Gladiators, Pirates and Games of Trust
Page 3
THE DICTATOR GAME
This is yet another version of the Ultimatum Game. Here, there are only two players, where the proposer, named the ‘Dictator’, has full control and the responder must accept anything that is offered – and is, in fact, an ‘idle’ player. According to the mathematical solution, the proposer should pocket the entire sum played for and go home. As you must have guessed by now, the standard economic assumptions are inaccurate predictors of actual behaviour. Very often the entire sum is not withheld: ‘dictators’ tends to give some of the money (sometimes they give substantial sums and sometimes they split the sum evenly) to the responder. Why do they do that? What does this teach us about human nature? What does it have to do with altruism, kindness, fairness and self-respect? Your guess is as good as mine.
Chapter 4
GAMES PEOPLE PLAY
In the following chapter we learn about several games that can be both fun and enlightening. We will expand our games vocabulary, gain some insights and improve our strategic skills. While we’re about it, we’ll become acquainted with someone I believe should be known as ‘Strategist of the Year’. Let’s play!
GAME 1
THE PIRATES GAME
‘You can always trust the untrustworthy because you can always trust that they will be untrustworthy. It is the trustworthy you can’t trust.’
Captain Jack Sparrow, Pirates of the Caribbean
A gang of pirates returns home from a hard day in the office, carrying 100 gold doubloons that are to be divided among the top five pirates: Abe, Ben, Cal, Don and Ern – Abe being the leader and Ern the lowliest member of the crew.
Although there’s a hierarchy of rank, the group is democratic, which is why the following principle is decided upon to determine the distribution of the booty. Abe suggests some distribution formula and all the pirates (including Abe) vote on it. If that formula wins the support of the majority of pirates, Abe’s idea is implemented and this is the end of the game; if not, he’s tossed into the ocean (even democratic pirates are unruly). If Abe is no longer with us, it’s now Ben’s turn to place a motion on the table. They vote again. Note that now there’s a possibility of a tie. We’ll assume that in the event of a tie vote, the proposition is dropped and the proposer will be tossed into ocean (though there’s another version in which in the event of a tie the proposer has the casting vote). If Ben’s proposition wins the support of the majority of pirates, his idea is implemented; if not, he’s tossed into the ocean and Cal will put an offer on the (shrinking) table. And so on.
The game continues until some suggestion is accepted by a majority vote. If this doesn’t happen, Ern remains the last pirate standing and pockets all 100 gold pieces.
Before you go on reading, please stop and think for a moment how this game should end, assuming that the pirates are greedy and smart.
The Mathematical Solution
Mathematicians resolve such question by ‘backward induction’, going from the end to the beginning. Let’s assume that we’re now at a point where Abe made a suggestion and failed, Ben’s motion was rejected and he’s no longer with us, and Cal didn’t fare any better. Don and Ern are the only two pirates left, and now the solution is quite obvious: D must suggest that E take the 100 doubloons, or else D might find himself swimming with the sharks (remember that a tied vote means the proposition fails), which shouldn’t last long. Being a clever pirate, Don suggests that Ern take the whole bag.
Don Ern
0 100
Pirate Cal, who is just as clever, knows that the above will be the final stage of the game (if it ever gets that far, which Cal hopes to prevent at all cost). Furthermore, Cal knows he has nothing to offer Ern because Ern’s interest is to get to that next stage no matter what. However, Cal can help Don improve his situation, compared with what would happen if he were left alone with E, and can make Don vote for him by offering him a single doubloon (in which case, Don will vote with Cal and together they are the majority). Thus, when we have three players, the coin distribution is 99 for Cal, 1 for Don, and 0 for Ern.
Cal Don Ern
99 1 0
Ben is naturally aware of these calculations. He knows there’s nothing he can offer that would improve Cal’s situation, but he could make Don and Ern offers they can’t refuse, going about it thus: Cal gets nothing, Ern ends up with one coin, Don takes two, and Ben lands the remaining 97 coins:
Ben Cal Don Ern
97 0 2 1
Now we arrive at the point where it should be easy to see how Abe should act (being the senior pirate, he’s very experienced in matters of loot distribution). Abe makes the following suggestion. He takes 97 coins; he doesn’t give a penny to Ben (who can’t be bought in any event); he gives 1 coin to Cal (which is better than the 0 coins he’d receive if Abe swims and it is Ben’s turn); Don gets nothing too; and Ern is given 2 coins (Ern’s vote is cheaper to buy than Don’s). This offer will be endorsed by a majority of 3 against 2, and the pirates will go on raiding ships until the seas run dry.
Abe Ben Cal Don Ern
97 0 1 0 2rn
This last distribution seems rather odd. Will we arrive at the same siutation if we try this with five maths students? How about an experiment with five psychology postgraduates? How would the psychologists go about working through the possibilities?
Are the players allowed to form coalitions and make deals? If so, what would this game look like? The mathematical solution always assumes that all the players are wise and rational, but is it wise to make this assumption? Is it rational? (I watched this game played a number of times, and never saw the participants reaching the mathematical solution. What does this mean?) The mathematical solution ignores important emotions such as envy, insult or schadenfreude. Can feelings change the mathematical calculation?
