(candidates from most favorite to least favorable)

18 voters: A, D, E, C, B

12 voters: B, E, D, C, A

10 voters: C, B, E, D, A

9 voters: D, C, E, B, A

4 voters: E, B, D, C, A

2 voters: E, C, D, B, A

Who should win this election? PBS Infinite Series explains why the mathematical answer depends on the method of counting the votes. It turns out the winner is dependent on the kind of voting system. Amazingly, each of the common voting systems in practice would result in a different winner! Watch the video for an explanation.

**Voting Systems and the Condorcet Paradox**

If you cannot watch the video, I give a detailed explanation in the rest of this post.

**Different voting systems**

We will consider 4 of the common ways that elections are determined to see which candidate would win.

**1. Plurality – most first place votes (America’s presidential election)**

The candidate with the most first-place votes is the winner. Candidate A has the most first-place votes (18). Notice candidate A only got about 32.7% of the votes, meaning that a majority of voters did not vote for candidate A. In a plurality system, the winner does not have to earn a majority of the votes, only more than any other candidate.

18: A, D, E, C, B –> candidate A plurality winner

12: B, E, D, C, A

10: C, B, E, D, A

9: D, C, E, B, A

4: E, B, D, C, A

2: E, C, D, B, A

**2. Two-round runoff – a majority wins, or top two face off (presidential election in many countries)**

The candidate with the most first place votes wins if the candidate gets over 50% of the votes. Otherwise, another election is held between the two candidates with the most first-place votes. With these preferences and honest voting, candidates A and B would get the most first place votes.

18: A, D, E, C, B –> second round

12: B, E, D, C, A –> second round

10: C, B, E, D, A

9: D, C, E, B, A

4: E, B, D, C, A

2: E, C, D, B, A

In practice, the second election is held with the top two candidates. In this example, we can guess what would happen if people voted honestly according to the rankings. Suppose that whenever someone ranks A above B, that person would cast a vote for A; otherwise the person would cast a vote for B.

It happens that all the votes would shift to candidate B.

18: A, D, E, C, B –> second round candidate A

12: B, E, D, C, A –> second round candidate B

10: C, B, E, D, A –> votes go to B

9: D, C, E, B, A –> votes go to B

4: E, B, D, C, A –> votes go to B

2: E, C, D, B, A –> votes go to B

With the above preferences, this would mean candidate B is preferred to A by 37 voters, so candidate B would win the election with 37 votes.

18: A

37: B –> second round winner

**3. Instant Runoff – a majority wins, or eliminate the last place until there is a majority (Academy Awards Best Picture)**

This is a more complicated version of the two-round runoff election in which each step only eliminates the last place candidate, and the further rounds take place from calculations of the ranked ballots.

The candidate with the most first place votes wins if the candidate gets over 50% of the votes. Otherwise, the candidate in last place gets eliminated and a new election is held with the reduced set of candidates. The process is repeated until some candidate gets a majority of votes.

It can get complicated, so let’s work through the example.

In the above election, candidate E got the fewest first place votes. Those votes are then automatically transferred based on those ballot’s second place choices. Out of these, 4 have B and 2 have C as their second choices.

18: A, D, E, C, B

12: B, E, D, C, A

10: C, B, E, D, A –> candidate C gets 2 more votes

9: D, C, E, B, A

4: E, B, D, C, A –> E eliminated, votes go to B

2: E, C, D, B, A –> E eliminated, votes go to C

18: A, D, E, C, B

12: B, E, D, C, A

10: C, B, E, D, A

9: D, C, E, B, A

4: B

2: C

Now candidate D has the least number of votes, so those 9 votes then go to the second preference C.

18: A, D, E, C, B

12: B, E, D, C, A

10: C, B, E, D, A

9: D, C, E, B, A –> D eliminated, votes go to C

4: B

2: C

18: A, D, E, C, B

12: B, E, D, C, A

10: C, B, E, D, A

4: B

11: C

Now candidate B has the least votes, and those 16 votes would go to C.

18: A, D, E, C, B

12: B, E, D, C, A –> B eliminated, votes go to C

10: C, B, E, D, A

4: B –> B eliminated, votes go to C

11: C

18: A, D, E, C, B

10: C, B, E, D, A

27: C

Candidate C would get a total of 37 first places votes now, and that is a majority of the votes. Thus candidate C wins this election.

