Cards numbered 1 to 6 are placed face down on a table. The cards are shuffled and we each draw a card.

You look at your card and see it is a 2. I look at my card and ask, “Would you like to trade cards?”

The person with the higher card wins the game.

What is the correct response, assuming perfect logical reasoning by both players?

As usual, watch the video for a solution.

Can You Solve The 6 Cards Game?

Or keep reading.

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Answer To The 6 Cards Game

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Many people think you should trade cards, and you will have a great chance of getting a higher card.

The reasoning goes like this. Since you have a 2, the other person could have any of the 5 other cards. Only the card 1 is lower while the other 4 cards are higher. Thus it seems like a trade is a good idea, and there is a 4/5 = 80% chance you will get a higher card.

But this naive reasoning is incorrect! In this game, I look at my card before I ask to trade. So you might wonder: for which cards would a rational game theorist like me offer a trade?

Iterated eliminated of strictly dominated strategies

We cannot naively use probability to solve this problem. We have to think about game theory, which is the science of strategy and interdependent decision making.

We need to ask: after seeing a card, when would a person offer to make a trade? To analyze this we can think about good and bad strategies.

First a definition. A strategy is strictly dominated if some other strategy is better, regardless of what others do. In other words, it is a bad idea.

Once a strategy is eliminated, both players will never play it. Thus we can analyze the game as if the strategy was eliminated. We then have a reduced game and we can again remove strictly dominated strategies. We iterate in each reduced game until we are left with some strategies. Any of the remaining strategies will be reasonable.

This process is called the iterated elimination of strictly dominated strategies (IESDS) and is depicted below.

Let’s solve the game with IESDS.

Imagine a player gets the card 6. Would the player trade it? Absolutely not! Since 6 is the highest card it always wins the game and you should always keep it. Trading a 6 is a strictly dominated strategy and no rational player would ever do it.

We can thus eliminate the strategy trading a 6.

We now have a reduced game of trading or keeping the cards from 1 to 5.

Now imagine a player gets the card 5. Would the player ever trade it? The answer once again is no! The only better card is 6, but no one with a 6 would ever trade it. Thus the only people willing to trade must have cards lower than 5. Hence, it makes no sense to trade a 5.

Thus trading a 5 is a strictly dominated strategy in the reduced game and we can eliminate it.

We then iterate the reasoning to eliminate trading a 4, 3, and 2 in each of the reduced games.

The only remaining strategy is to trade with a card of 1. And this strategy makes sense because all other cards are higher.

Therefore, if you have a 2, and a rational opponent asks you to trade, should you do? By IESDS, you can conclude the other person must have a 1. There is a 0% chance the other card is higher, and your answer must be no!

This mathematical game demonstrates an important life lesson: you must always consider the information and intention of others. As they say, if an offer sounds too good to be true, it probably is!

Reference

Chess grandmaster Maurice Ashley TED-Ed talk
https://www.youtube.com/watch?v=v34NqCbAA1c

I really like your YouTube videos and I wanted to send you one of my favorite riddles! In my opinion it is somewhat hard but the solution is very cool!

Alice and Bob are playing a game using a chess board. Alice starts by placing a knight on the board. Then they take turns moving the knight to a new square (one it has not been on before). Standard chess rules apply: the knight can only move in an “L” shape, 2 squares in one direction and one square to the side.

The first player who cannot move the knight to a new square loses the game. Who wins if both players play optimally, and what is the winning strategy?

I hope you give this problem a try if you do not know it already! Please let me know your solution.

I was not able to solve the problem, and it has a really neat solution. Can you figure it out?

Watch the video for the answer.

Can You Solve The Knight On A Chessboard Riddle?

Or keep reading.

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Answer To The Knight On A Chessboard Riddle

(This was transcribed quickly after I made the video–please let me know if there are any typos/errors and I will correct them, thanks).

How do you even analyze this game? Like what constitutes a “good move” anyway? Alice wants to make the best move anticipating what Bob does, and Bob has to move based on what Alice has done and what Alice might do.

It seems like this problem is impossible to solve. However, this kind of cat and mouse chase is solvable and is an example of a game theory problem (see my list of the best game theory books). In this particular game, when viewed with the right lens, magically becomes as easy as coloring by number.

The answer is Bob will win the game because if Alice has a move, then Bob always has a legal move. Furthermore, the game must end in a maximum of 65 moves, since each move visits a new square and there are only 64 squares on the board. Thus, Alice will eventually not have a move and Bob will win the game.

The setup to prove the result

Sebastian provided the following clever proof that utilizes a colored graph.

First, divide the chessboard into 8 different 4×2 regions as follows.