In any event, though Abe’s distribution 97, 0, 1, 0, 2 is mathematically sound, I advise him to show how magnanimous he is by offering his fellow pirates the distribution 57, 10, 11, 10, 12 (i.e. an extra 10 from his pot of 97). This should hopefully result in relative contentment among the crew and prevent a mutiny.
If you feel that the Pirates Game, which is actually a multi-player version of the Ultimatum Game, is odd, what will you say about the following game?
GAME 2
DEAD RICH MAN
A very rich old man passes away, leaving two sons, Sam and Dave.* The two brothers could never stand each other. They haven’t seen each other or talked for 10 years, and now they meet in their father’s house to hear his last will and testament read out.
The father’s lawyer opens the envelope and reads out the peculiar document. It turns out that the father has left his sons one million and ten thousand dollars and a set of possible distribution outcomes.
In the first option, Sam, the elder, may take $100 for himself right away, give a dollar bill to his younger brother, and give the rest to charity (which would be indeed be quite charitable).
Sam Dave
100 1
Sam is under no obligation to accept that instruction, and may instead pass the lead to his younger brother Dave. If Dave handles the money, he takes $1,000, Sam gets $10, and the remainder goes to charity. This is the second option.
Sam Dave
100 1
10 1,000
But now it is Dave’s turn to decline, if he wishes. He may let Sam decide on a better distribution method in which he takes $10,000, gives Dave $100, and the rest … you guessed it.
Sam Dave
100 1
10 1,000
10,000 100
Now, however (yes, you guessed again), Sam doesn’t have to accept this option and may pass the lead to Dave, who this time may divide the money, taking $100,000 for himself, giving Sam $1,000, while the ever-shrinking remainder goes to charity.
Sam Dave
100 1
10 1,000
10,000 100
1,000 100,000
This, of course, is not carved in stone. Dave may decide
to let Sam divide the money again, but in the following manner: $1 million for himself, $10,000 for his hated brother, with zero dollars going to charity.
Sam Dave
100 1
10 1,000
10,000 100
1,000 100,000
1,000,000 10,000
What do you think will happen now? Again, this question may be resolved by backward induction. Everyone can see that there’s no chance on Earth the game would reach the last (fifth) round and Dave would let Sam take a million, because this would diminish his personal gain from $100,000 to only $10,000. Sam knows that, and therefore there’s no chance that he’ll allow the game to reach the fourth round in which he receives only $1,000, instead of the $10,000 of the third round. Keep going now and see that the game will not reach the third round either … and not the second one. It’s very surprising, but under the assumption that the two brothers are of the same species, Homo economicus statisticus (that they are both calculating humans who look after only themselves), the game should have ended with the very first step – with Sam collecting $100 and giving Dave $1 and lots of money to charity (bad intentions may lead to a generous outcome and the brothers winning a heavenly reward perhaps). This is the mathematical solution – $100 for Sam and $1 for Dave, and a lot of money for charity.
Is this solution logical at all? You be the judge of that.
GAME 3
THE CHOCOLATE AND POISON GAME
This is quite a simple game, better known as Chomp. (The chocolate bar formulation of Chomp, which I will use here, is indebted to the late American mathematician David Gale.) It’s played on a chequered board, each box on the board made of chocolate, but the bottom-left box contains a deadly dose of poison. These are the rules:
The opening player marks an X in one of the boxes. He can choose any box he wishes.
X
Poison
Once that’s done, all the boxes to the right and up from the X will also be occupied and turn X.
X X X
X X X
Original X X X
Poison
Now it’s the other player’s turn to mark a remaining box with O. Once that happens, all empty boxes to the right and up from it are marked O too:
X X X
X X X
X X X
O O
Poison O O
Then the first player marks another X and makes this box and all the boxes to the right and up (if there are any) turn X; and the second player marks another O and makes this box and all the boxes to the right and up (if there are any) turn O; and the game goes on until one of them is forced to choose Poison, so that he loses and dies (metaphorically, of course).
You’re welcome to try playing this on a 7 x 4 board (7 rows and 4 columns, or vice versa).
If the game is played on a square (equal number of rows and columns), there’s a strategy by which the opening player will always win. Can you find it? Take three minutes to think.
Solution: Let’s assume that the game is played between Joan and Jill. If Joan is the opening player, she should stick to the following strategy and win. As her first move she must choose the box right above Poison, diagonally.
X X X X
X X X X
X X X X
X Joan X X X
Poison
Now all that she has to do is symmetrically follow her opponent: that is, she’ll play the same move as Jill, but on the opposite side of the board. The following picture explains better than words:
O X X X X
O Jill’s choice X X X X
X X X X
X Joan X X
Poison X Joan’s response X
How this game is won should be very clear now.
