What follows is a strategy for earning an expected profit of about $3.63$ Euro, and an argument that this strategy is optimal.
First we'll pick some positive integer $k$. We will switch dice after every roll number that is an odd multiple of $k$, and continue to roll until we get a 6. So we will roll the first die $k$ times, then the second die $2k$ times, then the first die $2k$ times, and so on.
With probability $1$ we will eventually roll a 6 and win the 20 Euro. To compute the expected cost, we consider two cases, depending on which die we pick up first.
First die is fair.
We expect to roll the fair die $6$ times. We expect to roll the loaded die
$$
\sum_{n\geq 0}\left(\frac{5}{6}\right)^{2nk}2k = \frac{2k\cdot(5/6)^k}{1-(5/6)^{2k}}
$$
times. The expected number of switches is
$$
\left(\frac{5}{6}\right)^k+2\cdot\sum_{n\geq 1}\left(\frac{5}{6}\right)^{(2n+1) k} = \frac{2(5/6)^k}{1-(5/6)^{2k}}.
$$
So if we start with the fair die, our expected profit is
$$
20-6-\frac{2k\cdot(5/6)^k}{1-(5/6)^{2k}}-5\cdot\frac{2(5/6)^k}{1-(5/6)^{2k}}.
$$
First die is loaded. We again expect to roll the fair die $6$ times. We expect to roll the loaded die
$$
k+\sum_{n\geq 1}\left(\frac{5}{6}\right)^{2nk} 2k=\frac{k\cdot(1+(5/6)^{2k})}{1-(5/6)^{2k}}
$$
times. The expected number of switches is
$$
1+2\cdot\sum_{n\geq 1}\left(\frac{5}{6}\right)^{2nk}=\frac{1+(5/6)^{2k}}{1-(5/6)^{2k}}.
$$
So if we start with the loaded die, our expected profit is
$$
20-6-\frac{k\cdot(1+(5/6)^{2k})}{1-(5/6)^{2k}}-5\cdot \frac{1+(5/6)^{2k}}{1-(5/6)^{2k}}.
$$
Expected profit Our overall expected profit is the mean of the expected profit if the first die is fair and the expected profit if the first die is loaded. The expression simplifies to
$$
14-\frac{1}{2}\frac{(5+k)\left(1+(5/6)^{k}\right)^2}{1-(5/6)^{2k}}.
$$
Numerically, I found that profit is maximized taking $k=9$. The expected profit using this strategy is
$$
\frac{4218321}{1160653}\approx 3.63\,\,\,\text{Euro}.
$$
Is this strategy optimal? Here's a sketch of a proof that this strategy is optimal.
Recall that the odds a given die is loaded is the ratio
$$
\frac{\text{probability that die is fair}}{\text{probability that die is loaded}}.
$$
The product of the odds that each die is fair is $1$.
During the course of the game, we only get information of one kind, namely that the die we just rolled did not turn up 6 (if it turns up 6, the game ends). Bayes' Theorem says that whenever a die turns up something other than 6, the odds that it is fair get multiplied by $\frac{5}{6}$ (and so the odds the other die is fair get multiplied by $\frac{6}{5}$). This means that assuming no 6's have been rolled, the odds that a die is fair depend only on the different between the number of times that die has been rolled, and the number of times the other die has been rolled.
When we are deciding whether to roll our current die, switch dice, or walk away, we need only consider the odds our current die is fair. Any costs already paid (sunk costs) should not be taken into account, as future payments and earnings do not depend on them. This means that possible strategies fall into two possible kinds:
- If the odds the current die is fair are below a certain threshold, leave the game. Otherwise roll.
- If the odds the current die is fair are below a certain threshold, switch dice. Otherwise roll.
The strategy described above is of the second kind, where the threshold is $\left(\frac{5}{6}\right)^k$. We choose $k$ to optimize profit among strategies of this kind. The only other possibility is a strategy where we roll the current die $n$ times (for some fixed $n$), then leave the game if we haven't rolled a 6. It's not too hard to write down the expected profit here as a function of $n$, find the $n$ that maximizes profit (if I computed correctly it is $n=5$), and see that this doesn't do better.