Let’s focus on just one 4×2 region. If you put the knight in the lower left hand square, there is only one legal move for the knight in this same 4×2 region: the square 2 spots up and 1 spot right. And from that square, the knight could only go back to the lower left hand corner. From each of these squares, the knight only has 1 legal move to stay in the same 4×2 region. And that turns out to be the case for every square in this 4×2 region: from any square, the knight only has 1 legal move while staying in the same region. Let’s color two squares the same color if the knight can move between the squares legally. We get the following coloring:

Now this principle applies to all of the 4×2 regions: from any square in a 4×2 region, there is exactly 1 square the knight can move to.

Now we are ready to prove Bob has a winning strategy.

Proof Bob can always win

Alice starts the game by placing the knight in some 4×2 region. Bob then moves the knight to the only legal square in the same 4×2 region.

Now Alice has no legal move to stay in the same 4×2 region. So Alice has to move the knight to a different 4×2 region. At that point, Bob again makes the only legal move to another square in that 4×2 region.

On each turn Alice has to move to an unvisited square in a different 4×2 region. And if Alice has a legal move, Bob always has a legal move within the same 4×2 region.

Furthermore, the game must end in a maximum of 65 moves, since each move visits a new square and there are only 64 squares on the board. Thus, Alice will eventually not have a move and Bob will win the game.

It’s quite an elegant solution to a fairly challenging problem!

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Sources

Sebastian sent me the problem and solution. There is also a way to approach it using the “knight’s tour graph” but I found Sebastian’s solution is more direct and easy to understand. The following links also consider boards of other sizes besides the standard 8×8 chessboard.

https://math.stackexchange.com/questions/897092/who-has-a-winning-strategy-in-knight-and-why

https://puzzling.stackexchange.com/questions/23982/the-knights-game

The major reasons are:

–I need more time to make math videos.

–Over the last 10 years game theory has become part of mainstream news coverage, decreasing the uniqueness and need for this column.

I still want to cover game theory, but for reasons explained below, I need another strategy besides the weekly column.

Longer explanation

In the last 10 years I have posted over 500 articles on my weekly column game theory Tuesdays. I have also made several game theory videos on YouTube.

I have received many positive comments and emails from students, professors, and many others who love the game theory column, which also became the basis for my book The Joy of Game Theory. I want to thank you for supporting the site all of these years.

I have also happily shared game theory content from other amazing people discussing game theory. For starters, I have compiled an annotated bibliography of the best game theory books which has helped many people learn about game theory. I have also shared many resources on the web.

When I started the column in 2007, no one was blogging about game theory regularly, and there were only a handful of videos on game theory on YouTube. Today, things like the prisoner’s dilemma and Nash equilibrium are commonplace in mainstream news articles and YouTube videos. There are even excellent free video courses such as William Spaniel’s Game Theory 101 (now a University of Pitt professor), Game Theory with Ben Polak (Yale professor), or Coursera’s game theory course (two professors from Stanford + one from University of British Columbia).

It’s great progress that most people have heard of the prisoner’s dilemma and other game theory concepts.

But unfortunately there are still big misconceptions about game theory. For example, a video on SciShow, which now has over 700,000 views, defines the Nash equilibrium incorrectly! The video instead used the definition for a “dominant strategy.” How is this possible?! SciShow is one of the most popular YouTube education channels with ample funding and access to academics. And they still got it wrong! All it would have taken is reading Wikipedia or sending an email to me to get the definition correct.

Sadly my post to explain the error did not receive enough attention. But I did leave a comment on the video. I never heard back from SciShow about the mistake. But I think they got the message: the video now does have an annotation to indicate the error. You can see it at 3:30 in the video SciSchow game theory annotation at 3:30 (you can only see annotations on desktop, so you cannot see it off your phone’s app).

The incident made me realize: something has to change. There is an audience that wants to learn about game theory, but I need a better way to reach the audience.

I have over 400 saved ideas for game theory columns in my drafts, so I could easily write posts for the next 10 years without any trouble. But game theory has taught me to evolve with the times. As I wrote in an early post on this blog, if you aren’t winning the game you are playing, consider changing the game.

I am thinking about the best way to reach people, whether it be by making videos about game theory, creating a game theory course, writing another book about game theory, or re-launching the column in another way.

I’m putting a the weekly column on pause until I can figure out a more effective strategy.

In the meantime I might post about game theory if I see a newsworthy topic, watch a good YouTube video, or read a game theory book worth sharing.

I’m open to suggestions and happy to hear feedback if you have any ideas (leave a comment or email me: presh@mindyourdecisions.com). I have gotten many requests to do a podcast, but I currently do not have the time to be able to make a good podcast.

Let me know! Thank you for supporting Game Theory Tuesdays for a decade!