complex, but it's the exact same idea.

Just watched every episode (so far)!!! New favorite channel!!!!!!!!

This is a nice video. So many more things need to be clear in my head before following this video without struggle that makes me wonder even more about the displayed topic. I am not a mathematician but this channel is so happy to watch... I was a fun since the very first video I saw. GREAT JOB!!!

Loved this video! Haven't seen any of the older ones... let the binge begin.

The knigth will take 168 steps, on average, since P(Being in the corner | Stationary distribution)=168. The rook's stationary distribution is an U(64) (uniform in 64 posibilities), since all the points in the board have 14 posibilities to choose from, so in a chess board with 14 rooks in every cell you will, on average, stay in equilibrium. So, since the probability is the same for every cell, it must be 1/64. Using the theorem P(Being in the corner | Stationary distribution)=64

Hello from Germany :D here are my answers: Knight on corner: 1/(2/336) = 168 Rook on corner: 1/(14/(14*8*8)) = 64

Knight in the corner would take on average 168 hops because the Stationary Distribution State is 1/168. The distribution on rooks are even. No matter where you are on the board, you can always move to 14 other places for a total of 896 spots. That gives each Stationary Distribution State a value of 14/896 or 1/64. It would therefore take on average 64 moves to return to its original spot. Just for fun I did the bishop. The distribution of a bishop depends on how far you are from the nearest edge. If you wanted to show the number of moves each position can make-it's 7 where you are on the edge, 9 where you are one space from an edge, 11 where you are two spaces from an edge, and 13 where you are three spaces from the edge. This gives you a total of 560. A bishop on the edge would have a stationary distribution state of 7/560 or 1/80. That means it would take on average 80 moves to return.

I really enjoy these videos. They go into much more depth than most maths videos online, and the challenges are well set (I might give the rook's Markov chain a go, actually. Intuitively, I'd expect the probability distribution to be constant across all squares - so, one sixty-fourth ?). Well done !

Knight: 168 moves (see video at 9:57) Rook: starting from any state a rook can go to 14 other states. Therefore we place 14 rooks on every tile. The stationary Distribution then is 14/(14rooks*64states) = 1/64. Giving us an Average time to return to any given starting tile of 64 moves.

Fantastic video! Very powerful information explained in an accessible way

For your questions: 1. Looking at the diagram from before, By computing 2/336, the stationary distribution at state (x) is 1/168. Therefore, it will take 168 moves on average. 2. A rook on any square will have fourteen possible squares, so each of the 64 squares will have the same stationary distribution value of 1/64. Therefore, the average number of moves for the rook is 64.

The rook should have totally uniform distribution due to the way it moves. Since the rook only moves EITHER horizontally or vertically, you can imagine you're altering one coordinate of it's 2D coordinate on the grid. For simplicity, we can argue that we write the coordinate as an ordered pair of integers from 0 to 7 as (horizontal, vertical) = (a,b), with (0,0) being the lower leftmost box. When moving the rook, you effectively change one coordinate, either a or b, to some different value in the range, while holding the others constant. If the rook moved 3 to the right, the new coordinate is (a+3,b), or if it moved two down, (a,b-2). Note that a vertical movement doesn't dependent on what the horizontal coordinate a is, so the motion is equivalent across all columns. Likewise, moving horizontally depends not on the b vertical coordinate, so all rows are equivalent. Well, combining those two fill the entire grid, so a rook on any in the grid has exactly the same number of squares it can move to if it were on any other square. The exact number is 14 possible moves on every square. Following the proof, the total number of moves is 14 move/box * 64 boxes = 896 total. So the stationary distribution at each box is 14/896 = 1/64. Now we take the reciprocal of the distribution, and its 64, and since each box is identical in this sense, on average it'll 64 random moves to bring the rook back where it started.

