John Nash
(With a Discussion of Game Theory and Its Applications Along with Mention of Truman Capote and Holcomb, Kansas; George Orwell, Bugsy Siegel and Virginia Hill, J. P. Morgan and John D. Rockefeller, and even William Shakespeare)
- "The correspondence of each n-tuple with its set of countering n-tuples gives a one-to-many mapping of the product space into itself. From the definition of countering we see that these the set of countering points of a point is convex. By using the continuity of the pay-off functions we see that the graph of the mapping is closed. The closedness is equivalent to saying: if P1,P2, ... and Q1,Q2, .... Qn, ... are sequences of points in the product space where Qn→Q, Pn→P and Qn counters Pn then Q counters P. Since the graph is closed and since the image of each point under the mapping is convex, we infer from Kakutani's theorem that the mapping has a fixed point (i.e., point contained in its image). Hence there is an equilibrium point."
Big John
- John Nash
- "This is trivial, you know."
- John von Neumann
Well, we won't get into the argument between the two Gigantic Gians of Game Theory. We'll just consider the following game.
Two players each have a coin. Each can decide to display the "Heads" or "Tails"1. Depending on what they show, the "payout" is:
Footnote
A point of clarity is that Players 1 and 2 are not flipping the coin where they couldn't control the fall. Instead, the players decide which side of the coin to show the other player but their choice is kept secret until the moment of play.
| Player 1 \ Player 2 | Heads | Tails |
| Heads | -1¢ , 2¢ | ½¢ , 1¢ |
| Tails | 0¢ , ½¢ | 1¢ , -1¢ |
A little explanation is in order. A negative number like -1¢ means the player has to pay that amount. A postive number means the player wins the sum.
So if Player 1 decides to show <Heads> and Player 2 also shows <Heads>, then Player 1 looses a penny. Player 2 wins 2¢.
But if Player 1 shows <Tails> and Player 2 shows <Tails>, then Player 1 receives 1¢ and Player 2 looses 1¢.
If they don't match?
Then if Player 1 shows <Heads> and Player 2 shows <Tails>, Player 1 wins half a penny and Player 2 wins 1¢. If Player 1 shows <Tails> and Player 2 shows <Heads> Player 1 breaks even and Player 2 win the ha'penny.
OK. How should they play the game?
Playing the Game
The - quote - "layman" - unquote - would say the best way is for Player 1 to play <Heads> and Player 2 to play <Tails>. This gives the payout of ½¢ for Player 1 and 1¢ for Player 2. With this play, everyone comes out ahead although Player 1 gets a bit short changed.
Game theorists, though, scoff at such naïveté. With a superior smile, they point out that the strategy we chose - succinctly written down as <Heads , Tails> - is not the best solution. That's because it is not a rational strategy.
Instead, the proper solution is to play <Tails , Heads> with it's payout of <0¢ , ½¢>.
Ha? (To quote Shakespeare.) But with that strategy Player 1 wins nothing and Player 2 only gets ½¢! How can that be better than <Heads , Tails> which pays out <½¢ , 1¢>?
But most of all, how can a play that gives a lower payout for both players be considered rational? We'd really like to know that.
I thought you would as Captain Mephisto said to Sidney Brand. It's very simple really.
First, let's grey out Player 2's columns and twiddle with Player 1's colors. This will make things clearer.
| Player 1 \ Player 2 | Heads | Tails |
| Heads | -1¢ , 2¢ | ½¢ , 1¢ |
| Tails | 0¢ , ½¢ | 1¢ , -1¢ |
You see immediately that if Player 2 plays <Heads>, then Player 1 should play <Tails>. Although he wins no money with <Tails>, he actually loses money if he plays <Heads>.
And if Player 2 had played <Tails>, Player 1 should still play <Tails>. There Player 1 wins a penny. But if he played <Heads>, he only wins half a penny.
So in summary, since ...
0¢ (Tails) > -1¢ (Heads)
... and ...
1¢ (Tails) > ½ (Heads)
... by playing <Tails>, Player 1 gets the better payout regardless of how Player 2 plays.
So Player 1 should always play TAILS!
And Player 2?
Again greying out parts of the table and tweaking Player 2's colors we see:
| Player 1 \ Player 2 | Heads | Tails |
| Heads | -1¢ , 2¢ | ½¢ , 1¢ |
| Tails | 0¢ , ½¢ | 1¢ , -1¢ |
And the preferred payouts are:
If Player 1 plays <Heads> then we see that for Player 2:
2¢ (Heads) > 1¢ (Tails)
... and so Player 2 should play <Heads>.
And if Player 1 plays <Tails>?
Well, then for Player 2:
½¢ (Heads) > -1¢ (Tails).
... and Player 2 should still play <Heads>.
Clearly if Player 2 sticks to <Heads>, then she2 wins the larger amount no matter what Player 1 choses.
Footnote
In game theory discussions, it's common to designate one player as "he" and the other as "she" to facilitate use of pronouns.
So the most rational strategy is:
Player 1 always plays <Tails>.
Player 2 always plays <Heads>.
But dang it! It still doesn't make sense! Why the hey is <Tails , Heads> considered rational, when both players get more if they had played <Heads , Tails>?
We will continyeh.
Why is the rational play so irrational?
Well, as an ear-pulling president once said, let us continyeh.
Toujours Jouer le Dominant
Of course, the problem is we've been bandying the word "rational" about and not asking what we mean. So naturally we should first dip into the dictionary. There we read:
rational: having reason.
And so we look up reason:
reason: a rational ground or motive.
Not much help. So we wonder. Are we even asking the right question?
Instead of going to Noah Webster, perhaps we should ask just what do mathematicians mean when they say something is rational.
And yes, one characteristic of mathematical disciplines is they use words from ordinary conversation but they change the meaning. That is, they invent their own definition that fits in better with the math.
He warned us.
People inventing private definitions is something that George Orwell warned us about. George even said that such definitions are often used in a consciously dishonest manner. That is, the speakers have a particular meaning but want the listeners to think they mean something else. Although we make no such charge toward mathematicians who are all, all honorable men3 and women, game theorists do tweak the definition a bit when talking about a "rational" strategy.
Footnote
"The Tragedy of Julius Caesar", Act III, Scene II, Mr. William Shakespeare's Comedies, Histories, & Tragedies, Edward Blount and William and Isaac Jaggard, 1623.
To a game theorist, a rational person is someone who makes selections that game theory says they should select. More specifically, a rational player has a definite preference for particular outcomes and follows a stipulated procedure when selecting what actions to take. There's nothing in this definition that says a "rational" strategy should always pay better than the irrational.
