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After reading the other question about the Monty Hall problem, and seeing the solution of it, before giving my perspective to the question, I would like to point out that, to explain the possible cases, we either fix the doors that I choose and Monty Hall opens, and permutate the values of the doors, i.e what is behind them, or we fix the values of the door, and permutate the door we and Monty Hall chooses.

Given that, let WLOG I choose the first door and Monty Hall opens the 3rd door, then we have the following cases;

1-> car 2-> goat 3-> goat

or

1-> goat 2-> car 3->goat

Since, we do know that the car is not behind the door 3.

So, there is only 2 option in one of them I win, and in the other I loose, hence the there is no difference in changing the door I chose first, hence the probability of the door 1 has the case is 1/2.

So my question is what exactly wrong with this argument ? Exactly which step of the argument fails to be true ?

Edit:

In this question, I was pointing out some of the problems that I general answer to Monty Hall problems contains, so it is different than the question that is claimed that this question to be duplicate of.

Our
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  • What argument? What are you trying to show? Your notation is not self-explanatory...there aren't, for instance, any probabilities attached to it. – lulu Jul 16 '18 at 14:55
  • Are you trying to say that after they show door 3 it is $50/50$? – Holo Jul 16 '18 at 14:57
  • @lulu See my edit. – Our Jul 16 '18 at 14:57
  • @Holo Yes, that is what I'm saying. – Our Jul 16 '18 at 14:57
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    The fact that there are only two options does not mean that they are equally likely. – lulu Jul 16 '18 at 14:57
  • @lulu I have never claimed such a thing either. I have just found all the possible cases, and considered each cases that I won if I stay with the door 1. – Our Jul 16 '18 at 14:58
  • But you ignore all the probabilities. How do you claim to show that the two options are equally likely without computing any probabilities? – lulu Jul 16 '18 at 14:59
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    "I have never claimed such a thing either." Yes, you did. When you claimed that there is no difference and therefore $50%$ chance of winning either way, that's exactly what you did. – Arthur Jul 16 '18 at 15:00
  • @lulu I'm just counting the cases where I win over the all cases, which in return gives me the probability of my winning; Since I'm considering each cases, there is no probability in each separate case; For example when we have car;goat;goat situation, the probability that I win is 1. – Our Jul 16 '18 at 15:03
  • To be clear: Monty opening door $3$ is evidence (not proof) for the statement "the prize is behind $2$". After all, if the prize were behind $1$ instead, there's only a $\frac 12$ chance he'll open $3$ whereas if the prize is behind $2$ it is certain that he will open $3$. – lulu Jul 16 '18 at 15:04
  • I understand that you are counting the (two) scenarios but they are simply not equiprobable. If the prize is behind $1$ there is only a $\frac 12$ chance that Monty opens $3$ but if the prize is behind $2$ it is certain that he will open $3$. That asymmetry is crucial. – lulu Jul 16 '18 at 15:05
  • @Arthur Ok, that is true, I can see, but nevertheless, I cannot understand why it is not true ? After all, there are just possible case, they are not cases that happens with some probability; For example, I flip a coin, and the possible cases are head, and tail; I'm just considering each possible cases. – Our Jul 16 '18 at 15:06
  • @lulu Please read my first paragraph in the question; you are both permutating the values of the doors, and the door that the Monty chooses, which lead over counting. – Our Jul 16 '18 at 15:08
  • As you do no computations of probability, your argument is incorrect or at best incomplete. That's all there is to it. – lulu Jul 16 '18 at 15:08
  • Imagine we did this $300$ times (each time you choose $1$). In $100$ cases the prize really is behind $3$ so we throw those out. In $100$ cases the prize is behind $2$ and in all of those cases Monty opens $3$, so all those cases signify. In $50$ cases the prize is behind $1$ but Monty opens $2$, so we have to discard those cases as well. Thus there are $150$ cases out of $300$ in which Monty opens $3$ and in $100$ of those the prize is behind $2$. – lulu Jul 16 '18 at 15:11
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    I have purchases a lottery ticket. WLOG there are two outcomes: that I have purchased the only winning ticket, or that I have purchased one of the two million losing tickets. Therefore there is a fifty-fifty chance that I will win! Yah! – Graham Kemp Jul 16 '18 at 15:23

2 Answers2

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You cannot do the "WLOG" and expect probabilities to stay equally distributed over your cases.

That is, you can say -- assuming that the car is behind a uniformly random door -- that you choose door 1, and then the three options are (car, goat, goat), (goat, car, goat) or (goat, goat, car) -- each with probability 1/3. That is fine.

