These sites got big a couple years ago, and there are some great articles at codinghorror.com about how they are a scam and basically a lottery.

This would be fine if the sites were regulated like a lottery. But in fact, they advertise on TV as legitimate auction sites.

Here’s a video that summarizes some of the issues with penny auctions:

Video: Don’t Waste Your Money – Penny auction scams

Some of the main objections to these sites are:

• the timers are different: the auction time keeps increasing as people bid
• it costs money to bid, and often you end up with nothing
• there may be costs just to sign up

Another issue to add is:

• these sites often offer free bids for people to sign up. These free bids increase the length of the auction (as the timer resets) which forces paying bidders to have to place more bids, generating extra revenue for the site

The bottom line: avoid penny auctions unless you are looking to do something similar to gambling and playing the lottery.

One of my favorite topics related to game theory is the subject of auction theory.

The results from auction theory are very interesting, but I have yet to cover much about it because the math can be quite intimidating.

Today’s problem is still challenging, but it should be within the reach of a math enthusiast. Here is the puzzle:

Alice wants to auction off a rare collector’s item. She knows the item is worth somewhere betweeen $500 and $1,000, but she has had trouble finding interested buyers.

A company offers to find interested participants at the rate of $10 per bidder. (So they’ll find one bidder for $10, and ten bidders for $100)

How many bidders should Alice tell the company to find?

A couple of points:

–Assume the bidders have valuations randomly drawn from the uniform distribution on [500,1000]

–Suppose Alice holds an eBay style auction and she will sell the item for a price equal to the second highest valuation of the bidders*

(*this is a standard result in auction theory, though technically it’s for one bid above the second highest valuation. An example: if bidders had valuations of $500, $600, and $700, the person who values the item at $700 would win the auction. The price he would pay in an eBay style auction with dollar bid increments is $601–just enough to outbid the person with the second highest valuation)

The puzzle is about two conflicting forces: Alice wants more bidders to bring her higher bids, but she faces a tradeoff in the cost of acquiring bidders.

Can you figure out the optimal number of bidders?

The answer is written below.

Spoilers below!
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Answer to the puzzle

Alice wants to maximize her expected auction profits. The equation for profits for n bidders is something like this:

Profit(n)= E(revenue n) – Cost(n)

The cost part is easy to figure out. Alice pays $10 per bidder, so her cost is 10n.

The harder part is figuring out the expected revenue for n bidders. What we want to know is the following. If we take n draws from a uniform distribution, what is the expected value of the second-highest draw?

This question is actually part of a larger topic in probability called order statistics. One can explicitly solve for the expected value of any distribution using the technique outlined in this lecture (pdf).

I will not go through the math here. But I will mention the order statistics for the uniform distribution are easy to visualize. What happens is that if you take n draws from the uniform distribution, the expected value of the n draws can be visualized as n points being evenly spaced on the interval.

Here is a picture to illustrate what I mean:

So the n points separate themselves along the interval. So you divide the interval into n+1 segments, and the points will be at the fractions 1/(n + 1) along the way for the minimum, then 2/(n + 1) along the way for the second lowest point, etc., until the maximum draw which has an expected value of n/(n + 1).

By this logic, the second highest draw is expected to be at (n – 1)/(n + 1) along the way from 500 to 1000. This means the second highest valuation is expected to be:

500 + 500 * (n – 1)/(n + 1)

This is our formula for expected revenue. So we can substitute this expression back into the formula for profits:

Profit(n)= E(revenue n) – Cost(n)
Profit(n) = 500 + 500 * (n – 1)/(n + 1) – 10n

Now we need to solve for the profit maximizing point. Let’s just use a shortcut and plug this into WolframAlpha:

We can see that the profit maximizing point is at n = 9 bidders, and Alice can expect $810 of profit.

The lesson is that more bidders is not always optimal: you capture much of the expected revenue from the first few bidders, and then the returns are diminishing (unless some bidder is a big outlier and you can extract money from him).

Extension: suppose Alice earned the highest valuation

As an extension, let’s imagine that Alice somehow was able to extract the highest bidder to pay his entire valuation. This is not an assumption used in theory, but let’s say it happens because of some irrational bidding war.

In that case, Alice would expect to earn slightly revenue (the term (n – 1)/(n + 1) becomes n/(n+1)), meaning her profit function is:

Profit(n) = 500 + 500 * n/(n + 1) – 10n

How will that change the number of bidders?

