A commenter in another thread wanted to talk about trolls, so I’m opening up a new post where they’ll be on topic.
The specific trolls I have in mind operate toll-booths (or troll-booths?), both of which you must pass through to get from Hereville to Thereville. The question is whether you, as a traveler, prefer to have both booths controlled by a single troll, or by separate trolls. (This is Problem 12 in Chapter 3 of Can You Outsmart an Economist?.)
(SPOILER WARNING!)
The answer from the book is that one troll is better than two.
Here’s why: Trolls are restrained from charging exorbitant fares only by the fear of losing business. That fear is doubly compelling for a troll who owns both tollbooths and therefore stands to lose your business at both of them. Therefore a single owner sets lower fares.
If this had been a textbook, it would have gone into a bit more detail. (Skip this part if you’re not in a textbooky mood.) Suppose it costs a dollar to let one traveler pass through a tollbooth. (If it costs two or three or ten dollars, then just multiply everything to follow by two or three or ten. If it costs zero dollars, some of what follows needs to be modified, but not by much. If it costs an amount that varies with the number of travelers, it needs to be modified a bit more.) Then a profit-maximizing troll who owns both tollbooths and remembers his college calculus will set a combined fare of P1=E/(E-1) where E is the elasticity of demand for passing through both booths. A pair of competing trolls who own the booths separately will set a combined fare of P2=E/(E-2). If E is a constant, then P1 is unambiguously lower than P2 (assuming E>2 so that the equilibria exist in the first place!) so one troll is better than two.
If E is not constant, so that it has value E1 at P1 and a different value E2 at P2, then we have P1=E1/(E1-1), P2=E2/(E2-2). To construct a counterexample where you prefer two trolls, you’d need P2 < P1, which implies 2E1 < E2. Thus, elasticity is a decreasing function of price (and hence an increasing function of quantity!), which suggests that our trolls have only half-remembered their college calculus, and failed to check the second-order conditions for a maximum. For those who enjoy this kind of thing, you might want to fill in the details: Is it possible, despite all of the above, to construct some weird demand function (possibly coupled with a weird cost curve) that will allow P2 < P1 even when all trolls are properly profit-maximizing? Have fun.
This is not just a theoretical issue, but a well-known issue in antitrust law, often called something like “dual monopolization.”
But not well-known enough. It is not rare for there to be a monopoly manufacturer of widgets as well as a monopoly distributor of widgets. Courts and agencies have to have explained to them with some frequency that preventing a merger between those two monopolies (or forcing a split) is not likely to improve things for consumers for this very reason.
This has implications for net neutrality, doesn’t it? Net neutrality advocates tell us that internet service providers have a monopoly, and that they will use that monopoly to direct internet users to subsidiaries or partner companies by throttling their competitors. Isn’t that analogous to a troll who already has one booth (granting access to the internet at large) trying to monopolize a second booth (say, granting access to streaming video)?
I think this argument is telling us there isn’t much to worry about, because the troll who monopolizes the second booth (ISP who monopolizes access to streaming) won’t do anything that makes consumers worse off.
Is this the same argument as you presented in your Snidely Whiplash post years ago? I assume so, because you make the dual toll booths analogy in the comments section of that post.
@GregS,
SL’s original post deals with two “chained” monopolies, as SSA explains.
In the example of net neutrality, you have one monopoly/oligopoly which acts as a gatekeeper for accessing a huge variety of competing content providers. Not the same thing at all.
So on the weird demand function thing. You can make P1=P2 (if I’m understanding this right) by having everyone willing to pay exactly the same price. Let’s say $100, if there is one troll he’ll charge $100, if we have two trolls then they’ll push up the price until it’s $100 between them.
This feels about as extreme as you can get in this direction, so if we’re looking for P2<P1 I suspect we'd need to look elsewhere. If I'm remembering my terms right this example has a stepwise demand function? So it's non-differentiable, does that make elasticity undefined at that point? Maybe another non-differentiable function is the way to find a counter-example.
Another way of looking at this problem is by imagining 3, 4, or n number of trolls.
So in the case of 2 trolls I’m slightly confused about exactly how they set their prices. Presumably they’re adjusting prices independently to maximise their own profit. Is it obvious (true?) that this game always has a unique equilibrium?