**4. Borda Count – score based on ranked votes (NBA voting for MVP)**

Each candidate gets a score based on how many times the candidate was ranked 1st, 2nd, 3rd, and so on on a ballot. A score system gives more points to being ranked higher, like giving 4 points to being ranked 1st on a ballot, 3 points to 2nd, 2 points to 3rd, 1 point to 4th, and 0 points to 5th.

Other scoring systems are possible too, such as 50-25-10-5-0 points for being ranked 1-2-3-4-5.

Imagine the election used the 4-3-2-1-0 point system.

18: A, D, E, C, B

12: B, E, D, C, A

10: C, B, E, D, A

9: D, C, E, B, A

4: E, B, D, C, A

2: E, C, D, B, A

We will count the number of votes each candidate got for each ranking and then find a total score. For example, candidate A earned 18 votes as 1st ranked and 37 votes as 5th ranked. Candidate A’s score is:

Candidate A: 18(4) + 37(0) = 72

We can similarly calculate the scores of the other candidates.

Candidate B: 12(4) + 14(3) + 11(1) + 18(0) = 101

Candidate C: 10(4) + 11(3) + 34(1) = 107

Candidate D: 9(4) + 18(3) + 18(2) + 10(1) = 136

Candidate E: 6(4) + 12(3) + 37(2) = 134

Candidate D has the highest score of 136, meaning D would win the election.

**5. Head to head competition (Condorcet winner)**

Is there a candidate that could beat every opponent, if the election were held with just two candidates? Such a candidate, if there is one, is called a Condorcet winner. Here are the preferences again.

18: A, D, E, C, B

12: B, E, D, C, A

10: C, B, E, D, A

9: D, C, E, B, A

4: E, B, D, C, A

2: E, C, D, B, A

Amazingly it turns out that candidate E would win every head to head election:

E vs A: 37 to 18

E vs B: 33 to 22

E vs C: 36 to 19

E vs D: 28 to 27

By this criterion, candidate E should be the winner. But by any other system, candidate E would not stand a chance!

**5 different winners!**

Amazingly, each election method produced a different winner!

Now, there are some caveats to the above discussion. These ranked lists were specifically concocted to demonstrate that each voting system would produce a different winner. It’s amazing that someone devised these preferences. But people’s voting preferences in real life might not be so varied to result in this paradoxical result.

There is also an important game theory consideration. The ranked lists were taken to be honest rankings of the voters. Then we assumed that voters would vote honestly according to the rankings. This most definitely does not happen! Many people vote strategically given the rules of a system.

In a plurality system like America’s presidential election, a voter that prefers an unpopular third-party would instead vote for one of the two candidates backed by the two major parties. Some voters will remain loyal to their party to avoid electing the other party’s candidate, so as to vote for the “lesser of two evils.”

In runoff elections, similarly, voters might strategically vote for a less preferred candidate in the first round, knowing their own top preference will be eliminated anyway.

In a Borda count, a voter might also artifically downvote or omit a candidate to help the voter’s preference get more relative points. Or a voter might omit a vote to make a statement (this is how LeBron James or Michael Jordan never won the MVP with all first-place votes unanimously).

And as for the Condorcet winner, this is a thought experiment and not something done in elections, because many times the Condorcet winner does not exist–there is no candidate that would beat all others in a head to head contest.

Elections are about trying to determine which candidate is best. But occasionally, the winner might reflect more about the voting system than about the voter’s preferences.

**Sources**

Voting Systems and the Condorcet Paradox | Infinite Series

https://www.youtube.com/watch?v=HoAnYQZrNrQ

AMS feature column on voting and elections

http://www.ams.org/samplings/feature-column/fcarc-voting-decision

Alice and Bob are on a game show. Each is secretly told a whole, positive number. They are told the two numbers are consecutive, but neither knows the other person’s number. For example, if Alice is told 20, she does not know if Bob was told 19 or 21. And if Bob is told 21, he does not know if Alice was told 20 or 22. The point of the game is to guess the other person’s number. The game works as follows.

–Alice and Bob cannot communicate with each other, and they are not allowed to plan their strategy either.

–The two are in a room where a clock rings every minute.