But in general the rational strategy does tend to steer you away from the worst case scenario. And in our game that's what happens.
Neither Player 1 playing <Tails> nor Player 2 playing <Heads> gets the best payouts possible. But those strategies always avoid the worst - loosing a penny. To game theorists this is not only a rational strategy but a particular type of rational strategy.
If one strategy gives a better payout than another no matter what, then the first strategy is said to strictly dominate the other. We see that for Player 1, playing <Tails> strictly dominates playing <Heads>. And for Player 2, <Heads> strictly dominates <Tails>.
A strategy which strictly dominates another strategy is called - what else? - a strictly dominant strategy.
On the other hand, a strategy that always gives the worse payout is called a strictly dominated strategy. For Player 1, playing <Heads> is strictly dominated by <Tails>. And for Player 2, playing <Tails> is strictly dominated by <Heads>.
And the basic rules of game theory are:
ALWAYS play a STRICTLY DOMINANT STRATEGY.
... and ...
NEVER play a STRICTLY DOMINATED STRATEGY.
So things are settled, nicht wahr?
As Big Jake said, "Not hardly."
Enter Big John
The game we have just shown:
| Player 1 \ Player 2 | Heads | Tails |
| Heads | -1¢ , 2¢ | ½¢ , 1¢ |
| Tails | 0¢ , ½¢ | 1¢ , -1¢ |
... is famous in the history of mathematics. It was a test to see if game theory predicted what human players would actually do. More particularly it was a test of whether two players would play a Nash Equilibrium.
A Nash Equilibrium is a strategy that after the game has been played, no player will look back and wish they had played differently. That such strategies exist in all finite games - that is games with a finite number of steps and a finite number of players - was proven in 1949 by Princeton mathematician John Nash.
According to game theory, Nash Equilibrium Points are strategies that rational players will prefer. And <Tails , Heads> - with its payout - <0¢ , ½¢> - fits this bill.
How so? Well, after the game is played, Player 1 will have lost no money. But if he had switched to <Heads> while Player 2 had stuck to playing <Heads>, then Player 1 would have lost a penny. So Player 1 would be glad they had played <Tails>.
And Player 2?
If she plays <Heads> while Player 1 plays <Tails>, she would win ½¢.
But if she had switched to <Tails> while Player 1 kept playing <Tails>, she would have lost a penny. So Player 2 would not have regretted4 her first choice of <Heads>.
Footnote
We should also mention that game theory has its own definition of "regret" as well. "Regretting" an outcome does not presuppose any emotional response lamenting the choice with gnashing of teeth and rending of garments. It simply means the players can recognize that they could have done better by making another selection.
And since both players did not regret their strategy, <Tails , Heads> is a Nash Equilibrium Point.
It's important to realize that if two players each have a strictly dominant strategy, their preferred mode of playing will be a Nash Equilibrium Point. On the other hand we will see that games without strictly dominant strategies still have a Nash Equilibrium.
Reality at the RAND
In other books and articles about this experiment, what we've been labeling as <Tails> and <Heads> are replaced with the more generic terms, <Defect> and <Cooperate> as follows:
| Player 1 \ Player 2 | Defect | Cooperate |
| Cooperate | -1¢ , 2¢ | ½¢ , 1¢ |
| Defect | 0¢ , ½¢ | 1¢ , -1¢ |
The labels were changed to emphasize that the better payout was from mutual cooperation, <Cooperate, Cooperate>5. But despite the payout advantages, full cooperation runs the risk of a player getting the poorest payout if the other player switches to <Defect>.
Footnote
In the actual test at the RAND Corporation, the strategy labels were even more general, just "1" and "2".
We've already shown that what we are now calling <Defect , Defect> and not <Cooperate, Cooperate>, was the Nash Equilibrium Point. So you would expect both rational opponents to defect.
But what happens when this game is actually played by real people?
The "real people" test was run at the RAND Corporation in 1950. Merrill Flood and his colleague Melvin Dresher set up the game between Armen Alchian, an economics professor at UCLA, and John Williams, an astronomer and mathematician who worked at RAND.
Throughout the session, Armen and John recorded comments. Before the first game, Armen noted that if John played <Defect> it was a guaranteed win. So he recognized <Defect , Defect> as the preferred play. John, on the other hand, simply said that he hoped Armen "was bright".
But the first play, was not <Defect , Defect>. Instead it was <Defect , Cooperate>. So Armen won 1¢ while John lost a penny.
Armen was perplexed and asked "What is he doing?!!" John in turn said "He's isn't [bright], but maybe he'll wise up.
The next play was the same and Armen mused "Trying mixed?" and John simply said "OK, dope."
The next two plays were <Defect , Defect> - the Nash Equilibrium Point. Armen broke even (0¢) and John won ½¢. But the next play was <Cooperate , Defect> where Armen lost 1¢ ("Perverse!" he commented) and John won 2¢.
But as the game progressed they moved toward favoring <Cooperate , Cooperate>. This was the layman's choice which gave both players a higher payout than the rational Nash Equilibrium. After 100 plays Armen and John had played <Cooperate , Cooperate> 60% of the time. Armen won 45¢ to John's 65¢.
The comments give us a good idea of their thinking. During the game Armen would comment how John would not "share". Evidently he felt that when he played <Defect> - which was a sure win for John - John should also occasionally play <Cooperate> so Armen would earn some money.
John, though, was clearly trying to force the play to <Cooperate , Cooperate> where both would get a better payout. If Armen didn't cooperate, then John would switch to <Defect> where Armen would get nothing and he, John, would get ½¢.
Although at the time game theory was only about six years old, Armen and John knew the rudiments. But nevertheless these two top-flight mathematicians largely avoided what was supposed to be the mathematically logical strategy, the Nash Equilibrium, a strategy that they played only 14% of the time.
Does this mean that game theory is horse hockey, bullshine, and balderdash?
Well, before passing such a harsh judgement, maybe we should go beyond the outcome of a somewhat contrived game with virtually no benefits or consequences to the players and move into the real world. Fortunately an example of such a - quote - "real world game" - unquote - exists.
The Rational Ramblings of Perry and Dick
Truman Capote
He gave us an example.
The Flood-Dresher Experiment was the first examples of what has become famous as the Prisoner's Dilemma. Today the Prisoner's Dilemma is more often taught as a problem when two partners in crime find themselves apprehended. Should they stay silent (<Cooperate>) or confess (<Defect>)?
Here we can find a real real-world example. It shows us that game theory sometimes - as the television commercials say - really, really works.