That said I think I have an example of a demand function where a single troll is cheaper. Imagine a world with only 3 possible bridge crossers (Billy Goats I guess) called S,M and L. S is willing to pay $7 to cross while M and L are willing to pay $10.
If there is a single troll X he has a choice between charging $7 or $10. If he charges $7 he collects 3*$7=$21. If he charges $10, then S doesn’t cross and he collects 2*$10=$20. So a single troll will charge $7.
Now a second troll Y appears. Y notices that X is charging $7 so he immediately charges $3 (pushing the total price up to $10 a crossing). X now collects $14 total and Y $6. X can’t increase his price (or his income goes to zero) and decreasing it to $4 to get S to buy would decrease his income from $14 to $12 so he doesn’t.
Admittedly I made a bit of an arbitrary assumption here that the trolls start from (7,0). Is that allowed? I don’t think it matters in this example but it might make a difference in others?
@Jonathan, #6: It sounds like in your example, the trolls post themselves serially, so that I have to go through each of them to cross the bridge one way. The way I understand the problem, the trolls control the separate ends of the bridge, and entering the bridge at either end are distinct decisions.
Thanks for the explanation. I feel a little thick, because I don’t know why the 2 trolls set the price at E/(E-2). Nevertheless, it is clear enough that given a flat or linear cost curve, the one troll will charge at most as much as the 2 trolls.
Perhaps I was thinking too much in terms of the real world. I was assuming different cost structures for the different operations. One can easily imagine the two types of outfits would have different cost tables. For instance, maybe the larger organization is inefficient with a small number of customers, but scales well with a bigger base. In reality, also, costs do not necessarily follow a curve. It could easily be the case that the cost per customer is moderate at 100 customers, low at 1000 customers and then increases again at 10,000 customers. For instance, with a bridge, the trolls would face larger costs every time they need to add a toll booth and hire a goblin to operate it. If a booth can support 100 customers, the marginal cost for customer number 100 is relatively low and the marginal cost for number 101 is relatively high.
Below is a sample of prices and costs for a Big Troll (one big enough to control both endpoints of the bridge) and a Small Troll (one that must share with the troll at the other end). I’m not sure how best to post a table, so used a horizontal format. I started from a price/quantity sample table, and modified it so Elasticity is at least 2 throughout the various price points.
Price 20, 18, 16, 14, 12, 10, 8, 6, 4, 2
Customers 1, 2, 3, 4, 6, 8, 12, 18, 30, 60
Revenue 20, 36, 48, 56, 72, 80, 96, 108, 120, 120
Big Troll:
Cost per customer: 4, 4, 4, 4, 4, 3, 3, 3, 3
Total cost 4, 8, 12, 16, 24, 32, 36, 54, 90, 180
Profit 16, 28, 36, 40, 48, 48, 60, 54, 30, -60
Small Troll:
Cost per customer: 6, 5, 5, 5, 4, 4, 4, 3, 3, 3
Total cost 6, 10, 15, 20, 24, 32, 48, 54, 90, 180
Profit 14, 26, 33, 36, 48, 48, 48, 54, 30, -60
With these sample costs, the profit optimizing price for the Big Troll is 8 and for the Small Troll it is 6. The Small Troll is less efficient, it just happens to have a sweet spot that are better for me, the consumer. It is easy enough to dismiss this with the objection that I made up these cost progressions, which I readily admit. That is precisely it, though. When I saw the puzzle in the book, my answer was “it depends on their costs.”
@Henri #7: Yes the way I understand the setup is that both booths are on the same bridge. Looking at it some more I still see it that way. Maybe someone else could clarify this.
“both of which you must pass through to get from Hereville to Thereville.”
This seems as Jonathan says – you must pass through both. If this is the case one troll would be the same as one toll booth.
Presumably if you were going from Hereville to Elsewhere instead of Thereville you may only need to pass through one of the booths. This would seem to complicate things.
Harold, Jonathan:
I reread the puzzle in the book, and it also makes it clear you have to pass through both trolls.
If the price elasticity can be different at the two booths, it seems to me the users must not know the price at the second booth before they encounter it. If they did, they would mentally add the prices together and treat it as one before deciding whether to pass.
