Modified Monty Hall

For the ease of notation, we can say that $W$ is the event that the mathematician picked the right door initially (i.e. he'd win if he doesn't switch). Also, we call $S$ the event that the host gave the chance to switch. Thus, we are interested in calculating $P(W \mid S).$ In order for the mathematician to be ambivalent between switching and not switching, this means that he must be equally likely to have chosen the correct door as to not have which means that we are solving $P(W \mid S) = \frac{1}{2}.$ Finally, we note the definition of conditional probability to get that $P(W \mid S) = \frac{P(W \cap S)}{P(S)}$ but we know that $P(W \cap S)$ is just the probability that he initially chose the right door because the host will always give him the chance to switch if he initially chose the right one. He has a one in three chance, so this probability is $\frac{1}{3}$ .

In order to calculate $P(S)$ we can use the Law of Total Probability to set it equal to $P(S) = P(I)P(S \mid I) + P(I^C)P(S \mid I^C) = \frac{1}{3} \cdot 1 + \frac{2}{3} \cdot p$ and putting this all together gives us that $\frac{1}{1 + 2p} = \frac{1}{2} \implies p = \boxed{\frac{1}{2}}.$