Our real life example of the Prisoner's Dilemma was described by the American writer Truman Capote in his book In Cold Blood. Truman told how sometime after midnight on November 15, 1959, two ex-convicts, Perry Edward Smith and Richard Eugene Hickock, entered the home of a wealthy farmer, Herbert Clutter, in Holcomb, Kansas. They planned to rob Mr. Clutter of the $10,000 contents of a safe and "leave no witnesses". Neither the safe nor the money existed and before leaving the house with less than $50, a portable radio, and a pair of binoculars, Perry and Dick murdered all four of the family.
With no witnesses, Perry and Dick figured that unless they confessed there could be no conviction for murder which at the minimum carried lengthy prison sentences. So they agreed to keep silent. Then the worst that could happen would be getting pinned with relatively short prison sentences for parole violations and passing bad checks.
But the absolutely worst thing would be if only one confessed. Then he could testify against the other - turn state's evidence - and get off scott free. The other would go to "The Corner"6.
Footnote
"The Corner" was the name given to the location of the gallows at the Kansas State Penitentiary. It was literally in the southeast corner of the prison warehouse.
As with most game theory problems, the Prisoner's Dilemma for Perry and Dick is best summarized in a table:
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | Light Sentence , Light Sentence | "The Corner" , Go Free |
| Confess | Go Free , "The Corner" | Stiff Sentence , Stiff Sentence |
... which in terms of <Cooperate> and <Defect> is:
| Perry \ Dick | Cooperate | Defect |
| Cooperate | Light Sentence , Light Sentence | "The Corner" , Go Free |
| Defect | Go Free , "The Corner" | Stiff Sentence , Stiff Sentence |
Although you can reason through the table, it helps to assign numerical payouts by ranking the outcomes:
| "The Corner" | = | 0 |
| Stiff Sentence | = | 1 |
| Lesser Charge | = | 2 |
| Go Free | = | 3 |
In some problems, the actual numbers are important. But here we simply take these somewhat arbitrary values where the higher number are better outcomes for the prisoner.
The table is now:
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3 |
| Confess | 3 , 0 | 1 , 1 |
As before you can prove there is a strictly dominant strategy. If Dick stays silent Perry should confess (3 > 2). If Dick blabs, then Perry should also confess (1 > 0).
And if Perry stays silent, Dick should confess (3 > 2). If Perry blabs, then Dick should also confess (1 > 0).
So regardless of what either does, each will get the better deal if he confesses. And indeed <Confess , Confess> is a Nash Equilibrium Point.
And here in this real life example we see the problem with going for the higher paying but less rational strategy. <Stay Silent , Stay Silent> - which Dick and Perry agreed to - is simply not stable or sustainable. It's just too easy for individual self-interest to be put above any collective benefits.
In this case both men left their bootprints in the Clutter home and when taxed with this extra evidence, Dick confessed. Once Perry learned that Dick "let fly" and "dropped his guts", he also confessed. So the well-planned cooperative strategy, <Stay Silent , Stay Silent>, broke down and sent both men to the Nash Equilibrium Point, <Confess , Confess>.
Dick and Perry were soon to learn of another problem with real life game theory. The players play in a manner that they think fits with the payouts. But neither Dick, Perry, nor the detectives called the shots. Instead the payouts were dictated by the county prosecutor, a young man named Duane West. And Duane's payout table was:
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | "The Corner" , "The Corner" | "The Corner" , "The Corner" |
| Confess | "The Corner" , "The Corner" | "The Corner" , "The Corner" |
And that's what they got.
The Short Cut
Trying to reason what strategy is a Nash Equilibrium can sometimes be confusing. Fortunately, there is a most simple short cut.
All you do is go through the table for each player and mark the "best" payout for a particular strategy. If you find a payout where both numbers are marked, you have a Nash Equilibrium.
To show what we mean we'll look at the Prisoner's Dilemma again. First here's the table:
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3 |
| Confess | 3 , 0 | 1 , 1 |
OK. Grey out everthing except Perry's payouts when Dick chooses <Stay Silent>.
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3 |
| Confess | 3 , 0 | 1 , 1 |
Obviously the best payout is <Confess>. So we asterisk the number 3.
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3 |
| Confess | 3* , 0 | 1 , 1 |
Now grey out all but Perry's payouts when Dick decides to <Confess>.
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3 |
| Confess | 3* , 0 | 1 , 1 |
So we see the best payout is <Confess> and so we asterisk the number 1.
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3 |
| Confess | 3* , 0 | 1* , 1 |
We now look at Dick's payouts. First we grey out everything except when Perry keeps silent.
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3 |
| Confess | 3* , 0 | 1* , 1 |
The best payout for Dick is clearly <Confess> and so we asterisk the number 3.
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3* |
| Confess | 3* , 0 | 1* , 1 |
Now we grey out everything except Dick's payouts when Perry confesses:
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3* |
| Confess | 3* , 0 | 1* , 1 |
And we see that the asterisk should go on the 1:
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3* |
| Confess | 3* , 0 | 1* , 1* |
Now is there a strategy where both payouts have an asterisk? Here there certainly is:
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | 2 , 2 | 0 , 3* |
| Confess | 3* , 0 | 1* , 1* |
You can confirm <Confess , Confess> is a Nash Equilibrium Point. If both men had confessed, Perry would have said, "Had I switched to <Stay Silent> and Perry stuck with <Confess>, I would have come out worse that had I confessed. Therefore I would have regretted my switch and wish that I had stuck with <Confess , Confess>.
And Dick likewise would have said, "Had I switched to <Stay Silent> and Perry stuck with <Confess>, I would have come out worse that had I confessed. Therefore I would have regretted my switch and wish that I had stuck with <Confess , Confess>.
Since <Confess , Confess> is a strategy that both men would not have regretted playing, it is a Nash Equilibrium Point.
So we see that simply by asterisking the best payouts, we found a Nash Equilibrium Point with much less effort than appealing to reason.
OK. Do we now have an infallible method for finding the rational strategy?
Once more we defer to Big Jake.
Pitching Pennies for Fun and Not Much Profit
In our original presentation of the Flood-Dresher experiment, we had the two players decide whether to show heads or tails. But let's go to the real game of pitching pennies. Each player flips their coin and how the pennies land determines their payout:
| Joe \ Josephine | Heads | Tails |
| Heads | 1¢ , -1¢ | -1¢ , 1¢ |
| Tails | -1¢ , 1¢ | 1¢ , -1¢ |
If they match, then Joe wins a penny and Josephine loses a penny. If they don't match, Josephine wins and Joe loses.