–After the clock rings, either player can call out a guess of the other player’s number, or they can stay silent.

–The game continues until Alice or Bob makes a guess. After the first guess is made, the game ends.

–Alice and Bob win $1 million each if the guess is correct, and they lose and get nothing if the guess is incorrect.

How should Alice and Bob play this game to have the best chance of winning? Each knows the other person is perfect at logical reasoning.

Watch the video for a solution.

**The Seemingly Impossible Guess The Number Logic Puzzle**

Or keep reading.

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**Solution To Seemingly Impossible Guess The Number Logic Puzzle**

At first it seems like Alice and Bob can do no better than random chance. If Alice is told 20, for instance, there is no way to know if Bob has 19 or 21. But since Alice can limit Bob’s number to 2 possibilities, she can at least have a 50% chance of guessing correctly.

Bob has the same issue. If he is told a number *N*, then he cannot be sure if Alice was told *N* – 1 or *N* + 1. If Bob guesses between the 2 possibilities, then he also has a 50% chance of guessing correctly.

It would seem Alice and Bob are stuck. Neither person has can do better than random chance, so regardless of who guesses, it would seem they are limited to a 50% chance of winning.

But remarkably they can do much better than random chance! They can actually increase their odds of winning to 100%. That is, they can win the game for sure! The trick is that they can use logic and the ringing clock to coordinate which player guesses.

**The strategy**

The answer lies in the subtle rule that the clock rings every minute. The clock essentially serves as a signal between Alice and Bob that allows each person to reason inductively.

One key detail is the two are given positive consecutive numbers. When Alice gets a number *N*, she usually has to consider Bob has *N* – 1 or *N* + 1. But this is not always true. Suppose that Alice gets the number 1. She would have to consider that Bob got 0 or 2. But since 0 is not positive, she knows that Bob must have gotten 2.

So if Alice gets 1, then she would know Bob has 2 for sure, and she would answer on the first ring of the clock. Similarly, if Bob got the number 1, he would know Alice must have 2, and he would answer after the first ring of the clock.

Now consider instead that Alice was given 2 and Bob was given 3. Alice would be wondering if Bob has 1 or 3. But Alice would think, “If Bob has 1, he surely will answer after the first ring of the clock. Therefore, if the clock rings and he does not answer, he must surely have 3 instead.” So the clock will ring once, and then after it rings a second time Alice will answer and guess Bob has 3. (If instead Bob was given 2 and Alice was given 3, then Bob would answer after the second ring and guess Alice has 3.)

This reasoning can be extended inductively. If Alice and Bob are assigned *N* and *N* + 1, then the player with the lower number will answer in exactly *N* rings of the clock and correctly answer the other person has *N* + 1.

They win every single time!

**The connection with common knowledge**

The puzzle illustrates the game theory concept of *common knowledge* which is distinguished from *mutual knowledge*.

An event is *mutual knowledge* if each player knows the event. An event is instead *common knowledge* if all players know the event, all players know that all players know it, and so on *ad infinitum*.

Here is how the two concepts work in the game. When Alice is given the number 20 (and Bob could have 19 or 21), it is mutual knowledge that neither player has the number 1, neither player has the number 2, or so on, up to neither player has the number 18. But that deduction is not good enough to solve the game.

That is where the clock ringing provides a helping hand. When the clock rings the first time, and no one answers, the fact that neither player has 1 transforms from mutual knowledge into common knowledge. This seems like a trivial distinction, but it is an important one that allows for the building up of logical deductions. Each time the clock rings, the set of excluded numbers becomes common knowledge to both players, which eventually allows the players to win for sure.

**Source of puzzle**

Impossible?: Surprising Solutions to Counterintuitive Conundrums by Julian Havil. The puzzle is called “consecutive integers.”

Link to Amazon: http://amzn.to/2rLqjNu

(I may get a small percentage of a sale)

Now consider a variation: player one wins if either Xs or Os make 3 in a row, and player two wins only if the board ends in a draw. Now who wins the game?

This seems like a fair game, but it is far from it! ScamSchool presented why in a video, so check it out.

Or keep reading for my explanation.

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**Answer To Who Wins Tic-Tac-Toe, If The Second Player Wins Only For Ties**

It seems like a fair game, but actually the first player can force a win by the third move. The problem is the second player has to block both Xs and Os from forming a 3 in a row, and the first player can force a “double attack” in just three moves.

