Alice draws a line connecting two of the points with a line segment. Bob and Alice move in turn, but the line segment cannot cross any of the previously drawn lines.

A player who cannot draw a suitable line loses.

What’s the winning strategy and who has it?

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The answer

The problem is easier to analyze when considering smaller cases.

Suppose there are 4 points around a circle numbered in clockwise fashion 1, 2, 3, 4. Let’s say Alice draws a line between the points 1 and 3. The points 2 and 4 are separated on opposite sides of the line. Bob’s line would intersect Alice’s, and so Bob loses on his first turn.

If there are 6 points, then Alice should draw a line that connects points 1 and 4. Bob can then connect points 2 and 3, or 5 and 6. Either way, Alice can connect the points that Bob does not pick, and then Bob cannot draw any more lines so he loses.

It turns out this pattern continues, Alice can win with 100 points if she connects the points numbered 1 and 51 on her first turn–I’ll leave it to you to fill in the details

Alice writes a “+” or “-” sign in front of one of the numbers, and then Bob and Alice take turns.

Once the four signs are written, the arithmetic expression is then evaluated. Bob gets points equal to the absolute value of the result.

What is the best strategy and the value of the game?

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My attempt at the solution

I took a long time to solve this problem because the solution was not apparent to me. I actually wrote out a gigantic game tree in a spreadsheet and solved for each players best response manually.

(So I should mention I am fairly sure this is the answer, but I could have made a mistake in the process–I’m sure someone will point out if there’s a flaw in my argument)

Because Bob moves second, he’s at an advantage. What Bob wants to do is move the total as far away from 0 as possible.

The best strategy is to make sure that the largest terms, 3 and 4, are of the sign. So whenever Alice writes a sign for one of those terms, Bob must immediately match the sign on the other. For instance, if Alice writes +4, then Bob must write +3 right away.

This will ensure that the expression gets 7 units away from 0.

Alice, for her part, can at least throw in one term that is the opposite sign from the pair 1 and 2, to lower the absolute value to 6.

Here is an example of gameplay:

Alice writes +3
Bob writes +4
Alice writes -2
Bob writes +1

Total of game: 6

Other combinations end up at 6 as well with strategic play on both sides.

The game is quite favorable to the second player, so they should definitely take turns going first.

Extension: What would happen if the game were played with numbers up to 6, 8, or 10? What about in general 2n?

I have yet to find an answer, but I bet there’s an elegant solution.

Two generals have 5 units each to deploy. Each person decides how many units to send to battle. The general who sends more troops will win, but it’s a draw if they both send the same number.

Each also has the option of “passing” which averts war and ends in a draw.

What’s the best way to play this game?

William Spaniel shows how to solve this game in his video series Game Theory 101. (video after the jump)

This is a nice explanation of the concept of using best responses to find Nash equilibria.

Video: Safety in numbers and best responses

If you liked this video, note that William Spaniel has written a great introduction to game theory which I have previously reviewed. Check it out.

You and another student are called up to the front of the classroom to participate in an auction.

Here is how the auction works:

–the prize money equals the sum of money in both of your wallets

–the winner is determined by an open outcry auction (suppose the professor announces selling prices that increase by $1 increments, and the game goes until someone says “I quit”)

–the winner has to pay the final price in exchange for the prize

Let’s say you are chosen for this game, and you are allowed to peek in your wallet before the auction begins.

You see that you have $10 in your wallet. How much would you be willing to pay in the auction?

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The answer

It is apparent you should be willing to bid at least $10. At that price you will at least break-even. But you should probably be willing to spend a little more so that you can profit sometimes too, if the other person has money but backs out early.

It turns out there is a general rule that is an equilibrium. If both people use the same bidding technique, then each should be willing to spend up to double the money in his wallet, and you should be willing to bid up to $20.

Why is that the case?

Let’s analyze the bidding behavior. Suppose you know the other person will bid up to twice the amount in his wallet. What’s the maximum amount you should be willing to pay?

To begin, we set up some variables. Let’s say your wallet contains x, the other person’s wallet contains y, so that the prize money equals x + y.

If you lose the auction, you walk away with nothing so there is nothing to analyze. So let’s consider the case that you win the auction at a price of p. What’s the maximum price you want to pay?

Notice that if the other person bids up to two times their wallet, 2y, then it must be the case that you won at 2y = p.