Here is the answer:

So Alice will only need to acquire 6 bidders, but she will earn nearly $870. This is 3 fewer bidders than above and she gets about $60 more.

This is, of course, exactly what we would expect: if Alice can extract more money from the bidders–the highest valuation instead of the second–she does not need as many bidders and she can earn more out of it.

This is common sense, but it’s useful to check the theory matches intuition.

My friend Aaron emailed me about another interesting problem from the show:

I’ve got a good problem for you to throw at your readership involving the Price is Right “Wheel” game. I assume you’re familiar with it but in case not…

–Three winners of earlier pricing games get to spin a wheel with all multiples of 5 from 5 to 100 inclusive [there are 20 numbers in all].

–The goal is to get as close to 100 as possible without going over.

–Contestants can choose whether or not to use a second spin which would “add” to the first spin but again, but if the total exceeds 100 the contestant is eliminated.

–The first contestant completes his turn (whether it’s 1 or 2 spins), then the second contestant goes (whether it’s 1 or 2 spins), and then the third contestant goes.

Question is: How high must Contestant 1′s first spin be in order for him to not make a second spin? What about Contestant 2?

I love this email, especially because this is a direct application of game theory. The game also brought to mind a few follow up questions, so there are my questions I pose to you.

1. What is the optimal strategy for each player?

2. What are the respective winning percentages? How much of an advantage is there to spinning second or third?

Also, you may know there are special bonus payments in the wheel game. If your spin total(s) reach 100, then you get a bonus payment of $1,000. This also entitles you to a bonus spin. If you get 100 on that, you win another bonus of $10,000. If you get 5 or 15 points (the numbers next to 100), then you get a bonus $5,000.

3. How do bonus payments affect the optimal strategy?

The first three questions deal with a theoretical model. This may not be exactly how people play in practice.

4. How do you think actual contestants play in comparison to the theory? Will they play close to optimal strategies, or will they play more conservatively or aggressively?

You may wish to try a two-person variant of the game to understand how the strategy works. The only suggestion I have is to use a spreadsheet or computer program because the number of strategies lends itself to numerical analysis.

Can you solve it? Aaron emailed me an answer for the two-person game, but the three-person game is much harder (admittedly I could not solve it and I resorted to finding a resource online).

Post your thoughts as usual. The answer is posted in my comment below.

Rather than run away, Neo chooses to fight Agent Smith head on. The scene begins with a duel which I find particularly interesting due to the strategies they take.

Here is a link to the video. The duel that I am going to discuss is roughly the first 30 seconds of the video.

Youtube video: The Matrix subway fight

The detail I find most interesting is the order in which the two shoot their bullets.

It appears that Neo takes the first shot, but then they mostly fire at almost the same time. This becomes clearer as Neo and Agent Smith run towards each other. When they jump in the air and approach point blank range, the clip moves into super slow motion and it is clear they fire their bullets at exactly the same time.

In the film the simultaneous firing has the beauty of symmetry and the timing was probably chosen for aesthetics.

But one can ask: does it make sense that Neo and Agent Smith fire at the same time?

At this point in the film, Agent Smith appears to be more skilled than Neo. If we were thinking about the strategy of a duel, should the more accurate shooter fire first, or should he wait?

What’s the right time to shoot in a duel?

A simple dueling game

Consider a duel between two players A and B.

For simplicity, suppose each only has one bullet. The game works as follows.

A and B start the duel at a distance t = 1 from each other. They approach each other at the same speed, and each has to decide when to shoot.

As they get closer to each other, their accuracy increases. At distance t, the player A has a chance a(t) of killing his opponent, and for player B it is b(t). Assume both players are aware of the other’s skill.

In this duel, missing your shot is very costly. If a player shoots and misses, then the other player keeps approaching until he gets to point blank range and shoots with complete accuracy.

What is the optimal strategy of this game? That is, at what point should each player shoot?

Solution: when player A should fire

The tricky part to the game is balancing when to shoot. If you fire too early, then your opponent kills you for sure. If you wait too long, then you can also get beaten if your opponent is a good shot.

We can think about when player A should fire by listing out the chance of surviving in the different possibilities of firing at point t.

