13
$\begingroup$

I am working on a model of optimal payout percentages in the gambling industry.

Because the nominal price of a \$1 ticket is always \$1, we use an effective price strategy where Q = \$1 in won prizes. If a game pays out 50%, the effective price is \$2, since that is what would need to be spent to win an expected \$1 in prizes. Pretty simple, right?

Well, I ran into this footnote in some research, and can't figure out how they got to the First Order Condition for Profit Maximization from the first equation:

"Let $C(Q)$ represent operating costs as a function of quantity units, where one quantity unit is defined as one dollar in expected value of prizes.

The lottery agency's net profits are given by

$$N = PQ - Q - C(Q)$$

where $P$ is the price charged for a quantity unit.

The first-order condition for profit maximization can be written

$$-E_{PQ} = P(1 - C')/[P(1 -C')- 1] $$

If marginal operating costs are $6$ percent of sales and the payout rate is $50$ percent, we have $P = 2$ and $C' = .12$, implying that the price elasticity of demand at maximum profit is $-2.3$.

For an increase in the payout rate to increase profits, $E_{PQ}$ must exceed $2.3$ in absolute value."

-[Citation] Clotfelter, Charles T, and Philip J Cook. "On the Economics of State Lotteries." Journal of Economic Perspectives: 105-19.

In the FOC equation, $-E_{PQ}$ is the effective price elasticity of demand. That normally would be found by taking the derivative of $P$ with respect to $Q$ in the first equation.

How did they end up where they did? There has to be something I'm missing.

I am having trouble understanding how that particular First Order Condition was reached- whether it was a result of some some derivative process on the Net Revenue equation, or if it is simply an external condition being applied.

Thanks!

$\endgroup$
  • 3
    $\begingroup$ Yay! MathJax works :-) $\endgroup$ – LateralFractal Nov 19 '14 at 2:18
10
$\begingroup$

The expression in question is in footnote $11$ of the referenced article. Reading the paper, we see that the decision variable here is "the payout rate", which is the reciprocal of $P$. So equivalently, we can solve the maximization problem with respect to $P$ (and not w.r.t. $Q$). More over, "price elasticity of demand" involves the derivative of $Q$ with respect to $P$, and not the other way around:

$$E_{PQ} = \frac {dQ/dP}{Q}P $$

and we expect it to be negative (higher price means lower payout rate leading to less demand for the quantity measure here, i.e. less "demand for prizes").

We can write the maximization problem as $$\max_{P}N = \max_{P}\left[P\cdot Q(P) - Q(P) - C(Q(P))\right]$$

The first-order condition is

$$\frac{\partial N}{\partial P} = Q + P\cdot Q' - Q' - C'\cdot Q' = 0 \tag{1}$$

Multiply throughout by $P/Q$:

$$Q\frac PQ + P\cdot Q'\frac PQ - Q'\frac PQ - C'\cdot Q'\frac PQ = 0$$

$$\Rightarrow P +P\cdot E_{PQ} - E_{PQ} - C'\cdot E_{PQ}=0$$

$$\Rightarrow -E_{PQ} = \frac {P}{P-1-C'} \tag{2}$$

This makes sense. Plugging in the values presented in the reference, we have

$$-E_{PQ} = \frac {2}{2-1-.12} = \frac {2}{0.88} \approx 2.27 $$

which is very close to the value resulting from the equation presented by the authors. I haven't been able, by whatever algebraic manipulations I tried, to replicate their formula, but eq $(2)$ is correct in any case. If a reconciliation comes up, I will update.

$\endgroup$
  • 1
    $\begingroup$ Fantastic. This is where I ended up,as well. Apologies for not including my previous work in the question (I'll have to remember to do that). $\endgroup$ – datahappy Nov 19 '14 at 3:40
  • $\begingroup$ I emailed the authors of the paper- if they respond at any point, I will add their reasoning as another answer...I will wait to mark you as the answer to give some other folks time to answer since we are in beta. :) $\endgroup$ – datahappy Nov 19 '14 at 3:42
  • 3
    $\begingroup$ Of course you should wait. We want more than one answer per question! $\endgroup$ – Alecos Papadopoulos Nov 19 '14 at 3:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.