Let’s work through the cases to see why. Suppose player one starts in the center square.

| |

————-

| X |

————-

| |

Player two can either play in a corner or an edge. Either way, player one can force a win in just a few moves. Let’s go through the cases.

**(a) Player two plays a corner**

Without loss of generality, we can make it the top left corner.

O| |

————-

| X |

————-

| |

Now player one can play in either adjacent corner to where player two played (not the opposite corner). In this case player one plays in the third row left corner.

O| |

————-

| X |

————-

X| |

Now player two has to block in the first row.

O| | O

————-

| X |

————-

X| |

Now player one can play in the third row middle column, creating a double attack.

O| | O

————-

| X |

————-

X| X |

Player two is stuck. Either player two blocks the three in a row of Xs, thereby making three Os in a row. Or player two moves elsewhere, allowing player one to make three Xs in a row on the next turn.

**(b) Player two plays an edge**

Without loss of generality, we can make it the top center edge.

| O |

————-

| X |

————-

| |

Now player one plays in an adjacent corner to player two’s move. In this case, the upper right corner is taken:

X| O |

————-

| X |

————-

| |

Player two has to block.

X| O |

————-

| X |

————-

| | O

And this allows player one to create a double attack by playing in the edge in adjacent to Xs in the corner and the center.

X| O |

————-

X| X |

————-

| | O

Just like in regular tic-tac-toe, player two cannot block both attacks and loses.

**Player one can also win by playing in a corner square**

Player one could play in the center, and it’s an easy win. But there are other easy ways to win too, like if player one starts out in a corner.

X| |

————-

| |

————-

| |

Now there are more cases to consider because player two can either play in (a) the center, (b) an adjacent corner, (c) an adjacent edge, (d) the opposite corner, or (e) an opposite edge. Let’s work each case.

In each case, I’ll demonstrate a strategy in which player one can force a win. (There might be other methods, but these are the paths I saw)

**(a) Player two moves in the center**

This is player two’s move.

X| |

————-

| O |

————-

| |

Now player one can pick an adjacent edge.

X| X |

————-

| O |

————-

| |

Player two has to block the corner where player one is threatening a three in a row.

X| X | O

————-

| O |

————-

| |

And now player one can play the other adjacent edge.

X| X | O

————-

X| O |

————-

| |

Now player two is stuck. If player two blocks with an O in the corner, then player two makes 3 in a row and loses. Otherwise player one will make three Xs in a row on the next turn. So this is a loss for player two.

**(b) Player two moves in an adjacent corner**

This is player two’s move.

X| | O

————-

| |

————-

| |

Now player one can play in the center.

X| | O

————-

| X |

————-

| |

Player two has to block in the corner.

X| | O

————-

| X |

————-

| | O

And now player one can create a double attack by playing in the left edge.

X| | O

————-

X| X |

————-

| | O

Once again, player two is stuck. Unlike regular tic-tac-toe, if player two blocks the row with an O, then player two makes 3 in a row and loses. So player two cannot win on this turn, nor can player two block the middle row and first column attacks simultaneously.

**(c) Player two moves in an adjacent edge**

This is player two’s move.

X| O |

————-

| |

————-

| |

Now player one plays in the center.

X| O |

————-

| X |

————-

| |

Now player two has block in the corner.

Then player one picks the other adjacent edge.

X| O |

————-

X| X |

————-

| | O

This is a double attack of Xs in the first column and the middle row and player 2 cannot block both.

**(d) Player two moves in the opposite corner**

This is player two’s move.

X | |

————-

| |

————-

| | O

Now player one takes an adjacent edge.

X| |

————-

X| |

————-

| | O

This forces player two to block in the corner.

X| |

————-

X| |

————-

O | | O

Now player one plays the other adjacent edge.

X| X |

————-

X| |

————-

O| | O

Player two has to block the first row.

X| X | O

————-

X| |

————-

O| | O

And now player one creates a double attack by taking the center.

X| X | O

————-

X| X |

————-

O| | O

If player two blocks either spot, then player two makes three Os in a row and loses.

**(e) Player two moves in an opposite edge**

This is player two’s move.

X| |

————-

| |

————-

| O |

Player one takes the center.