I think we can do this technique in a vastly more general setting than in the video (I mean for more kind of chess pieces and more chess table shapes). If the Markov chain looks as follows: For every arrow present, its reversed version is also present. From every state, the weight of every arrow that is going out of this state is the same. Then you can just do the same technique as in the video, you number the states by the arrows going out of there and you divide it by the total number of arrows (if we assume for the sake of simplicity that every state can be reached from every other state) and you get the expected time for the first return. Observe that this works for all kind of (chess?) pieces that have moving rules such that if they move somewhere, they can immediately return from that place. And this also works for all kinds of chessboards (even disconnected ones!) you only need that from any square, you can eventually get to any other square with your weird chess piece. This video is a nice buildup for a video about random walks on the integers, and random walks on the integers "on the plane" and "in 3d space" which might culminate in the sentence: A drunk person will eventually find his home, but a drunk bird might not. :-)

This channel is amazing! Have been binging on it all weekend. Having learnt all these topics back in college, this serves as a brilliant refresher. Love the fact that it goes into much more detailed explanation of the topics. I have seen it being applicable directly to my work. P.S. I think I have a crush on you too! So that helps in coming back for the videos. You are so smart

Love these videos! I feel like I've learned more from PBS Digital Studios than my university. Keep up the great work!

Assumption 1 hop per sec if knight starts from A1 it will take 168 sec if Rook starts from A1 then it will take 112 sec in the case of rook, we have 2 moves per corner so 4 set of 2 moves 3 moves along the edge so 24 set 3 moves and at rest square, it will have 4 moves hope it was helpful In case, If I am wrong please correct me Thank you

168 for the knight at the corner, 64 for the rook, also about 70 moves for the queen, 80 moves for the bishop, 84 moves for the king, and 0 moves for the pawn, as the pawn can only move forward.

Random hops on average each chess piece needs in order to return to their original spots (where initially placed at the beginning of a chess game) with an (otherwise) blank board: knight: 112 rook: 64 bishop: 80 king: 84 queen: 208/3 (69 and 1/3) pawn: pawns can't move backwards :( Also, the number of spaces a queen can move to from a given tile is the sum of the number of spaces a bishop and rook can make from the same tile. This make sense as the queen moves like a bishop and a rook!

Hmmm, the argument presented for coming up with the stationary distribution for the knight's chessboard suggests a much more general statement. Which is that, given a Markov Chain where each state has an equal chance to transition to its neighboring states, and the edges go both ways(as in, if X can transition to Y, then Y can transition to X) the stationary distribution at any state is its degree divided by the total number of edges in the graph. Neat.

Knight at corner 336/2*60 min Rook at corner(or at any other position) 14*64/14*60 min For any one of the two bishops 7*32/7*60 min as it can cover only the dark or light squares.

Markov chains have also been applied to American football. By using Markov chains to analyze the expected value for the number of points scored given 1) a position on the field, 2) the number of yards until a 1st down, and 3) a choice between punting, kicking a field goal, or going for the 1st down, researchers have concluded that football coaches punt far too often. One of many introductions to the subject: www.advancedfootballanalytics.com/index.php/home/research/game-strategy/120-4th-down-study

Can the knight cover every square on the board exactly once? If so, from how many starting positions?

really enjoy these videos, not sure whether I should be happy or sad that I got to focus more on the math than in the poison video :D

Knight in the corner: 168 moves, following the same logic as you did in the video. Rook: No matter where the rook is on the board, it has exactly 14 possible moves. Since this number is the same for every state, the stationary distribution becomes 1/64 for every state. This makes the average number of moves 64.

Knights rarely return to their original squares because they're eaten . But the reversal of direction is a characteristic that can aid strategy and tactics in that it is not eaten and thereby retains some of its value.

I love your delivery, you are the best host in all the PBS series (nothing to do with me being a mathematician hahahahah)

Another way to solve this would be through a system of equations. If we denote the average amount of moves it would take to return to B2 from B2 itself as u_b2, we would have u_b2 = 1 + (u_a3 + u_c3 + u_d2)/3. For A1 we would have u_a1 = 1 + (u_b3 + u_c2)/2 and so on. For squares that can reach B2, we don't get a u_b2 contribution. For A3 we get u_a3 = 1 + (u_b5 + u_c4 +u_c2)/4. We end up with a system of 64 equations and 64 unknowns. Then we can solve for u_b2 :)

Wow! You can explain complex notions in way to make them accessible to a general audience. I don't know if you take suggestions for future episodes; if you do, I'd invite you to consider an episode on Hopcroft-Karp algorithm. Great stuff anyway, keep up the good work!

