I recently listened to a podcast on which they talked about the Monty Hall problem, in which legendary game show host Monty Hall lets you choose between 3 doors, one of which has a car behind it, while the other two each conceal a goat. He then opens one of the two doors you haven't chosen, and lets you stick with your original choice, or change to the remaining unselected door. After your 2nd choice, the door you've now picked is opened and if the car is behind it, you win( presumably, the car). In the standard version of the problem, he always shows you a goat( because if he revealed the car behind a door you didn't choose, you'd know you lost since your first choice was wrong and your 2nd choice would be moot as only the other goat remained). So after the reveal, what's the probability that you win by switching in your second round?

Most people learn about the problem, note that two doors remain, one of which is a winner and one a loser, and so they rashly assume that there are two possibilities with one winner, OR they see it as two of four chances to win—two scenarios, each with two possible outcomes: if you picked right the first time, you win sticking to your choice and lose by switching; if you picked a goat, you lose by sticking and win by switching; 1/2 = 2/4 = 50% chance of victory with either strategy( stick vs. switch). One of two or two of four is a serious oversimplification of the problem, however.

Mathematicians and noted intellectuals frequently recognize three possible scenarios, rather than two, and thus six possible outcomes rather than four; and they count it as 2/6 = ⅓ outcomes in which sticking wins, ⅔ in which switching wins. They note that the opening of a losing door adds NO relevant information for the contestant, it only tells you that you would NOT have won by initially choosing that door, AFTER the possibility of your doing so has already been precluded. So they focus on the relationship of the inital choice( between 3 doors, with 1 being correct) to the second choice( stick or switch before learning whether first choice was correct). Because there are 2 goats and only 1 car, they recognize the 6 possibilities as:

Stick( to car); outcome: win. (×1)
Stick( to goat); outcome: lose. (×2)

Switch( from car to goat); outcome: lose. (×1)
Switch( from goat to car); outcome: win. (×2)

...but here's a twist: that's actually wrong, despite the smartest people in the world insisting on it. It's true, as the mathematicians say, that there are 3 initial possibilities for the condition of the system after the first choice: 2 in which the contestant chose a goat door and 1 in which the contestant chose the door hiding the car. It's ALSO true that Monty showing a goat between the contestant's choices doesn't change the winning strategy, which is based on whether the initial choice was correct. What they should have known better is that it DOES still matter that Monty shows a goat, because it adds a conditional variable to the sequence of events when the contestant's first choice is correct, meaning that the overall state of the system is NOT still limited to 3 possibilities immediately before the final choice.

It took me a minute to figure it out, but I realized the correct answer by imagining the 2 goats as different colors, grey vs. brown; I think that's also the easiest way to explain the truth. The mathematician-accepted answer makes the same type of reductive mistake as those mathematicians criticize, and my differently-pigmented goats lay it bare: there are three possibilities immediately after the first choice, but actually FOUR possible scenarios immediately before the second choice, meaning EIGHT possible outcomes depending on the final decision, i.e.:

Scenario A: Chose grey goat door, Monty reveals brown goat:

Choice 1: Stick( to grey goat); outcome: lose.
Choice 2: Switch( to car); outcome: win.

Scenario B: Chose brown goat door, Monty reveals grey goat:

Choice 3: Stick( to brown goat); outcome: lose.
Choice 4: Switch( to car); outcome: win.

Scenario C: Chose car door, Monty reveals brown goat:

Choice 5: Stick( to car); outcome: win.
Choice 6: Switch( to grey goat); outcome: lose.

Scenario D: Chose car door, Monty reveals grey goat:

Choice 7: Stick( to car); outcome: win.
Choice 8: Switch( to brown goat); outcome: lose.

…all of which still preserves the direct, unchanging relationship between the first choice and the correct final answer. So why would so many smart people make this mistake? That question actually needs to be broken down further:

Why do they bypass the right answer?

As to WHY they have trouble seeing/accepting the correct answer, I think it's because so many people arrive at the right answer( equal probability) in a way that is obviously wrong to some people( including mathematicians). Those who think it's a question of two possibilities with one being a win are wrong, as are those who think it's a question of two chances to win out of four possible scenarios. Those people calculate a 50/50 chance of winning regardless of strategy, but they’re basing it on an incomplete mathematical model of the system described in the word problem.