X| |

————-

| X |

————-

| O |

Now player two blocks the corner.

X| |

————-

| X |

————-

| O | O

Player one creates a double attack in the corner.

X| |

————-

| X |

————-

X| O | O

This is a loss for player two just like regular tic-tac-toe.

**Player one can also win for an edge**

It turns out player one can also win for edges, meaning player one wins no matter what the first move is.

Suppose the first move is an edge.

| |

————-

X| |

————-

| |

There are 5 possible responses from player two: (a) the center, (b) an adjacent corner, (c) an adjacent edge, (d) an opposite corner, and (e) the opposite edge.

The analysis is similar to how player one can win from playing in a corner. For completeness, I’ll go through each case.

**(a) Player two takes the center**

Here is the gameboard after one move.

| |

————-

X| O |

————-

| |

Player one can take an adjacent corner.

X| |

————-

X| O |

————-

| |

Now player two has to block.

X| |

————-

X| O |

————-

O| |

Now player one takes the edge adjacent to the corner.

X| X |

————-

X| O |

————-

O| |

This is a double attack in this game: either player two blocks, making three Os in a row which loses, or player two does not block and allows for three Xs in a row.

**(b) Player two takes an adjacent corner**

Here is the gameboard after one move.

O| |

————-

X| |

————-

| |

Now player one takes the opposite edge.

O| |

————-

X| | X

————-

| |

This forces player two to block in the center.

O| |

————-

X| O | X

————-

| |

Now player one can make a double attack in the opposite corner as illustrated:

O| | X

————-

X| O | X

————-

| |

Player two either has to block it (and lose with three Os), or player two does not block and allows player one to make three Xs.

**(c) Player two takes an adjacent edge**

Here is the gameboard after one move.

| O |

————-

X| |

————-

| |

Player one takes the center.

| O |

————-

X| X |

————-

| |

Player two has to block.

| O |

————-

X| X | O

————-

| |

Now player one makes a double attack by playing in the corner.

X| O |

————-

X| X |

————-

| |

Player two cannot block both attacks on the next turn, so player one can win.

**(d) Player two takes an opposite corner**

Here is the gameboard after one move.

| | O

————-

X| |

————-

| |

Player one takes the corner opposite player two’s move.

| | O

————-

X| |

————-

X| |

Player two has to block.

O| | O

————-

X| |

————-

X| |

Now player one plays takes the center.

O| | O

————-

X| X |

————-

X| |

Player two again has to block.

O| | O

————-

X| X | O

————-

X| |

Now player one makes a double attack in the last row.

O| | O

————-

X| X | O

————-

X| X |

If player two blocks either spot, player two makes three Os in a row and loses.

**(e) Player two takes the opposite edge**

Here is the gameboard after one move.

| |

————-

X| | O

————-

| |

Player one can take an adjacent corner.

X| |

————-

X| | O

————-

| |

This forces player two to block.

X| |

————-

X| | O

————-

O| |

Now player one can take the edge adjacent to the corner X.

X| X |

————-

X| | O

————-

O| |

This again forces player two to block.

X| X | O

————-

X| | O

————-

O| |

Now player one creates a double attack.

X| X | O

————-

X| | O

————-

O| X |

Player two is stuck: either spot makes for three Os in a row.

**Conclusion**

This variation of tic-tac-toe seems fair, but player one can actually force a win from any of the initial moves!

]]>Can you figure it out? Watch the video for a solution.

**Can You Solve The 6 Rectangles Puzzle?**

Or keep reading.

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**Answer To 6 Rectangles Puzzle**

At first, it seems like the problem does not have enough information because of how the rectangles are overlapping. But in fact you can solve for the perimeter!

The answer is 666, which is equal to the perimeter of 3 rectangles.

There are a couple of ways to derive the answer.

**Method 1: overcount and subtract**

One way is to start with the perimeter of 6 rectangles and then subtract out the overlapping segments.

When two rectangles overlap, we have to subtract two times the overlapping segment. The overlapping segments are:

Overlapping lengths

Top rectangle overlaps 1 length –> subtract 2 lengths

Bottom left rectangle overlaps 1 length –> subtract 2 lengths

Bottom right rectangle overlaps 1 length –> subtract 2 lengths

Overlapping heights

Middle left rectangle overlaps 1 height –> subtract 2 heights

Middle right rectangle overlaps 2 height –> subtract 4 heights

Bottom two rectangles overlap 1 height –> subtract 2 heights

We need to subtract a total of 6 lengths and 6 heights. As each rectangle consists of 2 lengths and 2 heights, this is equivalent to subtracting the perimeter of 3 rectangles.

