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.

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    What's wrong with the process (and what's really behind the whole Monty Hall paradox) is the fact that revealing a goat doesn't change the fact that there's a $1/3$ chance of $A$ containing a car. No matter what door you pick, there will always be at least one goat door you didn't pick, so by just having that fact confirmed (by revealing a goat), you haven't really recieved any information that changes anything about the door you picked. – Arthur Jan 14 '16 at 23:19
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    "If we have two doors, and know a car is behind one, then we have a 50/50 chance of choosing the right door." True, but only if that is all you know. If you have two doors, and one door is on a tiny doll house, and one is on a full size garage, you know the chance is not 50/50. If both doors were clear glass, again the chance is not 50/50. There was information used in the Monty Hall problem to reduce down to the set of two doors, and although now you only see two, one piece of information you have is that Monty had full knowledge of what was happening, and he would always open a goat door. – piCookie Nov 07 '16 at 20:42

9 Answers9


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: $$P(A_{\text{car}}|C_{\text{revealed}})\neq P(A_{\text{car}}|C_{\text{goat}})=\frac{1}2$$

Milo Brandt
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  • Well, this was one one the most convincing argument that I have seen so far. Thank you. – Our Jul 20 '18 at 02:48
  • **1.** *"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%."*. Can you elaborate this please? If the car isn't behind door A, then Monty can pick door A. Why didn't you mention A as a possible choice of door? **2. In your pen ultimate equation, where did the 1/3 spring from? 1/3 appears merely once in your whole answer!** – NNOX Apps Dec 31 '21 at 05:02
  • Feel free to edit your answer, instead of replying as a comment. – NNOX Apps Dec 31 '21 at 05:02

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!

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Kelvin Soh
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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.

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  • What is we see "switching door" as a "second guess"? If I give you two dates, and one is my bday, then you have a 50/50 chance, right? Then how is it different? Not switching = guessing 4/2, and switching = guess the other one. – OneZero Jan 14 '16 at 23:25
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    @OneZero Yes, switching doors corresponds to a second guess. If you give me only two dates, with no prior choices, then it's $50/50$. But you give me $365$ dates, and tell me to choose one of them. The probability that I'm right is $1/365$. You then eliminate $363$ possibilities, leaving me with $2$ choices, one of which is my original pick. Just because I now have only two choices left doesn't mean it's $50/50$, and the reason is that I made my choice back when there were $365$ options. – Mankind Jan 14 '16 at 23:32
  • Very nice way to put it. If someone isn't convinced, just try it 10 times with random friends and see how much you'll be right :-D – Ant Apr 08 '16 at 07:04

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 $50/50$.

David K
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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.

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    Can you elaborate on this? What's added? – OneZero Jan 14 '16 at 23:26
  • Have you seen an actual proof/ working out of the probabilities of the monty hall problem using bayes law and conditional probability? – flabby99 Jan 14 '16 at 23:29
  • Also I think it might potentially be helpful to imagine a situation where there is 100 doors, 99 of them have goats and 1 of them has a car. You choose a door. Monty opens 98 that have goats, and asks should you switch? – flabby99 Jan 14 '16 at 23:32
  • This answer doesn't answer the question. This should be a comment. – NNOX Apps Dec 31 '21 at 04:59
  • "What's wrong with this thinking process?" was the original question, which I attempted to answer. However, I did an egregious job of answering the question in the comments "Can you elaborate on this? What's added?", so here it is; What is added is that you have been revealed a goat in one of the two doors you didn't pick. To add context to that information - as you were more likely to randomly pick a goat from the three doors in the first place, being revealed another goat is great. You know where one goat is, and the other goat is likely behind the door you chose, so time to switch doors. – flabby99 Jan 01 '22 at 10:10

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.

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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.

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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.

Ronald Becerra
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  • **1. aren't your 2 tables identical? why did you present it twice? can you delete one of them?** **2.** Where does the $\color{Red}{1/2}$ hail from, in $\color{red}{1/2}*(99/100 + 1/100)$? – NNOX Apps Dec 31 '21 at 04:59
  • Feel free to edit your answer, instead of replying as a comment. – NNOX Apps Dec 31 '21 at 05:03
  • @NNOX Apps I have to reply. The two tables are not identical because one represents when 99 balls are still in the box, and the other when the other person pulled out 98 balls from there. Later, the 1/2 is for when the player would randomly decide to pick the ball in the hand or the one in the box, meaning he/she has 1/2 chance to select each. – Ronald Becerra Apr 02 '22 at 08:32

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.


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.


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  • Example 1: If you actually simulate it, by randomly choosing 1 door and opening 999998 empty doors, then it actually is almost certain that the prize is in the other door. – E. Z. L. Aug 06 '20 at 18:15
  • For 2: They aren't playing the game together, that is ridiculous. The original argument concerns a single player – E. Z. L. Aug 06 '20 at 18:18
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    That would be like saying 1+1 isn't 2 because 1+3 isn't 2 – E. Z. L. Aug 06 '20 at 18:19