• If player A fires first: Player A will survive only if he hits, which happens with chance a(t)
• If player A waits to fire: Player A survives only if player B misses, which happens with chance 1 – b(t)

Now we can reason out player A‘s strategy. Player A will want to fire first if his chance of hitting is greater than player B‘s chance of missing:

a(t) ≥ 1 – b(t)

But he must also be careful not to fire too early. He should always wait if his chance of hitting is smaller than player B’s chance of missing:

a(t) ≤  1 – b(t)

Putting those two equations together, we can see that Player A should shoot at the time when he is at distance t* where

a(t*) = 1 – b(t*)

Or alternately written,

a(t*) + b(t*) = 1

Solution: when player B should fire

We can do the same exercise for player B. Notice the same conditions are true:

• If player B fires first: Player B will survive only if he hits, which happens with chance b(t)
• If player B waits to fire: Player B survives only if player A misses, which happens with chance 1 – a(t)

Now we can reason out player B‘s strategy. Player B will fire first, if his chance of hitting is better than his opponent’s chance of missing:

b(t) ≥ 1 – a(t)

But he must also be sure not to fire too soon. He needs to wait so long as his chance of hitting is smaller than his opponent’s chance of missing:

b(t) ≤ 1 – a(t)

Putting those two equations together, we can see that Player B should shoot at the time when he is distance t* where

b(t*) = 1 – a(t*)

Or alternately written,

a(t*) + b(t*) = 1

Solution: they fire at the same time!

From the equations, you’ll notice that both players choose to fire at the same time! There is one specific distance which is optimal for both players.

This would not be surprising if the two players had the same accuracy level. But we solved this game using the assumption their accuracy levels were different.

So why do they end up shooting at the same time?

We can reason why this must be the case. If one person chose to fire earlier than another, say 5 seconds earlier, then he would be better off waiting. His opponent is not shooting for another 5 seconds, so he might as well wait a few more seconds to get closer and increase his accuracy.

As the equations show above, the right time to shoot is just when your chance of hitting equals your opponents chance of missing. And since one person’s failure is another person’s success, this means both players choose the same time when they are a distance such that their accuracy functions sum to a probability of 1.

The Matrix duel

While Neo does fire the first bullet, the two do appear to fire shots at almost or exactly the same time as they run towards each other.

But then again, it should come as no surprise that Neo and Agent Smith play smart strategies given their superhuman abilities.

Additional reference

The theory of duels is covered in much more detail in this game theory lecture at Yale.

Lecture 16 – Backward induction: reputation and duels

The relevant part about duels–he uses two players with wet sponges–starts about 25 minutes into the video.

Youtube video: Lecture 16 Game theory at 25 min 28 seconds

Is anyone favored to win this game?

What happens when the game is repeated?

The answers are after the break.

A matter of perspective

At first glance, the game seems favorable for Bill Gates.

He might think as follows. He could either have more or less money, so he has a 50/50 chance of winning the game. If he loses, then he would lose only the money in his wallet. If he wins, however, then he would win more money than what he has in his wallet.

That is, Bill Gates can wager the money in his wallet (say x dollars) and have a 50 percent chance of winning the money in Warren Buffett’s wallet, which is more than x dollars. This is obviously a winning gamble, and thus Bill Gates will want to play as he seems favored.

On the other hand, Warren Buffett thinks about the game. By symmetry, he can reason exactly the same thing. He feels that he has an even chance of winning more money than he has in his wallet, and thus he is favored by the game.

Both billionaires find the game advantageous. And yet this cannot be. The game is a zero-sum game (the profits of one are exactly the losses of another) and thus it is impossible for both to expect a profit at the same time.

How can we resolve the reasoning and resolve the apparent paradox?

Careful accounting

The setup is similar to a wager I previously wrote about called the necktie paradox. The wallet paradox can be resolved in a couple similar ways.

Resolution 1: fix the reasoning

Bill Gates and Warren Buffett are unlikely to reason so wrongly. They would see the error in the above logic.

The problem with the reasoning is that each is considering his wallet as being both the one with more money and less money at the same time. In reality, a wallet can either be the one with more money or it can be the one with less money.

The correct logic is:

–If I have the wallet with more money, and I play the game, then I will lose my more valuable contents

–If I have the wallet with less money, and I play the game, then I will win the more valuable contents of the other wallet

There is an equal chance of being in either situation. Therefore, the game does even out as either player can win or lose the wallet with more money. The game could end with Bill Gates winning half the time, and Warren Buffett winning half the time. Thus neither is favored and the game is fair.