So when educated people who know better subsequently calculate the same answer correctly, they likely experience an unpleasant sensation known as cognitive dissonance, which arises from trying to believe two conflicting things simultaneously. So they then fall victim to a nigh-uncontrollable neuro-cognitive phenomenon which drives most humans suffering cognitive dissonance to misapply logic, misremember facts, and essentially make any excuse to justify resolving the related issue in a way that seems emotionally acceptable, as quickly as possible—then never think of or reconsider it ever again. I've been unable to find an official medical/scientific term differentiating the sensation of cognitive dissonance from the reaction, so I've taken to calling the latter “the reflexive reaction to cognitive dissonance”, or r.r.c.d.. Because of r.r.c.d., when they get the same answer as the people who did the problem in a way they knew to be wrong, smart, well-educated people find a way to twist the problem so they get a different answer. Even Marilyn vos Savant( her real name!), who held the Guinness record for highest IQ ever recorded( before the category was retired), correctly identified a common mistake that people answering the problem make—but failed to correctly apply her own observation.

Why do they come up with the same wrong answer?

As for why they got the SAME wrong answer, perhaps it comes down to the existence of valid math describing the scenario incorrectly. The easiest way to illustrate this is to strip the values of 'win' and 'lose' from the outcomes, and the identities 'car' and 'goat(s)' from the door contents. Instead, we'll call the objects behind the doors x, y, and z; and from there we can lay out all possible versions of what was behind the first door chosen, what Monty revealed behind one of the 2 remaining doors, and whether the contestant chose to stick or switch; plus what was subsequently revealed behind their final choice( the outcome).

Sequence: 1st chosen, revealed, 2nd choice; outcome

A: x, y, stick; x
B: x, y, switch; z
C: x, z, stick; x
D: x, z, switch; y
E: y, x, stick; y
F: y, x, switch; z
G: y, z, stick; y
H: y, z, switch; x
I: z, x, stick; z
J: z, x, switch; y
K: z, y, stick; z
L: z, y, switch; x

As we can see, there are 12 possible sequences of events, out of which the final door opens on any particular contents in 4 of them, or ⅓ of the time. So, if only one possibility is a winner, that's the chance of winning and the chance of losing is ⅔. However, as Marilyn vos Savant pointed out, in the real-world scenario, Monty Hall knew which door the car was behind, and never opened that door, thus the possibilities wherein Monty's reveal is the designated winning choice don't actually come into play.*

Here's where the simplified table above is handy, because it allows us to quickly see that, for example, if x is the winning outcome, Monty's insider info will preclude four sequences: half of those in which the initial choice concealed y( E & F) and half in which the initial choice concealed z( I & J). When the initial choice by the contestant is wrong, only possibilities in which Monty opens the other wrong door( G, H, K, & L) are valid, BUT all of the sequences in which the initial choice concealed x( A, B, C, & D) will remain, because in that case Monty might open either of the other two doors. Thus we can see that eight sequences remain, half win, half lose; and half of the wins are stick( as are half of the losses), while the other half are switch.

The tricky thing here is understanding that the loss outcomes are non-fungible and must always be counted separately( e.g., by thinking of them as grey goat vs. brown goat instead of just a goat and another goat), not because losing over a particular goat makes any difference to the contestant or to the impact of their second-round choice( it doesn't), but rather because, despite being irrelevant to the final answer, giving Monty a choice when the contestant's first door is correct, that he doesn't have if it's incorrect, doubles the incidence of the former within the set of possible event sequences leading up to the second choice.

So the experts are wrong, there really is an equal chance regardless of strategy; BUT you were also wrong if you arrived at that 50% value via 1/2 doors winning or 2/4 sequences in which the final answer right, because it's actually 4/8 sequences( since 4 of the 12 total permutations would include Monty opening the door and are thus not actual possibilities).( Props to vos Savant for recognizing where the error arises, despite committing a very similar one there herself.)

—D.R.T.Y. whiz E.M.

*this might be where MvS made her mistake: since Monty always has a choice of 2 doors to open, and never opens the door with the car, her r.r.c.d.-stricken brain may've fudged that that as eliminating half of the 12 permutations—in fact, it only eliminates half of the possibilities in which the door with the car wasn't initially chosen, but since the car was chosen in one-third of them, only half of the other two-thirds are eliminated, NOT half of all permutations… then she could have combined her misunderstanding of this fact with her determination that 50% was a wrong answer and focused on the 3 initial choices to work out a way in which ⅔ vs. ⅓ seemed to be correct, all because of r.r.c.d.