@Henri #12
I’m pretty sure they do know both prices in advance (remember this is meant to be modeling a world where a consumer needs to buy two products). I think Steve is assuming a constant price elasticity here, not at different booths but at different total price levels.
I’m fairly sure my 3 goats example in #6 breaks the constant elasticity assumption (with 3 goats I’m not sure price elasticity is even defined?) and hence the conclusion.
@Harold #10: Thanks for clarifying.
and now I notice that the goats example in #6 tells me to prefer the single goat, so really doesn’t work.
Thinking some more in the 2-price type example I was trying you’d need the single troll to prefer the higher price and the two trolls to prefer the lower price, which I don’t think is possible.
Jonathan,
In the last paragraph, Steve asks us to imagine different price elasticities at the two points.
Won’t trolls smart enough to learn college calculus be smart enough to collaborate, effectively becoming a single troll?
I withdraw my previous collaboration comment, which was unworthy even of a troll.
And how about:
>give golden eggs to troll
>fee fie foe foo
Surely someone in this thread knows what that refers to.
@Advo #18
It has been a while since I played Colossal Cave, so maybe I forgot that part?
>>It has been a while since I played Colossal Cave, so maybe I forgot that part?<<
You're in a maze of twisty little cubicles, all alike.
Since we’re talking about trolls, has anyone read Orconomics?
It’s one of the best books I’ve read recently.
https://www.amazon.com/Orconomics-Satire-Dark-Profit-Saga/dp/B01N5H85E0/ref=sr_1_1?ie=UTF8&qid=1543404153&sr=8-1&keywords=orconomics
Thanks, Advo. That looks pretty interesting.
To paraphrase from the second book:
“The Collateralized Threat Obligations will be fine unless the undead army takes the city of XXX. But that’s really unlikely.”
Let’s say there are 1 million people willing to pay $1 to cross the bridge, and 1 person willing to pay $1 million. Then a single profit maximizing troll will charge $1 in order to collect $1,000,001. However, the only two-troll equilibrium prices are where P1+P2 = $1,000,000, since if both trolls are currently charging less than $1, the troll who is currently charging less can raise their price to $999,999 and increase their revenue by at least $499,999.
Of course, this equilibrium might not be discovered by continuously changing price. The revenue curve has two local maxima: P=$1 and P=$1,000,000, and price experiments will likely leave the trolls stuck in one or the other, depending on the initial price, whether there is one troll or two.
It’s possible to construct a scenario where with 1 troll, you end up in a low price equilibrium, and with 2 trolls, you end up in a high-price equilibrium. Suppose the demand curve is:
d(P) = 1000000 if P ≤ 1.000001
d(P) = f(P) = 1/(P-1) if P ≥ 1.000001
Since P*f(P) = P/(P-1) is a decreasing function of P, a single-owner pair of trollbooths initially charging any total price over 1.000001 will drop prices until the total price is 1.000001 (at which point demand is constant so there’s no reason to drop the price further).
On the other hand, suppose I control one trollbooth, and the other is charging price A. since (P-A)*f(P) = (P-A)/(P-1) is decreasing when A 1, if the other tollbooth price is currently >1, I will benefit by increasing prices. If my price is also >1, the other troll will do the same. So if the initial prices are, say, 2 at each trollbooth, we will both continue to increase prices without bound, and in fact there is no equilibrium price since the more each of us increases the price, the greater the incentive for the other to do the same. (As we do this, our customers become increasingly rare, but our total revenues stay pretty much constant as the price increase makes up for the customer rarity.)
If you do want this to reach an equilibrium in the two-troll case, you can add a third piecewise function when P is very large.
Okay, please disregard that last post. I confused myself and managed to construct some cases where you prefer one troll, which is the normal case.
Let’s suppose there’s a pair of trolls charging prices P1 and P2. Suppose further that if they were replaced by a single troll charging P=P1+P2, then the single troll could make more money by increasing price to P’ > P. Let P1′ = P’-P2, and imagine troll 1 in the two-troll case increasing his price to P1′ > P1. Then the total revenue of the two trolls is now higher (because P’ is an improvement in the single-troll case), and the portion of revenue that goes to troll 1 is also higher, so the first troll’s revenue is higher with P1′ instead of P1. Thus the two troll case is always worse for the customer than the one troll case, since any incentive to increase prices that exists in the one troll case also exists in the two troll case.
