An interesting thing to realize about the problem is that there actually is a way to make your odds the "obvious" 50/50, and that's to flip a coin to decide whe

An interesting thing to realize about the problem is that there actually is a way to make your odds the "obvious" 50/50, and that's to flip a coin to decide whether to switch doors at the end. The probability of a win on the two doors isn't equal, but the probability on the first door is 1/3 and the probability on the door you can switch to is 2/3, so using a coin toss to decide whether to switch would result in a 50% win rate. This is because the way the problem is structured gives you six choices, not three: [A, A], [A, switch], [B, B], [B, switch], [C, C], [C, switch]. 3 of these are wins and three of these are losses. If you choose from these six options randomly you will win 50% of the time, but if you only choose from the "switch" options you will win 2/3 of the time.