But lo! If you run through the exercise of asterisking the best payouts, you end up with a table like:
| Joe \ Josephine | Heads | Tails |
| Heads | 1¢* , -1¢ | -1¢ , 1*¢ |
| Tails | -1¢ , 1*¢ | 1*¢ , -1¢ |
So we see there is no strategy with both payouts asterisked.
But wait! You said that every game must have at least one Nash Equilibrium Point! Does this mean John Nash was wrong?
Well, no. The trick of asterisking the payouts identifies only the Pure Nash Equilibrium Points.
A Pure Nash Equilibrium Point is a Nash Equilibrium where the players stick to one of the listed strategies. However there are Nash Equilibria where the players mix their play between the available choices. If you mix your strategies and your play still fulfills the definition of a Nash Equilibrium, then this is called a Mixed Nash Equilibrium.
And Pitching Pennies as you usually play it - that is, you flip the coins - is indeed a Nash Equilibrium. You can see that by calculating the payoffs.
Assuming the coins are fair and heads comes up as often as tail, the expected payout for Joe is calculated thusly:
| Expected Payout | = | (0.5)(0.5)(1¢) + (0.5)(0.5)(-1¢) + (0.5)(0.5)(-1¢) + (0.5)(0.5)(1¢) |
| = | 0.25¢ - 0.25¢ - 0.25¢ + 0.25¢ | |
| = | 0¢ |
As for Josephine:
| Expected Payout | = | (0.5)(0.5)(-1¢) + (0.5)(0.5)(1¢) + (0.5)(0.5)(1¢) + (0.5)(0.5)(-1¢) |
| = | -0.25¢ + 0.25¢ + 0.25¢ - 0.25¢ | |
| = | 0¢ |
We do see that the game doesn't seem to be profitable. But on the other hand, can either player do better by switching if the other doesn't?
Well, let's have Joe play only <Heads> and Josephine sticks with her random flipping. Then Joe's payout is:
| Expected Payout | = | (1)(0.5)(1¢) + (1)(0.5)(-1¢) + (0)(0.5)(-1¢) + (0)(0.5)(1¢) |
| = | 0.5¢ - 0.5¢ + 0¢ + 0¢ | |
| = | 0¢ |
And if Joe had played only <Tails>? Well, he'd get the following payout:
| Expected Payout | = | (0)(0.5)(1¢) + (0)(0.5)(-1¢) + (-1)(0.5)(-1¢) + (1)(0.5)(1¢) |
| = | 0¢ + 0¢ - 0.5 ¢ + 0.5¢ | |
| = | 0¢ |
So we see that Joe gets the same payout - nothing - as long as Josephine keeps flipping her coin.
In fact, we can show any switch to any percentage of <Heads> or <Tails> will give Joe the same (zero) payout:
| Expected Payout | = | (PH)(0.5)(1¢) + (PH)(0.5)(-1¢) + (PT)(0.5)(-1¢) + (PT)(0.5)(1¢) |
| = | (0.5PH¢ - 0.5PH¢)
+ (0.5PT¢ - 0.5PT¢) | |
| = | 0¢ + 0¢ | |
| = | 0¢ |
... where PH is the percentage time Joe plays <Heads> and PT is the percentage time Joe plays <Tails>.
So as long as Josephine sticks to her strategy of flipping the coin, Joe can't do any better.
In the same manner, you can also prove that if Joe sticks to his strategy of flipping the coin, Josephine can't do any better. Since both players can't improve their lot by changing their strategy, pitching pennies using fair coins is a Nash Equilibrium.
Remember. All finite games have at least one Nash Equilibrium Point and some have more. They may be pure, mixed, or there may be both.
Equating Equilibria
A Nash Equilibrium is always a rational way to play a game. But not all Nash Equilibrium Points are created equal.
For instance, consider this game:
| J. P. \ Rocky | Heads | Tails |
| Heads | 1000000¢ , 1000000¢ | -1¢ , -1¢ |
| Tails | -1¢ , -1¢ | 1¢ , 1¢ |
Here we see the Pitching Pennies game as we imagine would be preferred by J. P. Morgan and John D. Rockefeller. When they get <Heads, Heads> they both end up winning 1000000¢ ($10000.00). Any losses are minimal.
And note they don't pay the big money themselves. They get the banks to do that. That makes this game significantly different than Joe and Josephine's game since it is not a zero-sum game.
So if J. P. and Rocky simply flip the coins, the payout's pretty good.
| Expected Payout (J. P.) | = | (0.5)(0.5)(1000000¢) + (0.5)(0.5)(-1¢) + (0.5)(0.5)(-1¢) + (0.5)(0.5)(1¢) |
| = | 250000¢ - 0.25¢ - 0.25¢ + 0.25¢ | |
| = | 249999.75¢ |
... and Rocky gets the same:
| Expected Payout (Rocky) | = | (0.5)(0.5)(1000000¢) + (0.5)(0.5)(-1¢) + (0.5)(0.5)(-1¢) + (0.5)(0.5)(1¢) |
| = | 250000¢ - 0.25¢ - 0.25¢ + 0.25¢ | |
| = | 249999.75¢ |
Despite the big payout, <Heads> is not a strictly dominant strategy. If you run through the payouts, you'll see that it doesn't give the better payout regardless of what the other player does.
<Heads, Heads>, though, is a Nash Equilibrium Point. You can show this by the asterisk trick.
| J. P. \ Rocky | Heads | Tails |
| Heads | 1000000¢* , 1000000¢* | -1¢ , -1¢ |
| Tails | -1¢ , -1¢ | 1¢* , 1¢* |
So <Heads, Heads> is a rational strategy.
And no, your eyes are not deceiving you. It's a surprise, but <Tails , Tails> is also a Nash Equilibrium. After all, J. P. and Rocky would have done worse if they had switched to <Heads> as long as their opponent stuck to <Tails>.
We see, then, that indeed not all Nash Equilibria are created equal.
Pitching Pennies You Can't Refuse
So what do you do if you find the Nash Equuilibria and none of them look that good? Well, if all you've done is find the Pure Nash Equilibria, you want to see if there is a Mixed Nash Equilibrium.
Although randomly Pitching Pennies produces a Nash Equilibrium by luck, most of the time you actually have to calculate the percentages. Although in principle the calculations are not all that difficult, they can be a bit confusing for the beginner.
Here's another variant on the Pitching Pennies game.
| Bugsy \ Virginia | Heads | Tails |
| Heads | -1000000¢ , 1000000¢ | 1¢ , -1¢ |
| Tails | 1¢ , -1¢ | -1¢ , 1¢ |
By asterisking the best payouts, we see that there is no Pure Nash Equilibrium:
| Bugsy \ Virginia | Heads | Tails |
| Heads | -1000000¢ , 1000000¢* | 1¢* , -1¢ |
| Tails | 1¢* , -1¢ | -1¢ , 1¢* |
There is, then, no pure rational strategy.
