There are a lot of ways to skin this cat. The most intuitive way I’ve come across is this:
1. If you choose door A, there is a 1⁄3 chance of the prize being behind that door. So there is a 2⁄3 chance that the prize is not behind that door. At this point we distribute the 2⁄3 chance evenly between doors B and C.
2. When the host reveals the the rotten avocado behind, say, door B, then door C picks up the entire 2⁄3 chance of having the prize.
When I apply Bayes’s Theorem to the problem, I essentially get the same thing. The prior probability of the prize being behind door A is 1⁄3, and the posterior probability of it being behind door A is also 1⁄3.
There are a lot of ways to skin this cat. The most intuitive way I’ve come across is this:
1. If you choose door A, there is a 1⁄3 chance of the prize being behind that door. So there is a 2⁄3 chance that the prize is not behind that door. At this point we distribute the 2⁄3 chance evenly between doors B and C.
2. When the host reveals the the rotten avocado behind, say, door B, then door C picks up the entire 2⁄3 chance of having the prize.
When I apply Bayes’s Theorem to the problem, I essentially get the same thing. The prior probability of the prize being behind door A is 1⁄3, and the posterior probability of it being behind door A is also 1⁄3.