I have a version of the Monty Hall Problem with a slight twist.
Classic Setup
Suppose there are 3 doors to choose from. You have to get the correct door. Once you choose one door, one of the wrong doors that you didn't choose is eliminated. Then you have the choice to switch or not.
In this case it is favorable to switch.
Twist
What if the initial probabilities of each door are not the same but p1, p2, and p3, such that p1 < p2 < p3. What would be the best strategy in this case?
Then always pick door $1$ and switch when given the option. Then you have the probability $p_2+p_3>\frac{2}{3}$ of being correct.