But hold on! Surely Bugsy would avoid playing <Heads> at all costs. So that seems rational.
Well, not really. Then Virginia could play <Heads> and win 1¢. If she played the game enough times, she'd get as much from Bugsy as if he had played <Heads>7.
Footnote
As to why Bugsy would agree to play this game with Virginia, you should consult her preliminary testimony before the Kefauver Committee.
But according to John, there must be a Nash Equilibrium. So there must be a rational strategy.
And indeed there is. It does, though, take a bit of calculating. Fortunately it has no more intellectual hurdles than middle school math.
First we now rewrite the table as:
| Bugsy \ Virginia | Heads | Tails | ||
| Probability (Virginia) | ||||
| σLeft | σRight | |||
| Heads | (Bugsy) | σUp | -1000000¢ , 1000000¢ | 1¢ , -1¢ |
| Tails | σDown | 1¢ , -1¢ | -1¢ , 1¢ | |
... where the σ values will refer to the percentages that Bugsy and Virginia will play their strategies. To avoid writing textual monstrosities like σBugsy, Heads, σBugsy, Tails, σVirginia, Heads, and σVirginia, Tails we simply write σUp, σDown for Bugsy and σLeft and σRight for Virginia.
What we do first is figure what percentage Bugsy should play <Heads> or <Tails> (i. .e, σUp and σDown). That way it doesn't matter how Virginia plays.
As is often the case with game theory, we break the calculations up to makes things easier.
OK. If Virginia is playing completely <Heads> her payout is determined only by Bugsy's as yet undetermined percentages. So for playing <Heads> Virginia receives:
| Expected Payout for Heads (Virginia) | = | 1000000σUp + (-1)σDown |
... and if Virginia plays Tails:
| Expected Payout for Tails (Virginia) | = | -1σUp + 1σDown |
To ensure that Virginia's payout for <Heads> is the same as <Tails> - so it doesn't matter which one she plays - we simply set the two equations equal:
| 1000000σUp + (-1)σDown | = | -1σUp + 1σDown |
With a little bit of middle school algebra:
| 1000000σUp + (-1)σDown | = | -1σUp + 1σDown |
| 1000000σUp + 1σUp | = | 1σDown + 1σDown |
| 1000001σUp | = | 2σDown |
Although we have one equation and two variables don't despair. If we express our percentages on a scale of 0 to 1, then we know that:
| σDown | = | 1 - σUp |
And we have:
| 1000001σUp | = | 2(1 - σUp) |
And with a little more middle school algebra:
| 1000001σUp | = | 2(1 - σUp) |
| 1000001σUp | = | 2 - 2σUp |
| 1000001σUp + 2σUp | = | 2 |
| 1000003σUp | = | 2 |
| σUp | = | 2/1000003 |
And so:
| σDown | = | 1 - σUp |
| = | 1 - 2/1000003 | |
| σDown | = | 1000003/1000003 - 2/1000003 |
| σDown | = | (1000003 - 2)/1000003 |
| σDown | = | 1000001/1000003 |
We have now calculated Bugsy's percentages to play <Heads> and <Tails> where Virginia will get the same payout if she plays completely <Heads> or completely <Tails>.
And we can put these in the table:
| Bugsy \ Virginia | Heads | Tails | ||
| Probability (Virginia) | ||||
| σLeft | σRight | |||
| Heads | (Bugsy) | 2/1000003 | -1000000¢ , 1000000¢ | 1¢ , -1¢ |
| Tails | 1000001/1000003 | 1¢ , -1¢ | -1¢ , 1¢ | |
To double check our numbers - always a good idea - we just crunch through the numbers:
For <Heads> Virginia's expected payout is:
| Expected Payout for Heads (Virginia) | = | 1000000 × 2/1000003 + (-1) × 1000001/1000003 |
| = | 2000000/1000003 - 1000001/1000003 | |
| = | (2000000 - 1000001)/1000003 | |
| = | 999999/1000003 |
And for Virginia playing <Tails>:
| Expected Payout for Tails (Virginia) | = | -1 × 2/1000003 + 1 × 1000001/1000003 |
| = | -(2/1000003) + 1000001/1000003 | |
| = | 1000001/1000003 - 2/1000003 | |
| = | (1000001 - 2)/1000003 | |
| = | 999999/1000003 |
So with our calculated percentages of Bugsy's, Virgina gets the same payout whether she plays <Heads> or <Tails>.
How about that?
Now here's the kicker. Using Bugsy's percentages that force Virginia playing <Heads> to be the same payout as <Tails>, we can prove that any percentage that Virginia plays will give her the same payout.
Let Virginia start pitching pennies with Bugsy. The only stipulation is Bugsy will keep to his preferred percentages. That is he will play <Heads> 2 out of 1000003 times and <Tails> 999999 out of 1000003 times.
But what if Virginia splits her play 50/50 between <Heads> and <Tails>. Well, then her payout will be:
| Total Payout (Virginia) | = | Payout<Heads> + Payout<Tails> |
| = | 0.5(2/1000003)(1000000) + 0.5(99999/1000003)(-1) + 0.5(2/1000003)(-1) + 0.5(99999/1000003)(1) |
And with the inevitable middle school arithmetic:
| Total Payout (Virginia) | = | 0.5(2/1000003)(1000000) + 0.5(99999/1000003)(-1) + 0.5(2/1000003)(-1) + 0.5(99999/1000003)(1) |
| = | 0.5(2)(1000000)/1000003 + (0.5)(99999)(-1)/1000003 + (0.5)(2)(-1)/1000003 + (0.5)(1)(99999)/1000003 | |
| = | (1)(1000000)/1000003 - (0.5)(99999)/1000003 - (1)/1000003 + (0.5)(99999)/1000003 | |
| = | (1)(1000000)/1000003 - (1)/1000003
+ (0.5)(99999)/1000003 - (0.5)(99999)/1000003 | |
| = | (1)(1000000)/1000003 - (1)/1000003 | |
| = | 1000000/1000003 - 1/1000003 | |
| = | (1000000 - 1)/1000003 | |
| = | 999999/1000003 |
Exactly what the Pure <Heads> and Pure <Tails> strategies earned!
This can, in fact be generalized where the 50% percentage (0.5) is substituted for any value. If x is the probability that Virginia plays <Heads> then 1- x must be how often she plays <Tails>.