Resolution 2: make a proper expectation

Another way to see the game is fair is to write out the math carefully. Suppose the more valuable wallet has x dollars.

To write the expecation, it is necessary to know two things.

First, what is the chance a player wins the bet? This is the chance the person’s wallet has more money. Assuming the game was not rigged, either Bill Gates or Warren Buffett could have more money than the other. It would be reasonable to assign a 50 percent chance on either being the winner of the game.

Second, what are the payoffs in the game? One case is when a player wins the game, in which case he wins x dollars from the other player’s wallet. The other case is when a player loses, so he loses the more valuable contents of his wallet of x dollars.

Putting these two facts together, we get:

The math demonstrates the game has a zero expected payoff to either player, and thus the game is fair.

Adding a twist with repeated play

Afterwards, it turns out both Bill Gates and Warren Buffett loved the game.

They add a twist as follows. They will play the game every day for a month. To manage the stakes, they agree to carry an average (mean) of $100. Is this game fair? What is the best possible strategy, given you know what your opponent is doing?

Winning strategy in repeated play

The game gets more interesting when repeated. Even when players agree to carry the same average money, the game is not necessarily fair. There will be times that one person will be favored to win.

Consider the example of Bill Gates and Warren Buffett agreeing to carry $100 on average. Suppose they limit their strategies to the following three:

Strategy A: carry $100 every time
Strategy B: carry $75 half the time, and $125 half the time
Strategy C: carry $75 two-thirds of the time, and $150 one-third of the time

We can evaluate how one strategy performs against another.

It turns out there is a rock-paper-scissors type of situation: strategy A is profitable against B, which is profitable against C. But then strategy C is profitable against A!

This cyclical nature gives a hint about the nature of the game. More on that in a bit. First, here is the math that confirms it.

Strategy A vs. B:

Strategy A means always carry $100. The way the game plays out depends on the realization of strategy B.

The wallets can either have amounts (100, 75) or (100, 125), each with probability 0.5.

In the (100, 75) case, the second wallet has less money, so employing strategy A means a loss of $100.

In the (100, 125) case, the first wallet has less money, so employing strategy A means a profit of $125.

The expectation of employing strategy A against B is thus: 0.5 ( -100 + 125) = 12.5 > 0. Thus strategy A is profitable against B.

Strategy B vs. C:

Calculating this expectation is just a bit more involved, as both strategies are probabilistic.

There are four possible outcomes, each with different probabilities of occurrence. Here is how strategy B fares:

(75, 75) –> happens 1/2 * 2/3 = 2/6, and is a draw
(75, 150) –> happens 1/2 * 1/3 = 1/6, and strategy B profits 150
(125, 75) –> happens 1/2 * 2/3 = 2/6, and strategy B loses 125
(125, 150) –> happens 1/2 * 1/3 = 1/6, and strategy B profits 150

Putting this together, the expectation is (2/6 * 0 + 1/6 * 150 + 2/6 * -125 + 1/6 * 150) = 8.33… > 0.

Thus we can see that strategy B is profitable when played against strategy C.

Strategy A vs. C:

There are two possible outcomes, each with different probabilities of occurrence. Here is how strategy A fares:

(100, 75) –> happens 2/3, and strategy A loses 100
(100, 150) –> happens 1/3, and strategy A profits 150

Putting this together, the expectation is (2/3 * -100 + 1/3 * 150) = -16.66… < 0

Thus it is seen that strategy A is a losing decision against C!

There is no one best strategy

The fact that strategy A is preferred to B, B is preferred to C, and C is preferred to A is a clue that the game might not have a single winning strategy. This hunch turns out to be true.

For any given strategy* with a fixed mean, one can find another strategy with the same mean that is profitable and is better. (*there are some technical restrictions that the strategy is well-defined)

The details and rigorous math can be found in this paper: The Wallet Paradox Revisited.

As a side point, this means the game has no Nash equilibrium. These types of games can be fun an interesting. Here is another game I wrote about with a similar flavor: the dollar auction game.

And I will close with something obvious but noteworthy. If someone asks you to play the game, you should definitely be skeptical. They probably have little or no money in their wallet and are hoping to make a quick buck.

It is only sensible to play the game in a fun party setting when both people are taken by surprise. It’s even better after a night of drinking where both people are less likely to remember exactly how much is in their wallet.