St. Pete's Casino is introducing a new dice game called "Guess and Win." After paying (not betting) a fixed ante of $x$ dollars, the player guesses what the next roll of a fair six-sided die will be.
If the player guesses wrong on the first roll, then he/she wins nothing. Otherwise, the player wins $6^{n+1}$ dollars, where $n$ is the number of consecutive correctly-guessed rolls before an incorrect guess. For example, a player who guesses correctly and then incorrectly wins $36.
Of course the expected winnings of this game are infinite ... but there's a catch. A player who guesses correctly on his/her cumulative $20^{th}, 40^{th}, \dots, (20n)^{th}$ roll gets cashed out. So if you've already rolled 16 times cumulatively, then your maximum winnings for the next round are $6^{4+1} = \$7776$, for a profit of $\$(7776 - x)$. Or if the next roll will be your cumulative $60^{th}$ roll, you can only win $36 next round.
The casino management want $x$ to be fixed, in particular not depending on the number $N$ of times one has already rolled. Also, they estimate that the probability that a player will continue to roll given a certain house edge $q$ (which, n.b., does depend on $N$) is given by the following equation: $$\mathbb{P}\Big(\mbox{continue playing}\mid q\Big) = \frac{0.99}{\max\{1, 50q\}}.$$ So there is a 66% probability that a player will roll again if, given his/her cumulative $N$ previous rolls, the house edge in the next round is $3\%$.
- Find the minimum integer $x$ that results in an expected net profit for the casino long-term (i.e., as the number of players increases without bound), or prove that no such $x$ exists.
- Assume that no player will roll more than 20 cumulative times. Find the minimum integer $x$ for which $\mathbb{P}(\mbox{casino profits}\mid\mbox{fixed player}) > 0.6$, or prove that no such $x$ exists.
- Does the answer to (2) change if the assumed limit on a player's cumulative rolls is dropped?