0

The Monty Hall problem can be stated as follows:

There are three doors on the set for a game show. Let’s call them A, B, and C. If you pick the correct door, you win the prize. After you pick some door, the host of the show then opens one of the other doors and reveals that there is no prize behind it. Keeping the remaining two doors closed, he asks you whether you want to switch your choice to the other closed door or stay with your original choice. What should you do if you want to maximize your chance of winning the prize: stay with your initial answer or switch — or would the likelihood of winning be the same either way?

The generally accepted answer to the problem: yes, you should switch.

Question: can the following argument be used to explain the answer to the problem?

Let's instantiate a new variable $I[nitial] = 1/3$, which denotes the probability to win, by choosing the door A at the start of the game. Suppose the host opens the door C. Now you are left with two doors: the door A and the door B. Should you switch? Let's examine what we might mean by switch precisely:

  1. Switch means to choose the door B or to choose to do nothing.

  • Case a) If you choose to the door B, then you are able instantiate the probability of winning as $F[inal] = 1/2$.

  • Case b) If you choose to do nothing, you are done with your line of reasoning.

  • Therefore, we should take the path which case a) offers since this is the only way to introduce $F$ by existential instantiation. We want to introduce the variable $F$ so that we can later conclude that $F > I$ and conclude that a switch is the right option.

  1. Switch means to be able to implicitly choose the door A again by sticking to your original answer or to explicitly choose the door B. After choosing either door the probability to win $F[inal] = 1/2$ is existentially instantiated, so in this case it does not matter which door you choose: both your implicit choice of A and explicit choice of B introduces a probability of $1/2$, which is greater that $1/3$.

Therefore, you should always switch regardless of you interpretation of switch since it always introduces a higher probability to win.

The Definition 2 of switch seems to be how most people interpret host's question when they first encounter the Monty Hall problem. Definition 1 is what makes this problem hard to grasp. My argument relies on the assumption that to introduce a probability, you must choose a door either implicitly or explicitly: given a sample space $S = \{A, B\}$, you must first introduce an event $E = \{A\}$ by implicitly choosing the door A by "staying with your initial answer" or an event $E = \{B\}$ by explicitly choosing the door B to introduce a higher probability $F = \frac{|E|}{|S|} = 1/2$. However, if you interpret "stay with your initial answer" as doing nothing, you do not introduce a higher probability of $1/2$ at all: there is no existential instantiation. Therefore, are only left with the probability of $1/3$, and, therefore, are not able to maximize your chance of winning.

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Vlad Mikheenko
  • 402
  • 4
    But the probability of winning if you switch is $\frac 23$ not $\frac 12$? – altwoa May 16 '25 at 07:59
  • In my reasoning, when the host opens the door C you essentially reduce the whole game to just two doors: A and B - the prize is guaranteed to be behind these doors. Before the host opened the door C, you've only instantiated a probability of $1/3$. a) Now, if you are able to choose either door, you instantiate a probability of $1/2$. b) If you for some reason are not allowed to choose the door A again, then your best option is to choose the door B since it instantiated a probability of $1/2$; if you do nothing, you don't get a higher probability. Therefore, you should always switch. – Vlad Mikheenko May 16 '25 at 08:12
  • 1
    "if you do nothing, you don't get a higher probability.." But $\frac12+\frac13<1$ so that cannot be correct. It is for sure that the prize is behind A or B. – drhab May 16 '25 at 08:16
  • 1
    If the host has picked randomly between B and C (i.e. without checking whether one of them hides the prize) and it appears that C does not hide a prize then by symmetry A and B both have probability $\frac12$ to hide the prize. If the host has used his foreknowledge (urging him to not reveal a prize) then A still has probability $\frac13$ to hide the prize so that door B has probability $\frac23$ to hide the prize. In both cases (the second is the real Monty Hall problem) the probabilities add up to $1$. – drhab May 16 '25 at 08:30

1 Answers1

1

I don’t think you fully understand the Monty Hall problem. One of the details of the game that you omitted is that the host is always going to open a door with nothing behind it, and is always going to pick a door that you did not choose.

So initially if you pick door A, you have probability $\frac{1}{3}$ of having picked the right door. When the host intentionally reveals a door with nothing behind it (that you did not choose), the probability that you intially picked the right door doesn’t change, because you gain no information on the first door that you picked ($\frac{1}{3}$ of the time, the door that the host will choose will be random between the other two doors, $\frac{2}{3}$ of the time, he will choose the other door with nothing behind it. You cannot distinguish between the two cases as a participant in the game until you decide to switch or not).

However, you gain information between the two doors that you did not pick, namely that in the case where you initially picked wrong ($\frac{2}{3}$ of the times), the door that the host didn’t open is the right door.

Therefore, if you switch, you have probability $\frac{2}{3}$ of winning, and if you don’t, you have probability $\frac{1}{3}$ of winning

Thomas Lacasse
  • 249
  • 5