What's wrong with this equal probability solution for Monty Hall Problem?

Question

I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective gives me equal probability.

Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.

Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)?
Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.

Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.

What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.

Answer 1

Let me try to guess your birthday. My guess is April $2$nd.

Ignoring leap years, remove $363$ dates that are not your birthday and that are not April $2$nd (because I chose that), so that we are now left with only two options. You offer me to switch to the date that I didn't pick, which I do.

That I was right in my initial guess is unaffected by the dates that you subsequently remove.

Answer 2

The flaw in your thought process is that in the $50:50$ case with only two doors where one is a goat and one is a car, that is the only information you have. In the other case where you have two goat doors and one car door, when you open a goat door, you gain some information by doing this. Thus you appear to have the same $50:50$ situation, but you have some added information.

Answer 3

Therefore, showing an open door with a goat reveals nothing new about which door has the car.

Two-thirds of the time, our initial pick will be wrong. In that case, Monty cannot randomly pick a door out of the remaining two to reveal. He must pick the door that has a goat. This gives the additional insight into which door has a car.

Conversely, one-third of the time, our initial pick will be correct so Monty can then open any of the two remaining two doors. Switching in that case will then lead us to the wrong door. The good news is, that only happens one-third of the time!

Answer 4

Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50\%$, whereas there is definitely a goat behind door $C$.

In particular, let $C_{\text{goat}}$ be the event that there's a goat behind door $C$, and $C_{\text{revealed}}$ be the event that door $C$ is revealed, and $A_{\text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{\text{car}}$ and $C_{\text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50\%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50\%$. Thus, one concludes that $$P(A_{\text{car}}|C_{\text{revealed}})=P(A_{\text{car}})=\frac{1}3.$$ The calculation you've done (annotated with a $\neq$ sign where things go wrong) is: $$P(A_{\text{car}}|C_{\text{revealed}})\neq P(A_{\text{car}}|C_{\text{goat}})=\frac{1}2$$

Answer 5

The idea of assigning $50/50$ probabilities to two alternatives makes sense only when the two alternatives are truly symmetric.

If you just start with two doors, you know there is a car behind exactly one, and nothing else has happened concerning those doors, then $50/50$ probabilities make sense.

But once you have chosen one of three doors and Monty has opened another door (having been forced by the rules of the game to open a door that wasn't yours and didn't have the car), you no longer have a symmetric situation. One of the remaining doors is an available choice because you already chose it once; the other is an available choice either (A) because you already chose the car and Monty got to take one of the "goat" doors away at random, or (B) because you chose a goat initially and Monty had to take away the remaining "goat" door, leaving the door with the car. Case (B) is twice as likely to occur as case (A), not $50/50$, and which door has the car is now completely a function of whether you're in case (A) or case (B), so it's also not $50/50$.

Answer 6

example 1 which shows there would be no advantage in shifting: lets say there are one million doors and I pick No 1. Monty reveals 999998 doors which are empty so now we have only 2 doors, one that I picked and another one. According to the paradox the possibility would be 1/999999 = almost 100% certain that the shift would have advantage. Obviously wrong.

Convinced?

example 2 which shows there would be no advantage in shifting: Lets say instead of one player we have 2 players and one of them choose door one and another choose door 2 and Monty opens door three. If the argument is true shifting would increase the chance for both players. Obviously not possible.

Convinced?

Answer 7

I'm amazed people still argue this.

Example 1 yes you would have an extremely high chance of winning if you selected the door that was filtered out 999,999 doors. If not, you are basically arguing that Door 1 had a 50/50 shot at winning a contest that had 1 out of 1,000,000 odds. The fact the game show host had to filter the losing 999,998 doors out of the set makes switching advantageous since that door is weighted by the 999,998 losing doors that were filtered out.

Example 2. This example isn't applicable to the stated problem. When you introduce a 2nd player the host no longer can filter 2 doors down to 1 door unless both players choose the same door. If they choose the same door then it's the same classic Monty Hall problem just 2 iterations happening simultaneously. If 2 players select different doors then Monty has to open the 3rd unchosen door. According to you, that is 50/50 right? This is your premise but you're forgetting another option.

