D&D Tavern Revenue Optimization

Question

I am trying to figure out the best way for my group to get our maximum profit from a tavern we just got. The rules for determining the amount of money have been changed slightly by our DM. They are: Roll a d100 ( 1-d10+1-percintile die (like a d10 but 10, 20,..,etc)) if the value is 10 or less then we have to pay the full amount of upkeep costs for the tavern (60 gold pieces). For 11-40 the tavern generated just enough to cover half it's expenses so we would owe 30 gold pieces. For 41-60 then the tavern would simply cover it's expenses. For 61-80 the tavern covers it's expenses and we will roll 6d6*6 profit. For 81-90 we will roll 6d8* profit. For 91+ we roll 6d10*6 profit. We also have the option of spending gold to promote the tavern; for each piece of gold we spend we can add 1 to the d100 roll. My first instinct was to simply spend 90 gold to make sure we will always get the highest tier of rolls. However, we can still roll poorly and lose money on our profit roll. Thus we will be spending too much on advertising. So what amount do we need to spend on advertising to be put in the highest profit bracket without spending so much that it cuts into our profit significantly. Perhaps a function can be built to be optimized? Or do I need to just run tons of Sims in Excel? Any help is appreciated! Thanks, Drew.

Answer 1

To summarize:

If the roll result is $1$ to $10$ you'll get $-60$ gold.

If the roll result is $11$ to $40$ you'll get $-30$ gold.

If the roll result is $41$ to $60$ you get $0$ gold.

If the roll result is $61$ to $80$ you get $6d6\cdot 6$ profit, expecting $6\cdot3.5\cdot6 = 126$ gold.

If the roll result is $81$ to $90$ you get $6d8\cdot 6$ profit, expecting $6\cdot4.5\cdot6 = 162$ gold.

If the roll result is $91$ or higher you get $6d10\cdot 6$ profit, expecting $6\cdot5.5\cdot6 = 198$ gold.

For every gold you spend, $1$ gets added to the roll result. I assume you spend the gold before rolling the die. Think of it like improving the $100$-sided die before you roll it by changing the numbers on the faces. But think of it like this: instead of increasing every face by one, you simply take the lowest face and increase it by $100$, putting its outcome into the highest profit bracket. The numbers on the die will end up being the same.

Now it's a simple matter of checking which faces are worth improving: Each face costs $1$ gold to improve, but lands only $1\%$ of the time, so the expected result when landing on the face should increase by at least $100$ gold from the upgrade to be worth the upgrade.

This is the case for faces $1$ to $10$ ($+258$ gold), $11$ to $40$ ($+228$ gold), and $41$ to $60$ ($+198$ gold), but not for the other faces ($+72$ gold, $+36$ gold, $+0$ gold respectively).

So to maximize your expected profit, you should invest exactly $60$ gold.

Answer 2

We can subtract $1$ from the roll and use only the first digit. There are then ten possibilities and we can tabulate them $$\begin {array}{r | r} roll & profit\\ 0&-60\\ 1,2,3&-30\\ 4,5&0\\ 6,7&6d6*6[126]\\ 8&6d8*6[168]\\ 9&6d10*6[198] \end {array}$$ where the square brackets have the expectation of the die roll. It costs $10$ to move up one number in the roll. Now we can compute the expectation with no advertising $$\frac 1{10}\left(-60+3(-30)+2(0)+2(126)+168+198\right)=46.8$$
If you spend $10$ on advertising, you remove the $-60$ and add in a $198$, which is clearly a good idea. It is worth $\frac 1{10}(258)=25.8$ Each additional $10$ you spend removes the current bottom profit and replaces it with $198$ You want to do that until the bottom profit is $98$ or greater, so you should spend $60$ on advertising. The expected return is then $$\frac 1{10}\left(126\cdot 2 + 168+198\cdot 7\right)-60=180.6-60=120.6$$