Game Theory Question

[viciouspenguin]

There are two firms in a market that produce an identical product. Each firm has either one or zero units to sell. The probability of having a unit to sell is q and the probability of having no units to sell is 1-q. There is a single consumer interested in buying only one unit of the good at a price not to exceed $1. If both firms have capacity available, it will buy from the lowest priced. If only one firm has capacity, it buys from that firm provided its price does not exceed $1. You must decide what price to charge for the good if you were to have a unit available to sell. Note that at the time of making this decision you do not know whether your competitor will have a unit of capacity to sell or not and what price it will choose.

What are the optimal price choices for the following 3 cases?

1. q = 1/4
2. q = 1/2
3. q = 3/4

[m_the0ry]

Question; q represents the probability of having a unit for both firms, or is the probability of one firm having a unit unknown?

[bigmonkey]

Interesting question. I'm finding it hard to work it out straight away so I'll just post my thoughts on it (the q=1/4 one). Firstly we don't know what happens if they charge exactly the same price and they both have it in stock...? I assume that the consumer flips a coin to choose who he buys it from, equivalent to the sellers selling half a product each.

If you charge a dollar then you have EV of at least 18.75c. That's the chance that you have it in stock and your opponent doesn't multiplied by your dollar.

It seems what you want to be doing is predicting what your opponent will charge then go 1c below that (let's assume you have to charge to the nearest cent). If your opponent charged $1 and you charged 99c your EV would be 24.75c. I think that is the highest prize.

I've done a table with values to the nearest 10c as any more is too much effort.

http://img509.imageshack.us/img509/3489/gametheorychartqq4.png (http://imageshack.us)

The yellow fields represent strategies that both agents know that neither player will take because they are dominated by strategies above them. But 80c isn't dominated by 90c because there are some possibilities where you get paid more for 80c than for 90c.

Now with these 3 strategies available each this is what we are left with: If I go 100 I want my opponent to go 100. If I go 90 I want my opponent to go 100. If I go 80 I want my opponent to go 100. Also if I go 100 he ought to go 90. If I go 90 he ought to go 80 and if I go 80 he ought to go 100.

Hence there is no equilibrium here. But if we believed that he will randomly choose between these three strategies then we are better off going with one dollar. It basically becomes like stone, paper, scissors then with him thinking you'll go 100 so he goes 90 so you go 80 so he goes 100 etcetera.

I'm not that knowledgeable of game theory, so I'd be interested if anybody does this problem a shorter way than me. Or if they can make some equation and observations based on my chart to do q equalling the other values. Also, at which point to the nearest cent does the strategy become dominated?

[HP]

I am assuming if both firms choose the same price, the consumer flips a coin to make a choice

For any positive q:

It would appear there is a co-operation/competition effect between the two firms resulting in only one Nash Equilibrium at both of them setting a price of 0

[HP]

however I'm not sure about that

[HP]

okay i've changed my answer:

There is a Nash Equilibrium only at setting the price at 0 for each of the firms, for any q except q = 1/2

In the q = 1/2 case, both firms should choose a price P from the uniform distribution over [0,1]

my answer is subject to further change...

[HP]

update on my next answer lol

when q > 1/2, The only Nash Equilibrium is both firms selling for a price of 0

when q = 1/2, Both firms should choose a price from the uniform distribution over (0,1)

when q < 1/2, not sure yet....

edit: nope this ain't right either. allow me to start over

[HP]

Alright my latest new answer:

each firm chooses a mixed strategy. If p(x) represents the probability density function of the price a firm chooses, p(x) is the following:

(1-q)/(q*x^2) when (1-q) < x < 1

and p(x) is 0 everywhere else

This is my final guess!! At least until some one else posts in this thread

and yeah this is for 0 < q <= 1

When q = 0, it would seem both firms choose a price of 0

edit: I'm talking about Nash Equilibriums obv

edit2: wrote a fraction down wrong

[HP]

btw that should read:
[ QUOTE ]

and yeah this is for 0 < q < 1

When q = 1, it would seem both firms choose a price of 0

[/ QUOTE ]

[bigmonkey]

I didn't really understand what your eventual answer was. What does the probability density actually tell us about the strategy they should employ? Also rational agents don't always have to obey a Nash equilibrium do they?

[janneman]

I don't really know anything about Nash equilibria, but if q=3/4 then by picking price=1 your company will make 1/4 on average. So regardless of any assumptions it seems that the price has to be at least 1/4 for it to be a better strategy than this trivial one.

Assume your opponent's price is uniformly distributed in [0,1]. Call your price P. On average you will make:

(1-q)P+q(1-P)P=P-qP^2
(parabola with maximum at P=1/(2q))

so if q=1/4: P_optimal=1
if q=1/2: P_optimal=1
if q=3/4: P_optimal=2/3

I don't know however if you're allowed to make this assumption (i guess not if there's game theory involved).

[HP]

[ QUOTE ]
I didn't really understand what your eventual answer was. What does the probability density actually tell us about the strategy they should employ?

