Why does the Monty Hall problem not apply when the contestant picks the second door?

So, we all know the Monty Hall problem. Where there are 3 doors with one winning door.

The contestant picks a door, then the host removes a non-winning door from the remaining two. Then the contestant is given the option of switching doors.

The originally selected door has a 33.33% chance of being the winning door, but the remaining door has a 66.67% chance of being the winner.

However, if this is done Deal or No Deal style, and the contestant gets to pick a second door in hopes that it's a non-winning door the probability changes.

Let's say the contestant successfully chooses a non-winning door, and is left with the same proposition as the Monty Hall problem. Where he has the option to swap the originally selected door with the one remaining door.

Now, the probability is 50% for each door.

Why? Why does it matter when the door is removed by the host (who knows which door is a winner), compared to by the contestant who removed the proper door by chance?

• There is no information gained in the Deal or No Deal style game. In the Monty Hall problem, you gain information when a door is removed (because you know that door did not contain the prize), so the conditional probability of winning changes. There is no change in conditional probability in the Deal or No Deal style game because no information is gained about either door.
– kccu
Jan 8 '16 at 18:39

What matters is that in the case where the contestant didn't pick the prize door initially, the host uses his knowledge to avoid picking the prize door to open.

On the other hand, if the contestant picks a door to open at random, there's a risk that it's the prize door he picks. When that happens, he doesn't get any useful chance to switch -- and this only happens in the case where the original rules would have led to switching being a benefit.

Thus, letting the contestant pick the second door would remove some switching opportunities that would be beneficial, but would not remove away any opportunities that would lose.

Therefore, the chance of winning by switching becomes smaller overall: It drops from $2/3$ to $1/2$.

• I think it clears it up, but where I still get tripped up is if we run this experiment 1,000 times, and remove the times where the second door was the winning door (roughly 33% of the time), and only analyze the times where the winning door was one of the final two, the probably still changes to 50/50. Is that because we lose 33% of the trials? Jan 8 '16 at 18:46
• @Javalsu: Yes -- because those cases you remove are not ones that can't happen, but ones where the host would open a different door instead -- an in all of those cases a switching contestant would win and a non-switching contestant would lose. Removing them from consideration means ignoring exactly the cases where a consistently-switching contestant gains a lead over a consistently-staying one. Jan 8 '16 at 18:49

We can look at this problem in terms of the "information" that the contestant has; specifically, what information the contestant has about each door.

For simplicity, I'll use "A" as the name of the door the contestant first selects, "B" as the name of the door that Monty or the contestant then opens, and "C" as the name of the remaining door.

In the standard Monty Hall setup, where we know in advance that after the contestant makes his/her initial choice, Monty always will open a non-chosen, non-winning door, the facts that Monty opened door B and that door B had no prize give us no information about door A. That's because both of these facts were guaranteed to be true regardless of whether the prize was behind door A.

But knowing that Monty was certainly going to open a non-winning door, Monty's action does give us quite a lot of information about each of the doors B and C. Obviously, it tells us that the prize is not behind door B, which we did not know before. In fact, before Monty's action, we would assign the probabilities $\frac13,\frac13$ to the possibility of a prize behind each of the doors B and C; after Monty opens a door, the probability assignments are $0, \frac23$ respectively.

In the case where the contestant opens a door other A, not knowing if the prize is behind that door, we do not know in advance that door B would not reveal the prize. The chance that the door B reveals the prize is much lower in the case where the door A is the prize door than in the case where it is not (probability $0$ versus $\frac12$); so the fact that B is not a prize door tells us that the probability the prize is behind door A is greater than we would have estimated previously. In fact, the probability the prize is behind door A increases from $\frac13$ to $\frac12$. The probability that the prize is behind door C also increases from $\frac13$ to $\frac12$.