If I understand things correctly, the knight from the corner should take an average of 168 moves to get back to the same corner. A rook can move to 14 other squares from any square on the board. If you put 14 rooks at every square that is 896 rooks, and each square has 1/64th of the rooks. The rook then starting at any square, has an average of 64 moves before it returns to the original square.

The answer for the first question for every square was already given to us @10:38, just remove the "1/". 168. The rook answer is nearly as simple but more interesting. Every square has 14 spots the rook can move to. So 14 * 64 squares divided by the 14 moves of the square in question. 14 * 64 / 14 = 64. It's interesting cause it will always equal the same number of squares the chessboard has, even if it's not a square board!

I did this for every piece to see which one would take the longest to get back to its starting point and what square it would start on. 1st place goes to the knight starting in the corner with 168 moves 2nd was the king starting in the corner with 127 moves 3rd was the knight starting on the four spaces adjacent to the corners with 112 moves also with 112 moves is a pawn on the top or bottom row(I let the pawn be able to move forward and backwards in the scenario but does not go side-to-side or diagonal)

Man I still sad this series ended like other RU-vid series it actually does a good job at explains it.

it would take 168 moves on average for a knight placed in the corner to return to the same position (1 divided by 1/168) it would take 64 moves on average for the rook to return to the initial position. Each step the rook can move to 14 different positions. Hence, each field has 14 assigned. This makes 8 * 8 * 14 = 896 in total leaving 14 / 896 = 1/64. Following the same procedure, the rook returns on average in 1 divided by 1/64 = 64 moves.

1) 168 - the probability is 2/336 = 1/168, so because of the theorem, the average is 168 steps 2) 64 - every place can get to 14 other places and 14 others can get to it, so the distribution is exactly even - 1/64, and 1/(1/64) = 64

The movements of the rooks are completely symmetric, in the sense that, in every given path, you could swap some rows and columns and it would still be a valid path. Thus, the stationary distribution has to be uniform across the board, so for it to sum 1 it's 1/64 on every cell: thus it takes an average of 64 moves

Can anyone explain the process for finding the Stationary Distributions of the second radio station example?

I'm studying this in university this semester!! With the great Giorgio Parisi! This is so cool: to see an application of what i studied in chess!!

Knight from corner: 168 hops. (1/(1/168)). Rook from any position: 64 hops. (All locations have 14 possible moves, 14*8^2 = 896, 14/896 [or 14/(14*8^2)] = 1/64, 1/(1/64) = 64).

The knights 8x8 middle configuration square actually marks the area of its competitive advantage as it has more versatility in those cells in the square . Knights on the rim tough to win . But having done some notation I've found some interesting repetitive numerical sequences in the files for the Knights square. For example 123 44 321 a sort of count up count down sequence . Mabe that doesn't relate to Markov chains but it does show a kind of equilibrium. Somewhat palindromic for the palladin.

Rook, queen and bishop: 64, but that is true regardless of the spot on the board. Though they each have a different number of spaces to which they can jump, that number is invariant across the board. The rook always has fourteen, the bishop always seven, and the queen always 21. Because of the invariant nature of the movement options, the variable ends up cancelling out. In fact, it can be proven fairly simply that if a piece has the ability to cross the entire board either diagonally, vertically, horizontally, or all three, on any square board with side lengths N, then the average number of hops a randomly hopping piece will take to return will always be N^2 pieces. We can generalize even further to demonstrate that a piece has the board crossing ability on a cubic chess board which has M dimensions with side length N, in any direction such that all tiles are connected by an edge or vertex, then the piece will return to it's starting position an average of N^M hops. While I can't show the math easily in my head, the proof is in the fact that with that board crossing ability, your piece can always move to N-1 spaces in all valid directions for that piece. So for example, the equivalent of a queen on a cubic chess board could move N-1 spaces in the vertical, horizontal, lateral, and on the 45 degree diagonal of all three axes, regardless of its position on the board

These videos are soo great! I wish they came out more often!