However, when you then say "Without loss of generality, Monty opens door 3", you are excluding case 3, and implicitly treating it while you are treating case 2. If you continue you then have to add the probability of that situation occurring to the treatment of case 2. There are two options left: (car, goat, goat) and (goat, car, goat), but the first has probability 1/3, while the second has probability 2/3.

Edit: As a more general point, you seem to be using the common argument, "there are $n$ options, and therefore each occurs with probability $1/n$". This argument is dangerous: there's no reason to expect it to hold in general, but you often need it as an implicit assumption, because people forget to specify e.g. that a die or a coin is fair.

Consider the following similar argument: you throw two fair 6-sided dice, and your roll is the sum of the two numbers. The possible outcomes are $\{2,3,4,5,6,7,8,9,10,11,12\}$. You might be tempted to conclude that, since there are 11 possible outcomes, the probability of rolling 2 is $1/11$ -- but if you've played enough Settlers of Catan you know that to be false!

Mees de Vries
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  • Nicely explained (+1). – Mark Viola Jul 16 '18 at 15:04
  • But, the set ${2,3,4,5,6,7,8,9,10,11,12}$ is not the sample space of the dice, whereas in this case, I'm directly looking into the case where we considering the each possible cases for the doors. – Our Jul 16 '18 at 15:13
  • @onurcanbektas, it could be the sample space for the dice, if all you are interested in is the outcome of their sum. And you are explicitly not looking into each of the possible cases for the doors; that is what the "WLOG" is doing for you. – Mees de Vries Jul 16 '18 at 15:15
  • Exactly. Saying that you open door 1 and Monty opens door 3 loses generality, because that can only happen in the specific case there is not a car behind it. – Graham Kemp Jul 16 '18 at 15:17
  • @GrahamKemp Then think like this; I'm in the contest right now, and I have chosen the door 1, and Monty opened the door 3, what about now ? I mean this situation is occurring right now, so I'm not saying "WLOG", I'm just saying what I'm seeing. What is the chance of my winning if I stay with 1 ? – Our Jul 16 '18 at 15:25
  • There was and still is a probability of $1/3$ for the car being behind the door you selected. Monty's reveal cannot have changed the probability that you chose door hiding the car. He has only revealed that if the car is not behind the door you chose, then it must be behind the remaining door. – Graham Kemp Jul 16 '18 at 15:36
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We have $$\begin{array}{c|c|c}\#1&\#2&\#3&\mbox{Result at }\#1&\mbox{result otherwise}\\\hline c&g&g&win&lose\\\hline g&c&g&lose&win\\\hline g&g&c&lose&win\end{array}$$We can't assume that the host open door $3$ because then you say "the third option(g,g,c) can't happen", what the host does is taking the column of $\#2$ or $\#3$ "away", but the results won't change.

Update:

You have to note that the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50\%$ for each door. you have $\#1$ your choice and host open $\#2$ to be $1/3\cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $\#1$ your choice and host open $\#3$ to be $1/3\cdot1/2=1/6$, together you gave $1/3$.

But if you chose incorrect there is $100\%$ that the host will open a particular door: $\#1$ your choice and host open $\#2$ to be $1/3\cdot1=1/3$ and and host open $\#3$ to be $1/3\cdot1=1/3$, together it is $=2/3$.

Holo
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  • Ok, but we do know that Monty will never open the door with the car, so yes that option will never happen. – Our Jul 16 '18 at 15:09
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    @onurcanbektas but the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3\cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3\cdot1/2=1/6$, together you gave $1/3$. if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3\cdot1=1/3$ and and host open $#3$ to be $1/3\cdot1=1/3$, together it is $=2/3$. – Holo Jul 16 '18 at 15:26
  • Well that was convincing :) I advise you to turn that comment into an answer. – Our Jul 16 '18 at 15:31
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    And that is where generality is lost. We can only say WLOG that you select door 1 - and the car will be behind a door with equal probability for each (1/3). If the car is behind door 3, Monty cannot open it; he must open door 2 instead - and switching will win. Just as if the car is behind door 2, Monty must open door 3 - and switching will win. Yet if the car is behind door 1, Monty may freely select either door 2 or 3 - and whichever the case switching will lose. So switching will win in two of the three rows. – Graham Kemp Jul 16 '18 at 15:32
  • @onurcanbektas I added it to the answer – Holo Jul 16 '18 at 15:32