[/ QUOTE ]

well, suppose q = 1/4

my answer becomes:

g(x) = 3/x^2 for (3/4)>x>1

g(x) = 0 everywhere else

It looks like:
http://img162.imageshack.us/img162/9237/picuu7.png

So if we break it down to blocks of 5 cents, we set the price:

75c-80c : 25.0%
80c-85c : 22.1%
85c-90c : 19.6%
90c-95c : 17.5%
95c-100c: 15.8%

[ QUOTE ]
Also rational agents don't always have to obey a Nash equilibrium do they?

[/ QUOTE ]

Quite right no they don't

I am assuming the other firm is rational, and they expect us to be. I find the problem much more interesting when we are assuming this, and I'm assuming viciouspenguin does too

I didn't mention this before, but if both firms use this strategy, your EV will be:

1-q (given you have a unit to sell)

So I guess, taking into account the chances you don't produce a unit, the EV is:

q(1-q)

[HP]

[ QUOTE ]
So I guess, taking into account the chances a firm doesn't produce a unit, the EV is:

q(1-q)

[/ QUOTE ]

Which would imply, if it doesn't cost the firm anything to build a unit, they should each set q to be 1/2 to maximize profits

nice problem viciouspenguin!

[HP]

[ QUOTE ]
Which would imply, if it doesn't cost the firm anything to build a unit, they should each set q to be 1/2 to maximize profits

[/ QUOTE ]

actually, as long as the the cost of making the unit is less than 25c, each firm should always set q=1/2

if it's more than 25c to make, it's not profitable unless they co-operate (not a Nash Equilibrium)

[tabako]

Quoting myself from the crosspost in OOT...

[ QUOTE ]

I don't think there are any NE in pure strategy.

For q = .5, I think the mixed strategy p = U~[.5,1] might work, though I am not sure how to verify if it does.


[/ QUOTE ]

HP, why are you talking about firms setting q? The question is asking what the optimal price level is for a given q (that is the same for both firms). Also, for any q =! 1, there is no pure strategy NE as the BR don't overlap. I am almost positive that the answer will have to be some sort of probability distribution, as otherwise the other firm would want to slightly underbid your price (or choose price = 1 if the price you chose was low enough). The best response functions never cross one another.

Here is xorbie's post on the same:
[ QUOTE ]

at q = 1/2, price of 1 has EV of .25, thus nobody should ever charge < .5. however, (.5,.5) gives each an EV of < .25, so it is not a NE. BR(.5) = 1. however, BR(1) = .99 and this continues down to BR(.51) = .5. thus no NE, not for any of these.

you may want to verify, this was just a quick sketch.


[/ QUOTE ]


Can anybody help me figure out whether the strategy of randomly choosing a number between 1/2 and 1 is a NE if played by both firms? (for q = .5)

[bigmonkey]

[ QUOTE ]


It looks like:
http://img162.imageshack.us/img162/9237/picuu7.png



[/ QUOTE ]

Thanks. The graph helps. I was basically looking for something that supported my table. I figured there would be some kind of preference for the lower prices in that strategy range even though there's slightly more EV given your opponent acts randomly in that range by going higher.

[HP]

[ QUOTE ]
The question is asking what the optimal price level is for a given q (that is the same for both firms).

[/ QUOTE ]
I answered this question, for any given q. I posted a graph of the answer for when q = 1/4 above

I can tell you the answer for any other value of q if you like. You simply use that general formula I posted

[ QUOTE ]

HP, why are you talking about firms setting q?

[/ QUOTE ]
After solving this for the general case of q, you can see that if the firms got to choose their own q, it would maximize profits if they chose q = 1/2. This was not what the OP asked, but I thought it was interesting to note

[ QUOTE ]
I am almost positive that the answer will have to be some sort of probability distribution, as otherwise the other firm would want to slightly underbid your price

[/ QUOTE ]
My answer is a probability distribution, not a pure strategy. So yes, I agree

edit: also, given that he crossposted in OOT, I am beginning to suspect I may have just done someone's homework. Oh well, it was interesting anyway

[tabako]

Sorry I missed it among all of your posts.

Can you explain how you got to p(x) = (1-q)/(q*x^2) when (1-q) < x < 1

[HP]

[ QUOTE ]
Can you explain how you got to p(x) = (1-q)/(q*x^2) when (1-q) < x < 1

[/ QUOTE ]
Do you want to know how I figured it out, or just why it is correct?

[tabako]

How you figured it out.

[HP]

[ QUOTE ]
How you figured it out.