The other way to estimate this is to use a sampling method. i.e. you start from a state, count the number moves it took you to get back to start state. Perform this sampling till you reach ergodicity. With this can easily calculate the mean to get the expected number of steps.

Really cool video with a deep explanation, thanks!

I'd like to ask if the channel or anyone could provide subtitles for the videos cuz I think that there's a lot of people including myself that'd be very benefited from it given the fact that english is not my first language (and of many people neither) and, specifically here in Brazil, there's literally no other place to find stuff like this, excluding some books of hard access. I'd be very grateful if possible and thank you for this amazing series!!!

For each piece in their classic starting positions in chess, The rook would take 64 moves (average) to return to its original position. The knight takes, of course, 112 moves. Each bishop would take 43.14 moves The king would take 91.2 moves The queen would take 30.38 moves. I did not calculate the pawn. But if it is strictly a pawn, it will never return to its original position. If we allow it to change pieces once it reaches the back, it could return to its original position.

Excellent video! Thanks for all the hard work.

Love your explanation.... You are an angel... Great job and God Bless you

For the rook, no matter where on our chessboard our rook is, it can move to any one of 14 squares at random, with uniform probability density; therefore we can expect 64 rook moves no matter where our rook starts.

Concerning the Rook question... everyone's probably checking to see what happens when it has an equal chance of each space... I've decided to skew the odds and see what happens. A Rook on any given space is more likely to move to a closer space according to the rules below: - The closest space to the CurrentState has a 7! weight - The second closest space to the CurrentState has a 6! weight - The third closest space to the CurrentState has a 5! weight - etc. Calculate the weight of all possibilities from a single CurrentState to calculate the odds of any movement. If my calculations are correct, the average number of moves for a Rook starting on a corner to return to that same corner is 109.1013... . On the other hand, a Rook starting on a center square would take only 54.7451... moves. Of course... that's only IF my calculations are correct. I mostly wanted to pose such a puzzle.

The Rook takes an average of 64 steps. Each square has an equal amount of possible steps of 14. Giving each a state a stationary distribution of 14/(8*8*14).

What will be the average number of steps would take for a knight to return to its starting point in a 3-dimensional chess? or multi-dimensional? Does stationary distribution apply to higher dimensions?

I love this, I love your jumper, I love how you move your hands, I love how you explain stuff so seemingly casual, I want to be like you, but it seems so faaaar... :( like I'm 20... I have no idea what to do after my degree... (cries)

168 moves on average for the knight in the bottom left. The rook regardless of its position has a possibility of 14 moves. Therefore the stationary distribution for every spot is 1/64, so the average number of moves would be 64.

PBS, bring this channel back!

KML(9 years old): The answers are: Challenge 1: 168 average number of moves if the knight is in the corner. (336/2) Challenge 2: 64 average number of moves for a rook in any position (including the corner). The number of possible squares a rook can go to from any place is 14. Using your method, 14* 64 (the number of boxes) is 896. 896 divided by the number of squares you can go to in your current position (14) is 64.

If each step has a random, exponentially distributed time length (instead of all steps being counted as one time unit like in this video) the time taken to arrive to a certain state is given by something called a phase-type distribution. These distributions have some interesting properties, e.g. "The set of phase-type distributions is dense in the field of all positive-valued distributions, that is, it can be used to approximate any positive-valued distribution" (Wikipedia)

I'm a simple man. I hear a mathematician swear, I upvote.

Each place in the board has the exact number of options to go (14) so in total there is 14*64 pieces so the distribution for any position is 1/64. So the average moved from any position are 64

My suggestions are 50% math and 50% learning colors and shapes for toddlers. What is even happening?

Great channel. Hope to see more content you pick good subjects. I also hope more people get to know about it !

If the Knight were to start on one of the four corner squares, it would take an average of 168 hops to return. The Rook would take an average of 64 hops to return to it's original square. The Rook has 14 possible moves from any given square. Total number of rooks is 896. The reciprocal of 14/896 is 64.