Things become much more complicated when the game is played on a rectangle, but still it can be proved that the opening player can have a winning strategy. The problem is that the proof does not specify this winning strategy. Mathematicians call this kind of proof a ‘non-constructive proof of existence’.
GAME 4
NO GAME FOR OLD MEN
One of the most precious skills I acquired at grammar school, back in my hometown of Vilnius, Lithuania, was playing strategic games, on paper, in class, without being caught by my teachers. I was fond of the ‘infinite’ version of tic-tac-toe (or ‘noughts and crosses’). The game often helped me to survive boring classes I had to attend.
I imagine most of us are familiar with the classic version of tic-tac-toe with a 3 x 3 grid, which is fascinating up to the age of six. Older children (and adults) normally end this game with a tie, unless one of the players falls asleep halfway through (which makes sense: it’s a boring game after all).
In the infinite version, however, the game is played on a board with a limitless grid and the goal is to create a sequence of five Xs or Os which, as in the original game, can be vertical, horizontal or diagonal. Players take it in turns to mark the grid with either an X or an O (according to prior agreement) and the first to complete a quintet wins.
In the drawing on the left the X player has already won the game.
In the drawing on the right it’s the O player’s turn, but she can do nothing to prevent the X player from winning. Do you see why?
Back at school, I used to believe that I’d invented the game, but in due course I realized this was not the case. I discovered a very similar game that’s been very popular in Japan and Vietnam for many years, known as Gomoku. Go in Japanese means five. Although Gomoku is sometimes played on the same board as the ancient game of Go, the two games are not related. Go is an ancient Chinese game that’s even mentioned in Confucius’s Analects, but it was introduced to the West by the Japanese and thus is known by its Japanese name.
Although I have plenty of experience playing the infinite version of noughts-and-crosses in endless classes or breaks (breaks are less fun because playing is allowed), I’m still not sure about the optimum winning strategy for the player who starts the game (the X player), or whether this game is always tied (or, actually, never ends) when played by two fine players. I am, however, willing to bet that a winning strategy exists.
When I retire and have plenty of time, I’ll try to find the winning strategy for the starting player.
Still, to be completely honest, I must say I have not played it for decades and was reminded of it while writing this book. Since my plans to revisit its strategic aspects are very long-term, you are welcome to go ahead, find it, and save me time and effort.
GAME 5
THE ENVELOPE IS ALWAYS GREENER ON THE OTHER SIDE
Imagine the following. I’m presented with two cash envelopes and told that one of them contains twice as much as the other. I may choose and take for myself whichever envelope I want.
Suppose I choose an envelope, open it, and find $1,000 inside. I’m pleased at first, but then I start wondering about the content of that other envelope, the unchosen one. Of course, I don’t know what’s in it. It could be $2,000, which means I made a bad choice, or it could be $500. I’m sure you can see the problem. Reflecting on this a while, I reach the following conclusion: ‘I’m not happy because the average of the potential money in the unchosen envelope is larger than the sum in my hands. After all, if it contains either $2,000 or $500 with equal probabilities, the average is $1,250, which is more than $1,000. I know my maths!’
In truth, anything I find in my envelope will prove Murphy’s Law which states that ‘Anything that can go wrong, will go wrong.’ The unchosen envelope will always be better than mine, on average. If I found $400 in mine, the other could contain $800 or $200, and the mean would be $500. Thinking like that, I can never choose right. The unchosen gain will forever be 25 per cent better than mine. So would I change my mind if that option is offered to me before I examine the contents of the other envelope? If I do that, I start a never-ending loop. So why did such a simple choice get so complicated?
In truth, the story I just told you is a famous paradox that was first pres
ented by a Belgian mathematician named Maurice Kraitchik (1882–1957), except that his story was about neckties. Two men argued about whose tie was nicer. They asked a third person, the leading Belgian expert on neckties, to referee and he agreed to the request, but only on condition that the winner would give his tie to the loser as a consolation prize. The two tie owners considered this idea briefly and agreed to it, because they both thought: ‘I don’t know whether my necktie is nicer. I may lose my necktie, but I may win a better one, so the game is to my advantage. Therefore the wager is to my advantage.’ How can both competitors believe they have the advantage?
In 1953 Kraitchik offered another version of the story, involving two other quarrelsome Belgians. They didn’t wear ties, because they were so stuffed with Belgian chocolate that they couldn’t breathe. Instead they challenged each other about the contents of their wallets and decided that the one who turns out richer and happier would give his wallet to his poorer rival. If they are tied, they go back to their chocolates.
Again, both believed they had the upper hand. If they lost the bet, they would receive more money than they had to give if they won. Is this a great game or what? Try playing it with people you don’t know on the street, and see what happens. In 1982 Martin Gardner made the story popular in Aha! Gotcha, one of the finest, simplest and most amusingone of the finest, simplest and most amusing books ever written about smart thinking.
Barry Nalebuf (Milton Steinbach Professor of Management at Yale School of Management), a leading Game Theory expert, offered the envelope version of this story in a 1989 article. You may find it surprising, but even today this game has no solution that all statisticians agree on.