So our middle school algebra leads us to:
| Total Payout (Virginia) | = | x(2/1000003)(1000000)
+ x(1000001/1000003)(-1) + (1-x)(2/1000003)(-1) + (1-x)(1000001/1000003)(1) |
| = | x(2/1000003)(1000000)
- x(1000001/1000003) - (1-x)(2/1000003) + (1-x)(1000001/1000003) | |
| = | x(999999/1000003)
+ (1-x)(999999/1000003) | |
| = | x(999999/1000003)
+ 999999/1000003 - x(999999/1000003) | |
| = | 0
+ 999999/1000003 | |
| = | 999999/1000003 |
Ha! No matter what percentage Virginia plays between <Heads> or <Tails>, as long as Bugsy plays <Heads> 2 out of 1000003 times and he plays <Tails> 1000001 out of 1000003 times, then Virginia will always receive 999999/1000003¢.
So she can't do any better and so will have no regrets as long as Bugsy sticks to those percentages.
But we're only halfway done. A Nash Equilibrium is where both players have no regrets. So what percentages should Virginia play to ensure Bugsy can't do any better?
If you guess the same as Virginia, you'll be right. But we'll just run through the calculations for the sake of completeness.
First we calculate how much Bugsy earns if he plays <Heads> using Virginia's percentages:
| Expected Payout (Bugsy Plays <Heads>) | = | -1000000σLeft + 1σRight |
When Bugsy plays pure <Tails>, he'll earn:
| Expected Payout (Bugsy Plays <Tails>) | = | 1σLeft + (-1)σRight |
Again we set the two equations equal:
| -1000000σLeft + 1σRight | = | 1σLeft + (-1)σRight |
And as before:
| σRight | = | 1 - σLeft |
And we have:
| -1000000σLeft + 1(1 - σLeft) | = | 1σLeft + (-1)(1 - σLeft) |
And working through the algebra:
| -1000000σLeft + 1(1 - σLeft) | = | 1σLeft + (-1)(1 - σLeft) |
| -1000000σLeft - σLeft | = | -(1 - σLeft) - (1 - σLeft) |
| -1000001σLeft | = | -2(1 - σLeft) |
| -1000001σLeft | = | -2 +2 σLeft |
| -1000001σLeft - 2 σLeft | = | -2 |
| -1000003σLeft | = | -2 |
| σLeft | = | -2/-1000003 |
| σLeft | = | 2/1000003 |
So Virginia's percentage for playing <Heads> is the same as Bugsy's!
And her percentage for <Tails> is also the same:
| σRight | = | 1 - σLeft |
| = | 1 - 2/1000003 | |
| = | 1000003/1000003 - 2/1000003 | |
| = | (1000003 - 2)/1000003 | |
| = | 1000001/1000003 |
Which gives us the final table:
| Bugsy \ Virginia | Heads | Tails | ||
| Probability (Virginia) | ||||
| 2/1000003 | 1000001/1000003 | |||
| Heads | (Bugsy) | 2/1000003 | -1000000¢ , 1000000¢ | 1¢ , -1¢ |
| Tails | 1000001/1000003 | 1¢ , -1¢ | -1¢ , 1¢ | |
As far as Bugsy's payout, we calculate that as:
| Total Payout (Bugsy) | = | Payout<Heads> + Payout<Tails> |
| = | (2/1000003)(2/1000003)(-1000000) + (2/1000003)(1000001/1000003)(1) + (1000001/1000003)(2/1000003)(1) + (1000001/1000003)(1000001/1000003)(-1) | |
| = | (-4000000/1000006000009)
+ (2000002/1000006000009) + (2000002/1000006000009) - (1000002000001/1000006000009) | |
| = | (-1000006000001/1000006000009)
+ (4000004/1000006000009) | |
| = | -1000001999997/1000006000009 | |
| = | -999999/1000003 |
A negative payout - and as you expect, it corresponds exactly to the amount that Virginia won!
We see then that if both Virginia and Bugsy play <Heads> 2 out of 1,000,003 times and <Tails> 1000001 times out of 1,000,003 times, Virginia will win 999999/1000003¢ and Bugsy loses 999999/1000003¢.
With this strategy both players can do no better as long as the other sticks to the specified percentages. They will look back and realize there was nothing they could have done to get a better deal.
Again we have a rational strategy and a Nash Equilibrium.
The Irrational Rational
The sharp eyed reader may have noticed a wee bit of a problem with playing this particular "rational strategy". And that's because you need to play it at least 1,000,003 times. Otherwise you couldn't switch to <Heads> 2 times out of 1,000,003 (you can't get heads once out of 500,001½ times).
Even playing "only" 1,000,003 times wouldn't work. The flipping must be random which means you could have sequences longer than 1,000,003 flips where you might not get <Heads> at all. Or you might get hundreds of <Heads>.
What we would need then is an "electronic" coin which would randomly and on the average land <Heads> 2 out of 1,000,003 times and <Tails> 1,000,001 out of 1,000,003 times. But you'd still have to play the game many times in succession - say 1000 times - to make sure the statistics are reliable.
That means you'd end up flipping the coins at least 1000003000 times! More than a billion flips!
At fifteen seconds per flip for eight hours a day with fifteen minute breaks every two hours and an hour for lunch, to play the Nash Equilibrium Point you need to play the game for 1756 years.
And at the end of the game, Virginia would have won 999,999/1000003¢ per flip - that is 0.9999960000119999640001079996760009(etc.,etc.)¢. Bugsy would have lost the same amount.
We see, then, that for all practical purposes, our rational Mixed Nash Equilibrium strategy earns Virginia earns 1¢ per flip. So within the framework of human endurance, this rational strategy is essentially the same as playing the pure strategy of <Tails, Tails> - which is not a Nash Equilibrium and not a rational strategy.
So after all that work, we've ended up with a rational strategy no one can play. Bugsy might as well hand over all the money to Virginia. Which, of course, is exactly what he did.
Perry and Dick (Again)
We mentioned that you can rank the payouts using arbitrary numbers. That's true regarding picking out Pure Nash Equilibria. But if you have Mixed Nash Strategies, the actual values of the payouts can affect the percentages the players use in their mixed strategies.
And changing the values can change what strategy predominates. This makes sense. For instance, sometimes the worst case scenario isn't that bad, and it may very well be worth the risk to choose a less rational but far better paying strategy.
In the Flood-Dresher Experiment if the men strayed from the rational strategy, they had the risk of loosing a few pennies. Not so with Dick and Perry. Their worst case scenario was the worst case scenario.