What you're forgetting is if neither player selected the correct door. In this case, Monty would have to Open player 1 or 2 door. This would then force that particular player to move as he now knows for a fact he will lose without switching. The player who's door was opened would switch obviously but he now has a 50/50 shot instead of 1/3. The player who's door wasn't opened should switch as this had no impact to him other than the typical Monty hall problem, i.e he now has 2/3 chance.

Answer 8

You can better see it imagining you had a box with $100$ balls, from which $99$ are black and only $1$ is white, which is what you want. You grab one randomly and keep it in your hand without seeing it. Do you agree that you are $99/100$ likely to have picked a black ball, so in $99$ out of $100$ attempts (on average) the white ball would still be in the box?

In case you agree, let's keep that ball in your hand and now suppose that another person deliberately pulls out $98$ black balls from the box. With "deliberately" I mean that that person sees what he is pulling; there is no risk that he removes the white ball by accident. In this way, there are only two balls remaining, one in your hand and one in the box, and one of them is necessarily the white.

What do you think is the probability that the white ball is which is in the box? If you say $50$%, what happened with the $99$ out of $100$ attempts in which it was still in the box? The revelation of the $98$ black ones didn't move it from the box to your hand.

Before the revelation of the $98$ black balls, the cases are:

                                        Hand       ||          Box
                                     =============================================
1) In 99 out of 100 attempts ->        1 black     ||    98 black ones and 1 white
2) In 1 out of 100 attempts ->         1 white     ||        99 black ones

So, when the other person removes the $98$ black balls from the box:

                                        Hand       ||          Box
                                     =============================================
1) In 99 out of 100 attempts ->        1 black     ||         1 white
2) In 1 out of 100 attempts ->         1 white     ||         1 black

So, it is true that you always end with two balls, one white and one black, but the important thing is that they are in two different positions (hand or box), and those two positions depend on the first selection. Moreover, that first selection determines that the white ball will end more frequently in the "box" position than in the "hand" position.

The way you are thinking the Monty Hall problem is like since you are always going to end with two balls, it would be the same if you started with both in the box and you had to grab one. But it is not the same. One thing is the probability to get the correct one when you randomly pick from two, and another different thing is the probability that the correct is already set in one position or in the other.

Note that if you randomly decide if you will pick the ball in the box or the ball in your hand, like flipping a coin, then you will get the white $50$% of the time. But that does not mean that it is $50$% of the time in the hand and $50$% in the box. It is because the extra times that you guess right picking the one from the box are compensated with the extra times you guess wrong picking the one from the hand. The $50$% $= 1/2$ is the average of the two cases:

$$1/2 * 99/100 + 1/2 * 1/100$$ $$= 1/2 * (99/100 + 1/100)$$ $$= 1/2$$

But if you always pick the ball that is in the box, your chances are:

$$1 * 99/100 + 0 * 1/100$$ $$= 1 * 99/100$$ $$= 99/100$$

In Monty Hall it occurs the same. Since there are two incorrect doors and a correct one, it is like if the $3$ doors were $3$ balls in the box, $2$ blacks and $1$ white. The initial selection is like when you start grabbing one ball randomly, and after the revelation the switching door is like the other ball that was left in the box. In $2$ out of $3$ attempts you pick a wrong door (like you would pick a black ball $2$ out of $3$ times) so in $2$ out of $3$ attempts the correct one will be the other the host leaves closed.

Answer 9

The error in your argument is that you don't start with two doors, you start with three.

If you pick one door out of three, you have a one in three chance of having picked the right one ($\tfrac{1}{3}$), and a two in three chance of having picked the wrong one ($\tfrac{2}{3}$).

If you have picked the wrong door, Monty shows you which door has the car behind it, by opening the other door. You win a car by switching.
Conversely, if you have picked the right door, switching makes you lose the car.

Since there's a $\tfrac{2}{3}$ chance of picking the wrong door at first, there's a $\tfrac{2}{3}$ of winning the car after switching. The chance of picking the right door is $\tfrac{1}{3}$, so there's a $\tfrac{1}{3}$ chance of losing the car after switching.