We want to solve for the highest price p you will pay such that the prize money x + y exceeds the price you win at p. The condition we want is:

x + y = x + 0.5p > winning price of p
if and only if p < 2x

In other words, if the other person bids up to double, you should also be willing to bid up to double and win profitably.

In real life, of course, things are not so simple. You have to consider the other party may use a different bidding strategy (say bidding up to a lucky number). But still, this is a reasonable rule of thumb and at least good enough to use for a game theory class.

This type of game is called an “almost common value” auction: the prize is a common value to all parties, but each party has good information on his own wallet to glean a bit more information.

As further reading, there are some neat applications of this auction in Telecom written about in Paul Klemperer’s 1997 paper Auctions with Almost Common Values: The Wallet Game and its Applications.

If you’re never seen the show, here is how it works. Each of two contestants independently chooses to split or steal the final prize. If both choose split, then the prize is divided evenly. If one chooses split and the other steal, the person who steals gets the entire prize. If both choose steal, however, then both walk away with nothing.

Here’s the normal form representation of the game:

How should you play this game?

One contestant had an amazingly brilliant strategy that I will discuss below.

The wrong way to play the game

Contestants are allowed to discuss strategy before picking split or steal.

Both realize that split gives a fair 50 percent share to each side, but each also sees the advantage of back-stabbing and stealing the prize.

The discussion usually involves the following strategy. Each person tries to convince the other person to split, and they promise to do the same.

I discussed an example of this in a previous post: strategy in Golden Balls.

In that episode, both were promising they would split the prize, but then one person decided at the last minute to steal all the money. She said she was not proud of the decision, but she herself did not want to be cheated.

So trying to split the money in a conventional way doesn’t work. Is there a better strategy?

Why it’s bad to promise you will split

First, I want to explain why the strategy of splitting does not work. When you promise the other person you will split the prize, you are trying to change the game.

You are telling them that instead of looking at the original payoffs, they should only consider the game under the assumption that you are going to split the prize. So you are telling them to consider the following game:

Do you see what’s wrong with your strategy? If you promise them that you will split the prize, they are faced with a very tempting option to steal. If they split, they will only get 50 percent. But if they steal, they will get the entire 100 percent.

And therein lies the problem: if you promise you’ll SPLIT the prize, then you pretty much are telling them not to worry about the mutual steal option. This makes it a very good idea for them to STEAL the prize.

Clearly it’s a bad idea to promise that you’ll split the prize. Is there another way out?

How to beat the Prisoner’s Dilemma

There’s a remarkably devious way to get cooperation: you must tell them that you will STEAL the prize!

How does this strategy play out? You should watch the following clip to see the negotiation:

Brilliant strategy in Golden Balls

The action proceeds as follows. One contestant, Nick, immediately announces

I want you to trust me. 100 percent I am going to pick the steal ball. I want you to choose split, and I promise you that I will split the money with you [after the show].

The other contestant is completely stunned by this strategy, and the audience finds it amusing too.

The next 2 minutes is a funny exchange between the two. Nick keeps explaining why he is going to steal, and the other is dumbfounded by this terrorist-like action. He wonders, why can’t they both split?

The host reminds them the plan is risky, as there is no legal requirement for the money to be split after the show is over.

Nick is called an idiot and the other contestant just can’t believe he expects cooperation. Nick has taken control of the game, and he has not acted nice. Why should the other person cooperate?

Nick promises over and over that he is an honest person and that he will definitely split the money after the show.

At the 5:21 mark in the video, they reveal their choices. It turns out that both of them choose to SPLIT after all! Thus both end up with a 50 percent share of the money.

Here’s why Nick’s strategy was so brilliant. Nick was credibly explaining that he was going to steal. This changed the game into the following payoffs:

On the one hand, the other contestant could steal and destroy the prize money. On the other, he could split and hope that Nick kept to his word. In other words, Nick has transformed the game so that the weakly dominant best response is to split!

The other contestant is happy at the outcome, but he shouts “Why did you put me through that?” as he really had to struggle over his decision, only to learn Nick would cooperate after all.

I think Nick has shown a brilliant way to beat the Prisoner’s Dilemma in a one-shot game. His statement that he would steal is a credible threat, and it changed the game so the other contestant found split to be the appealing option.

This might not work in a repeated game, as Nick would have burned his reputation by not keeping to his word.

But in a one-shot game, it was a smart way to assure that both go away with a split of the prize.