Up to the day he died (literally), Dick claimed he was innocent of murder. He had never intended to kill the Clutters, he said, it was Perry. In his tape recorded confession, Dick had even suggested a deal:
| Dick: | Is there anyway I can get out on a manslaughter charge because I never pulled a trigger. |
Harold Nye, a detective from the Kansas Bureau of Investigation, responded:
| Harold: | Of course that would have to be worked out with the County Attorney, Dick. I couldn't tell you that. |
We see, then, that Dick would be willing to plea to a lesser charge. So we can revise our payouts using what could be the actual prison sentences.
In those days life without possibility of parole was rare. So our estimates for the sentences are:
| Both Keep Silent (Conviction Only for Bad Checks and Parole Violations) | = | 5 |
| Manslaughter with Confession (Dick's Choice) | = | 10 |
| Murder Conviction After Turning State's Evidence (Only One Confesses) | = | 30 |
| Murder Conviction with Both Confessing | = | 99 (Life) |
| "The Corner" | = | [To Be Determined] |
Since it's customary to indicate the more favorable outcome with the larger (more positive) number, we make the sentence numbers negative. So our payout table becomes:
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | -5 , -5 | [TBD] , -10 |
| Confess | -30 , [TBD] | -99 , -99 |
TBD = "The Corner" ("Payout" To Be Determined)
The question is how do we equate "The Corner" in terms of the years of a prison sentence?
If you go to The Corner you're there forever. So it should be equated to an infinite number of years in prison8. But infinity is not a number, it's a concept. So how do you proceed?
Footnote
Officially, though, a "visit to the Corner" was consdered as a release from prison. Both Dick and Perry's release date in their prison files is listed as April 14, 1965.
For Pure Nash Equilibria that's not a problem. Infinity is rated as being larger in magnitude than any number. So we immediately see there are no strictly dominant strategies but there are two Pure Nash Equilibria: <Stay Silent , Stay Silent> and <Confess , Confess>.
| Perry \ Dick | Stay Silent | Confess |
| Stay Silent | -5* , -5* | -∞ , -10 |
| Confess | -30 , -∞ | -99* , -99* |
But is there also a mixed strategy? And if so, how do you calculate it?
Infinity may not be a number but we can see what happens as The Corner payout approaches (negative) infinity. First calculate the Mixed Nash percentages treating The Corner as being -100 years (1 year more than "Life"). Then we gradually decrease the value and see how the percentages change. If the percentages converge to a number as the number of "years" in The Corner increases without bound, then we have the Mixed Nash Equilibrium.
It can be shown that our calculations apply to both Dick and Perry. And what we get is:
The percentage for <Stay Silent , Stay Silent> starts out low - which means <Confess, Confess> is more important. But as "The Corner" approaches an eternity of imprisonment - certainly a theologically sound solution - the percentages converge to 1.
So what had been a Mixed Nash Strategy converges to the pure strategy <Stay Silent , Stay Silent>.
But - and it's a big "but" - there's more to consider. The <Stay Silent , Stay Silent> strategy - Nash Equilibrium though it is - has the risk of The Corner should the other switch to <Confess>.
So it seems that by proving that The Corner was something to be shunned no matter what, we've shown that it's something worth risking!
Forget it. In the end, the best strategy is that both should confess, the classic Prisoner's Dilemma solution.
Johnny's Solution
Oh, yes. One more thing.
We have to mention we've left out something that you rarely read in books or articles that talk about the Flood-Dresher Experiment. And that's Johnny von Neumann's solution.
Johnny Von Neumann
He gave us a solution.
John von Neumann was one of the greatest mathematicians of the 20th century (perhaps the greatest). In 1944 he published Theory of Games and Economic Behavior co-written with his colleague Oskar Morgenstern. This book expanded the ideas from an article Johnny wrote in 1928 inspired by the game of poker. So John and Oskar are widely regarded as the inventors of game theory. Merrill, by the way, had been one of Johnny's students at Princeton.
The Flood-Dresher Experiment was a test for the Nash Equilibrium as a rational strategy for non-cooperative games. But since Armen and John W. actually avoided it, Merrill concluded a Nash Equilibrium was not a realistic solution.
Merrill, though, did ask John N's opinion of the game's outcome. John said the discrepancy was that Armen and John W. were not playing a single non-cooperative game. Instead they had played a large multi-move game in which there was a lot of interaction between the players.
Although it was later shown that a multi-move non-cooperative game is rationally played at the Nash Equilibrium, John hit the nail on the head by saying there was too much interaction. Far from being a non-cooperative game, the play is best described as a multi-step game where an initial lack of cooperation moved toward near total cooperation.
Cooperative games were investigated by Johnny and Oskar. If the two participants are friends (as Archen and John W. were), cooperation is likely. The difference in cooperative and non-cooperative games is how the final payout is made. In some games, a player with an advantage can make a specified side payments to the other. They can simply split the difference or divide the take.
For instance, consider the Flood-Dresher Experiment where the money is divided equally between the two players. For what we can call the Cooperative Flood-Dresher Experiment, we revise our table to where the payout is the average of the original payouts.
So what was originally:
| Player 1 \ Player 2 | Defect | Cooperate |
| Cooperate | -1¢ , 2¢ | ½¢ , 1¢ |
| Defect | 0¢ , ½¢ | 1¢ , -1¢ |
... which by divvying up the payout becomes:
| Player 1 \ Player 2 | Defect | Cooperate |
| Cooperate | ½¢ , ½¢ | ¾¢ , ¾¢ |
| Defect | ¼¢ , ¼¢ | 0¢ , 0¢ |
From here on out, it's the same as before.
First of all, for Player 1 <Cooperate> is now the strictly dominant strategy. That is:
½¢ (Cooperate) > ¼¢ (Defect)
... and ...
¾¢ (Cooperate) > 0¢ (Defect)
On the other hand, now Player 2 does not have a dominant strategy. We see that because:
¾¢ (Cooperate) > ½¢ (Defect)
... but ...
¼ (Defect) > 0¢ (Cooperate)
In this case, though, you don't need to solve simultaneous equations to determine the best Nash Equilibrium.
Remember the fundamental rule of game theory. Always play the strictly dominant strategy.
So we know that Player 1 will always play his dominant strategy of <Cooperate>. Player 2 knows this, too. So she will always pick the strategy that gives her the better payout while Player 1 plays <Cooperate>. So Player 2 also plays <Cooperate>.
But is <Cooperate , Cooperate> a Nash Equilibrium? Well, if you asterisk the high values it certainly is:
| Player 1 \ Player 2 | Defect | Cooperate |
| Cooperate | ½¢* , ½¢ | ¾¢* , ¾¢* |
| Defect | ¼¢ , ¼¢* | 0¢ , 0¢ |
And there's no reason for the players to regret their choice.
