Hey guys, this might be a bit more math and book keeping than most people want to do, but I'm trying to figure out the optimal amount of money to spend on advertising for the tavern to minimize loss/guarantee profit. I know how to slowly do the math by hand on paper, but if I did I'd end up spending hours just mapping out probabilities of dice rolls, which is less fun than it sounds. Does anyone know a program I can plug these numbers into to get an optimal answer, or has anyone plug these numbers themselves?

Manor profits are calculated using the "Running a Business" chart from the DMG, with a base d100 + 10 roll, + 1 for each GP spent on advertising. Maintenance costs are 60 GP.

I dusted off my R programming and came up with an answer: Max expected profit occurs at 31 GP of advertising per tenday, and is ~20 GP (19.8 GP) per tenday.

After two years of operation, the tavern should finally start to bring in a profit compared with its start-up costs. In terms of real-world economics and return on investment, that's not bad.

EDIT: If anyone cares, I did 10,000 random draws on the business chart per GP in advertising expenditure from 0 to 100 gold. Never put in more than 80 GP of advertisement, of course; it doesn't provide any benefit.

EDIT2: If you're interested, here is the R script I used (R is a statistics programming langugage):

N=10000;
MaintPerDay=6;
NumDays=10;
AdMax=100;
Profit=array(rep(0,(1+AdMax)*N),c(1+AdMax,N));
MProfit=rep(0,1+AdMax);

for (ads in 0:100) {
  for (i in 1:N) {
    Roll=sample(1:100,1);
    Result=Roll+NumDays+ads;
    if(Result <= 20)
    {
      Profit[ads+1,i] = -1.5*MaintPerDay*NumDays;
    } 
    else if (Result <= 30)
    {
      Profit[ads+1,i] = -1.0*MaintPerDay*NumDays;
    } 
    else if (Result <= 40)
    {
      Profit[ads+1,i] = -0.5*MaintPerDay*NumDays;
    } 
    else if (Result <= 60)
    {
      Profit[ads+1,i] = 0;
    } 
    else if (Result <= 80)
    {
      Profit[ads+1,i] = sample(1:6,1) * 5;
    } 
    else if (Result <= 90)
    {
      Profit[ads+1,i] = (sample(1:8,1) + sample(1:8,1)) * 5
    } 
    else
    {
      Profit[ads+1,i] = (sample(1:10,1) + sample(1:10,1) + sample(1:10,1)) * 5;
    }
    Profit[ads+1,i] = Profit[ads+1,i] - ads;
  }
  MProfit[ads+1] = mean(Profit[ads+1,]);
}

Instead of simulations I made closed-form computations and found out the best advertisement investment is 30gp, not 31gp. I also got the following plot:

Here's the code in R:

getEC <- function( ae = 0){  # ae : advertisement expenses

bon   = 10 + ae                       ## Bonus to the table
pv    = c( 20,10,10,20,20,10,10)/100  ## Probability of each of the 7 outcomes, with no bonus
iv    = c( -90 , -60, -30, 0, 3.5*5, 9*5, 5.5*3*5)  - ae ### Expected profit for each outcome

### Change the probability of the 7 outcomes to account for the bonus
		pv[1]    = max(20 -  bon      , 0) ;
if (bon > 20){  pv[2]    = max(10 - (bon - 20), 0) }
if (bon > 30){  pv[3]    = max(10 - (bon - 30), 0) }
if (bon > 40){  pv[4]    = max(20 - (bon - 40), 0) }
if (bon > 60){  pv[5]    = max(20 - (bon - 60), 0) }
if (bon > 80){  pv[6]    = max(10 - (bon - 80), 0) }
	        pv[7]    = 10 + bon

return( sum( pv * iv/100 ) ) # return expected profit for a given ae

}
 data = rep(NA,91)

for (ae in 1:91){
data[ae] = getEC(ae-1)
}

plot(0:90,data, xlab="Advertisement Expenses (gp)", ylab = "Expected Profits (gp)", main="Expected Profits per Tenday by Advertisement Expenses" )