Thus we subtract the perimeter of 3 rectangles from the 6 rectangles and we are left with the perimeter of 3 rectangles.

The shape has a perimeter of 3 rectangles, which equals 3(222) = 666.

**Method 2: re-arranging the rectangles**

Notice that if the top rectangle “slides” to the left in the diagram the perimeter of the shape does not change. We can also slide the two bottom rectangles.

In this shape, we can easily count there are 6 lengths and 6 heights, so the perimeter is equivalent to 3 rectangles. The perimeter of this shape is 3(222) = 666, and that is equal to the perimeter of the original shape.

**Source**

St. Marguerite Twitter

https://twitter.com/smddurham/status/859579294108053504

It’s double overtime in the NBA finals. There’s one second left on the clock, your team is down by 1, you have possession under your own basket with no time-outs. You have to throw a full-court pass just to attempt a reasonable shot. What is your best play?

That was the situation the Phoenix Suns faced in the 1976 NBA Finals against the Boston Celtics. In what has been called the greatest basketball game, a player on the Phoenix Suns came up with arguably the smartest strategic decision, worthy of a game theory case study.

The analysis is after the jump. You can watch a video of the last two possessions of the game. The part with 1 second starts at 5:50.

**1976 NBA Finals Double Overtime**

**Considering your options**

Typically sports teams do not have time to deliberate during a game. In this case something unusual happened.

After the previous basket, Boston fans thought the game was over and they stormed the court in celebration. The referees, however, decided the game was not over and put 1 second back on the clock. It took several minutes to clear fans from the court, allowing the Phoenix Suns time to strategize.

The Suns had little chance to win. While full-court buzzer beaters do happen, they are essentially “miracle” plays that coincide with a defensive breakdown of the opposing team. The coach might have been drawing up a play to get off a 60 or 80 foot jumper.

That is when a player on the Suns, Paul Westphal, came up with a better idea.

**Strategic technical foul**

If the Suns had played out the final possession, they almost certainly would have missed the shot and lost the game. Westphal suggested Suns should deliberately call a time-out they did not have to incur a technical foul!

The penalty gave Boston a free throw, which they made to extend the lead to 2. But being down by 1 or 2 points did not really matter–the Suns had to make a 2-point shot to avoid losing (there were no 3-pointers in 1976).

What the intentional technical foul did is it allowed the Suns to inbound at midcourt. It is much, much easier to draw up a play from midcourt with one second than to try and heave a 60 or 80 foot shot across the entire court. And sure enough, the Suns inbounded to Garfield Heard who sank an 18-foot basket to tie the game, forcing the game into a third overtime.

(Now if you watch the video closely, there is something fishy about the timekeeping. Garfield Heard grabs the ball facing away from the basket, turns around to face the basket, and then gets a shot off. It seems unlikely anyone can do all of that in 1 second.)

**Game theory lessons**

The Suns’ decision illustrates several concepts of strategic thinking.

First, winning moves are not always obvious. The strategy to use a timeout deliberately was available for any team. But coaches rarely have time to think outside the box. They have to focus on major decisions like choosing the starting lineup, managing foul trouble, and using time-outs effectively. In fact, if the Suns had an extra timeout, they could have just used that to advance the ball to midcourt. It was only in desperation, with the fortuitous event of fans storming the court, that gave them extra time to brainstorm.

Second, exploiting loopholes is a short-term strategy. The next year the NBA changed the rule specifically to avoid a repeat: teams could not advance the ball to midcourt with a technical foul. Since flagrantly exploiting loopholes often leads to rule changes, you should be sure to utilize loopholes only in dire circumstances. The Suns were smart to use the strategy during a Finals series in a situation where it really mattered.

Although the Suns ultimately lost the game and the series, they deserve credit for thinking strategically and making perhaps the smartest move ever in an NBA Finals. They did not get a championship ring for their smart thinking, but they deserve to be recognized as a legendary example of game theory.

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