There is also a Mixed Nash Equilibrium. This is where both players randomly spit their play 50/50 between <Cooperate> and <Defect>. The payout then is 0.375¢ apiece. The proof will be left to the reader.
What shocks! shocks! the students of game theory is that Johnny von Neumann pooh-poohed the importance of the Nash Equilibrium. He felt it was a trivial finding and did not really go beyond his and Oskar's famous minimax solution.
A minimax - no, it's not a new Austin Powers villain - is a strategy where a player plays to get the best possible payout regardless of what the other person does. Equivalently it's the lowest possible payout you can force on your opponent.
The minimax solution applies to zero-sum games (what one player wins, the other loses) between two players. But you can run through the principle using the Flood-Dresher Experiment. In our original table this was:
| Player 1 \ Player 2 | Heads | Tails |
| Heads | -1¢ , 2¢ | ½¢ , 1¢ |
| Tails | 0¢ , ½¢ | 1¢ , -1¢ |
So the goal is to find the strategy where the player gets the highest possible value no matter what the other does.
For Player 1, the best he can do is win a penny in the <Tails , Tails> strategy. However, that requires Player 2 to play <Tails> as well. If Player 2 switches to <Heads> Player 1 wins nothing. So <Tails , Tails> is not a minimax.
For Player 1 playing <Heads, Tails> has a payout of ½¢. But again, dang it, it also requires Player 2 to play <Tails>. If Player 2 switches to <Heads> Player 1 looses a penny.
So the most Player 1 can be guaranteed to make regardless of what Player 2 does is to break even - that is, "win" 0¢ - when the play is <Tails , Heads>.
But is this also the worst payout Player 2 can force on Player 1? Yep, it is. Player 1 would loose 1¢ if he plays <Heads> instead. So by sticking to <Tails> he will only break even at the worst. To force Player 1 to this payout, Player 2 must play <Heads>. Otherwise Player 1 will win a penny.
So <Tails , Heads> is the worst that Player 2 can force on Player 1 - the same as the best Player 1 is guaranteed to make. <Tails , Heads> is a minimax.
By now you've realized that all this is simply looking for the Nash Equilibrium when both players have a strictly dominant strategy. So we understand why Johnny - in 1950 one the world's top mathematicians and a worldwide celebrity to boot - resented a young whippersnapper assistant professor claiming he had done something new. Young John, though, saw (correctly) that his proof was a generalization of the senior John's minimax theorem. All minimax's are Nash Equilibria but not all Nash Equilibria are minimaxes.
History has vindicated John Nash and the importance of his discoveries. You'll read that some game theorists say the minimax solutions are now mostly of historical importance. But discussions of Nash Equilibria abound and in 1994 John Nash won the Nobel Prize in Economics.
And yes, you'll read ad nauseum that there isn't really a Nobel Prize in Economics. Everyone knows it's the Sveriges Riksbank Prize in Economic Sciences in Memory of Alfred Nobel. It's that prize, you read, that John won in 1994 along with John Harsanyi and Reinhard Selten "for their pioneering analysis of equilibria in the theory of non-cooperative games."
But the Nobel Prize Committee still lists the prize along with the others in Physics, Chemistry, Medicine and Physiology, Literature, and Peace. The Economics Prize is awarded at the same ceremony and the winners get the same amount of cash. The only difference is who's putting up the dough. So it is OK to call the Sveriges Riksbank Prize, etc., etc., the Nobel Prize in Economics.
Some may balk at calling the Flood-Dresher Experiment a cooperative game. Neither Armen nor John W. had made any explicit agreement before they began to play and did not split their winnings.
John W. clearly understood that with no cooperation and by playing <Defect>, he was guaranteed a positive return. But he realized he and Armen could win even more money if they worked cooperatively. As he wrote in his comments:
I can guarantee myself a gain of 5 [i. e. 0.5¢], and guarantee that Player AA [Armen Alchian] breaks even (at best). On the other hand, with nominal assistance from AA [emphasis added], I can transfer the guarantee of 5 to Player AA and make 10 for myself too. This means I have control of the game to a large extent, so Player AA had better appreciate this and get on the bandwagon.
From the way Armen and John played we can see that the payouts are not necessarily the same as the utility of a strategy. If there is a reason for playing a strategy that seems irrational it may be that the overall utility to the players is higher or lower than the payouts would indicate. Obviously game theory will struggle to predict actions if the utilities are wrong.
With his high payouts, John Williams did control the game and so could influence the manner of Armen's play. So John and Armen had to ramble around a little before they were smoothly driven to playing the Cooperative Nash Equilibrium.
So perhaps we should call the game a Smoothly Driving Little Nash Rambler.
References and Further Reading
"Equilibrium Points in n-Person Games", John Nash, Proceedings of the National Academy of Sciences, January 1, 1950 Vol. 36, No. 1, pp. 48-49, Communicated by Solomon Lefschetz, November 16, 1949
Prisoner's Dilemma, William Poundstone, Doubleday, 1992.
A Beautiful Mind, Sylvia Nasar, Simon and Schuster, 1998.
"Politics and the English Language", George Orwell, Horizon, April, 1946, pp. 252 - 265; Reprinted in In Front of Your Nose, 1945 - 1950, Collected Essays, Journalism and Letters, Vol. 4, pp. 127 - 140.
Game Theory and Postwar American Literature, Michael Wainwright, Springer, 2016.
Game Theory 101: The Complete Textbook, William Spaniel, CreateSpace, 2011
Game Theory 101 MOOC, William Spaniel, Video Lectures, 2011
Some Experimental Games, Merrill Flood, RM-789-1, The Rand Corporation, 1952.
The Compleat Strategyst, John Williams, Dover Publictions, 1954.
On the Early History of Experimental Economics, Alvin Roth, Department of Economics, University of California - Santa Barbara.
"Introduction to Experimental Economics," John Kagel and Alvin Roth, Handbook of Experimental Economics, Princeton University Press, 1995, pp. 3-109.
"In Game Theory, No Clear Path to Equilibrium", Erica Klarreich, Quanta Magazine, July 18, 2017.
"John Williams, 55, A RAND Researcher", The New York Times, November 22, 1964.
Theory of Games and Economic Behavior, John Von Neumann and John and Oskar Morgenstern, Princeton University Press, 1944.
"John F. Nash, Jr.", The Nobel Prize, 1994.
Good Evening, Everybody: From Cripple Creek to Samarkand, So Long Until Tomorrow: From Quaker Hill to Kathmandu , Lowell Thomas, William Morrow and Company, 1976, 1977.