The above proof used no characteristics of the demand function or assumptions about an equilibrium, so applies even to weird demand functions, even if no equilibrium exists.
David D @25:
The proof shows the one troll would charge the same as the two trolls, not less.
It also ignores their costs. The trolls are not looking for their revenue maximizing price, they are looking for their profit maximizing price.
My proof only shows that two trolls would charge at least as much as one troll, not that they would charge the same. I proved any beneficial price increase in the one troll case would have a matching beneficial price increase in the corresponding two-troll case. However, a price increase which is harmful in the one-troll case may still correspond to a price increase in the two-troll case which is beneficial to the troll making the change, leading to the two trolls charging more.
You are of course correct that I was disregarding costs. But a similar idea works even in the presence of costs. Without costs, we can show that if a price increase of P at one tollbooth improves the total revenue, then whichever tollbooth makes the price increase will see an increase in revenue. With costs, even if the two tollbooths have different cost functions, we can show that if a price increase of P at one tollbooth improves the total profit, then at least one tollbooth will see a profit increase if that tollbooth increases its price by P.
Suppose there are two trolls currently making revenue R1, R2 and paying costs C1, C2, respectively, and a price increase of P by either troll increases the total shared profit (so it’s an improvement in the monopoly case). Let’s suppose a price increase of P by troll 1 changes the revenue/cost of each troll to R1′, R2′, C1′, and C2′, respectively, and a price increase of P by troll 2 changes the revenue/cost of each troll to R1″, R2″, C1′, and C2′. The new costs are the same either way, since the same total cost results in the same quantity demanded. Since the price increase is an improvement in the monopoly case, we know R1″+R2″-C1′-C2′ > R1+R2-C1-C2, and since the troll who increase their price increases their revenue share, we know that R1′ > R1″.
If R1′-C1′ > R1-C1, then troll 1 has a profit increase resulting from his price increase of P. Otherwise, R1’+R2-C1′-C2 = (R1′-C1′)+(R2-C2) ≤ (R1-C1)+(R2-C2) = R1+R2-C1-C2 < R1"+R2"-C1'-C2' < R1'+R2"-C1'-C2', and hence R2-C2 < R2"-C2', meaning troll 2 has a profit increase resulting from his price increase of P. Either way, at least one troll benefits if that troll increases their price by P.
So even when we include costs in the equation, the customer can never be better off with 1 troll than two.
@David D #27:
“My proof only shows that two trolls would charge at least as much as one troll, not that they would charge the same” Yes, but the exercise is to show that two trolls would charge more than one. Steve did it above for the revenue side.
“The new costs are the same either way, since the same total cost results in the same quantity demanded.” I don’t see why the costs are the same. It would be true with a flat cost curve, but then that is an assumption we need to call out and it leaves us with the other cases.
My post answered steve’s final question, “Is it possible, despite all of the above, to construct some weird demand function (possibly coupled with a weird cost curve) that will allow P2 P1 given reasonable niceness assumptions about the demand curves, and P2 can equal P1 for some demand curves, but my above proof shows that P2 ≥ P1 in general (if equilibrium prices exist so that P1 and P2 are well-defined).
“‘The new costs are the same either way, since the same total cost results in the same quantity demanded.’ I don’t see why the costs are the same.”
My assumption is cost is a function of quantity, and quantity is a function of total price, so the costs to each troll are the same whether the prices are (P1+P, P2) or (P1, P2+P).
The blog software seems to have eaten part of my post (presumably due to use of less than and greater than symbols). The first paragraph was meant to read,
My post answered steve’s final question, “Is it possible, despite all of the above, to construct some weird demand function (possibly coupled with a weird cost curve) that will allow P2 < P1 even when all trolls are properly profit-maximizing?” I agree P2 > P1 given reasonable niceness assumptions about the demand curves, and P2 can equal P1 for some demand curves, but my above proof shows that P2 ≥ P1 in general (if equilibrium prices exist so that P1 and P2 are well-defined).