Does Monty Hall logic apply to this real world situation?
No, this is not a Monty Hall problem. If your wife only knew the contents of #3, and was going to reveal it regardless, then the odds were always 50/50/0. The information never changed. It was just delayed until after your original choice. Essentially, you NEVER had the chance to pick #3, as she would have immediately told you it was wrong. (In this case, she is on your team, and essentially part of the player). #3 would be eliminated regardless: "No, not that one!"
Imagine you had picked #3. Monty Hall never said, "You picked a goat. Want to switch?"
If he did, the odds would immediately become 50/50, which is what we have here.
Monty always reveals the worst half of the 2/3 you didn't select, leaving the player at 33/67/0.
tl;dr: Switching in this case has no effect, unlike the Monty Hall problem, where switching doubles your odds.
The reason this is different is that Will's wife knew the content of one, and only one taco, and that it was a bean one, and Will knows that his wife only knew the content of one bean one. (Monty is different because he knew all doors.)
Here's why:
Unlike the MH problem, which has one car, and two goats that can be treated as identical, the Will's Wife Problem has one Steak, One known bean, and one unknown bean, so the beans need to be considered differently.
That's important because MH's reveal gave players a strong incentive to switch, but gave them NO new information by revealing a goat in an unpicked door - whatever the player picked, he could show a goat, so no new info is provided by the reveal. But that's not the case for Will's wife:
Since she only knows the content of the known bean taco, her revealing that it's not one of the ones you picked actually changes what you know about your odds. Because she would have behaved differently if you'd picked the known bean one - she'd have said, "the one you picked is bean".
Without her info, you only had a 1/3 chance of having the steak, but by showing you that you didn't pick the only bean one she knew about, it means you already know you have a 50% chance of having the steak.
And since you also have a 50% chance of having the unknown bean taco, it's irrelevant if you switch or not.
In the MH problem, the key fact is this: Since the reveal in no way changes what you know about your odds, there's only a 1/3 chance that you START with the car. And since you:
- Win by staying in all cases where you started with the car, and
- Win by switching in all cases where you didn't start with the car...
In the MH problem, switching doubles your odds (from 1/3 to 2/3), but in this case, switching has no impact (since it's 50% either way).
The other posts provide several intuitive ways to think about this. I will simply compute the conditional probabilities to show that indeed this is not a monty hall style problem.
A priori probabilities $$P(\text{meat in A}) = 1/3$$ $$P(\text{meat in B})= 1/3$$ $$P(\text{meat in C}) = 1/3$$
After your wife tells you that C is beans
(The notation $P(A \mid B)$ means probability of event "A" given information "B"
$$P(\text{meat in A} \mid \text{beans in C}) = \frac{P(\text{meat in A} \textit{ and } \text{beans in C)}}{P(\text{beans in C})} = \frac{1/3}{2/3} = \frac 12$$ and similarly $$P(\text{meat in B} \mid \text{beans in C}) = \frac 12$$ $$P(\text{meat in C} \mid \text{beans in C}) = 0$$
So you can see that $A$ and $B$ have the same probability of having meat once you have the additional information of "beans in C". So there is no advantage to switching and the probability is 50%.
For a comparison with the Monty hall case, I refer to Bram28's answer: https://math.stackexchange.com/a/2796175/66711