[/ QUOTE ]

thought so just wanted to make sure so i don't waste any effort. this will take me a while to write out, in the middle of it now

[HP]

Okay, so say our opposing firm's probability distribution is g(x)

Also, let's assume for now that there is no 'pure strategy' components in g(x) (this intuitively makes sense for the reason you mentioned before). Let's assume it's a nice looking probability density function. In other words:

http://img338.imageshack.us/img338/348/pic3ob5.png

Okay so say we choose a price P. P can be between 0 and 1. Our EV then is:

http://img524.imageshack.us/img524/5843/pic2rk4.png

now, for the opposing firm to be using a strategy from a Nash Equilibrium thing, the following must be true:

http://img524.imageshack.us/img524/7317/pic1pi2.png

where C is some constant. If this were not the case, then a certain value of P would would become the best choice for us, and hence our opposition's strategy is exploitable, and not part of a Nash Equilibrium thing

Also it should be noted our EV could actually be less than C, if we choose a P such that:

http://img405.imageshack.us/img405/8776/pic4to9.png

for any possible nonzero delta. But let's not worry about that just yet

okay so back to this equation:

http://img524.imageshack.us/img524/7317/pic1pi2.png

with a bit of alegbra:

http://img405.imageshack.us/img405/9544/pic5ea4.png

and now if we take the negative derivative of both sides with respect P:

http://img524.imageshack.us/img524/3324/pic6bw6.png

the left side reduces to just g(P). Differentiating the right side, we get:

http://img510.imageshack.us/img510/1015/pic7fp2.png

Now, we know g(x) is 0 when x>1. Intuitively, we want to be able to choose as high a P as we can. So we can assume that g(x) will have the form:

C/(q*x^2) when x<1

0 when L<x

Now, if we know what C is, we know L, or vica versa because the integral of the probability distribution must equal 1

You can show mathematically if you choose a P > L, your EV will equal C. However what happens if you choose a P <= L, trying to undercut his whole range?

If we undercut the competitor's whole range, we can not have an EV greater than C (as this would imply the competitor's strategy is not part of a Nash Equilibrium thing)

If your strategy is to undercut the competitor, you should set your price at P=L. This will maximize your EV.

Well, if we choose L, our EV is simply L (since we will always have the lowest price)

so, L has to be less than or equal to C (or else the competitor's strategy is not a Nash Equilibrium thing). Our competitor wants an L as high as possible (because this means he gets to choose a higher price on average and make more money from the consumer) so let's set L=C

and now, because the integral of the probability distribution has to equal one, the following must be true:

http://img405.imageshack.us/img405/2923/pic8mm2.png

solving this for L we find that L equals (1-q)

so finally, this gives us g(x):

http://img58.imageshack.us/img58/9482/pic9lj0.png

I think there may be some minor logical holes in my method (I'm not strong in math), however I think my answer is correct in hindsight, so you might philosophically say my process was a glorified guess and check

[janneman]

[ QUOTE ]
now, for the opposing firm to be using a strategy from a Nash Equilibrium thing, the following must be true:


[/ QUOTE ]

Upto this point your argument is right i think (for finding the Nash equilibrium of the game at least, i m not sure that is the question though)

However here you make a mistake, the lefthandside does not have to be constant over the whole range, just locally at our P so either P=0, P=1 or d/dP of the lefthandside is zero, but only at our P. If the lefthandside was really constant it would be zero by taking P=0.

[HP]

I believe this is covered with this part of my post:

Also it should be noted our EV could actually be less than C, if we choose a P such that:

http://img405.imageshack.us/img405/8776/pic4to9.png

for any possible nonzero delta.

[HP]

[ QUOTE ]


so finally, this gives us g(x):

http://img58.imageshack.us/img58/9482/pic9lj0.png

[/ QUOTE ]

err, that should be a '1-q' there for the range of x

(1-q)<x<1

[tabako]

Thanks

[janneman]

[ QUOTE ]
I believe this is covered with this part of my post:

[/ QUOTE ]

It clearly isn't. First of all first you define C to be the E.V. (and give the right definition). Then you say our E.V. can be less (not true) if your integral is zero for all delta. If P=0 your integral is 1 for delta=1 so C is still zero if it is constant.

Notice that this is not just a formality, your computation depends on C being constant which it clearly is not.

I think your calculation (when done right) will give us the optimal P for any distribution g. However it will not give g.

[HP]

[ QUOTE ]
It clearly isn't. First of all first you define C to be the E.V. (and give the right definition). Then you say our E.V. can be less (not true) if your integral is zero for all delta. If P=0 your integral is 1 for delta=1 so C is still zero if it is constant.

[/ QUOTE ]

Like I said I'm not good at math, and a miscommunication may have arisen from me writing down the wrong thing

In the P = 0 case, the integral can be zero, given a small enough delta. What I was trying to say, is if it can possibly be zero, then it's possible your EV is less than C

This is because you've chosen a P that your opponent would never have chosen. Maybe I should have just said that.

If you choose a P that your opponent would never have chosen, it means you are choosing a P that is not part of a Nash Equilibrium strategy, hence you are not guaranteed to have an EV of C

[HP]

btw, I can tighten my restrictions further. If the following can be 0 given the right choice of a non-zero delta:

http://img259.imageshack.us/img259/1734/piccw4.png

then P is not part of a Nash Equilibrium strategy, hence you are not guaranteed an EV of C

This is to avoid confusion in the case P is exactly on the edge of the domain of g(x)