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I am trying to solve the Monty Hall Problem with $3$ doors ($2$ goats, $1$ prize) with a simulation of the rules

  1. You initially pick a door randomly from the 3 possible doors.
  2. Monty hall does not know anything about the prize door and the goats door.
  3. He opens a random door except the one you have chosen ( The door he opens may even be the prize).
  4. Gives you the option to switch.
  5. Should switching be preferred.

My simulations give the $\mathbb{P}$robability as $\dfrac{2}{3}$ in this case. The event is $\mathbb{P}(C_2|G)$, where $C_i$ is the probability when the prize is behind $i$, $G$ is the event that a goat is revealed behind the third door.

Even if the 2nd condition is changed to Monty hall every time shows a different door from the chosen one but always the goat door, even then the $\mathbb{P}$robability is $\dfrac{2}{3}=\dfrac{1}{1+\frac{1}{2}}$

are my two probabilities correct

SMK
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Upstart
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  • Do you mean the 2/3 as the probability you get the car, on the condition that when offered to switch you choose switch? [If so it would seem to make no sense in cases where monty shows you the car] – coffeemath Aug 13 '23 at 10:45
  • Suppose the car is in 1st door, and I chose the 2nd door. Monty randomly chose 1st door to reveal. Gives me an option to switch, I switch to the 1st door. I win. – Upstart Aug 13 '23 at 10:51
  • It makes not much sense if the door with the prize can be opened. Good for the candidate to win without guessing , but a serious show must avoid this nonsense. – Peter Aug 13 '23 at 11:43
  • Then what else does a statement mean " the host chooses a different door at random" so at random he might as well choose the prize door to be revealed. So to avoid this we would have to discard that round of guessing. – Upstart Aug 13 '23 at 11:55
  • You stated My simulations give the $\mathbb{P}$robability as $\dfrac{2}{3}$ in this case. The probability of what event? The way you have stated step 5, you have not defined an event, so it is not clear what you are asking. – Lee Mosher Aug 13 '23 at 12:09
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    It would be an event if Step 5 were replaced by "You choose not to switch". It would be a different event, with a different probability, if Step 5 were replaced by "You choose to switch to the unopened door". And it would be still different if Step 5 were replaced by "If Monty reveals the prize then you choose to switch to the door he opened but if Monty does not reveal the prize then you choose to switch to the door he didn't open". – Lee Mosher Aug 13 '23 at 12:11
  • What I modelled was actually "If Monty reveals the prize then you choose to switch to the door he opened but if Monty does not reveal the prize then you choose to switch to the door he didn't open". – Upstart Aug 13 '23 at 12:24
  • Well, the game is easy to analyze if Monty Hall can also open the prize-door. I think you have no problem to do this , right ? So , what exactly is your question ? – Peter Aug 13 '23 at 17:23
  • Another question in this case is whether the showmaster would allow the switch if the prize-door was revealed. This depends on the purpose of the game. Shall the candidate win to increase the viewing figures , or shall the candidate lose because it would be extremely strange to let him win in this scenario. – Peter Aug 13 '23 at 17:27
  • Finally : You need no simulation for a game with just $3$ possibilities : 1. The candidate decided originally right. Then, switching loses whatever door is opened and not switching wins. 2. The candidate was wrong and the showmaster opens another wrong door. Then , switching automatically wins and not switching loses. 3. The candidate was wrong and the showmaster opens the right door. Well, this case is obvious. – Peter Aug 13 '23 at 17:30
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    If when he reveals the car you are allowed to switch to that car, and when he reveals a goat you switch to the other closed door, it is obvious that you will win as long as the car is in any of the two doors that you didn't initially pick, which happens 2/3 of the time. – Ronald Becerra Aug 13 '23 at 19:13
  • @RonaldBecerra - switch. don't switch, it matters not. In the bog-standard Monty Hall problem, switching was how you took advantage of Monty Hall's knowledge of which door held the prize. But in this version, Monty doesn't know (or at least reveal) any more knowledge than you have. If a goat is revealed, switching has no impact on the odds. – Paul Sinclair Aug 14 '23 at 16:30
  • In the actual program, when Monty opened an unpicked door/curtain, that choice was eliminated. Instead of "there's the big prize, do you want it?", it was "tough luck!" – Paul Sinclair Aug 14 '23 at 16:30
  • @PaulSinclair Yes, I was not denying it. But I was referring to the OP's comment that he was modelling: "If Monty reveals the prize then you choose to switch to the door he opened but if Monty does not reveal the prize then you choose to switch to the door he didn't open" That way, it is obvious that he will get the car 2/3 of the total games, but that will be because that proportion is also including those in which the car is revealed, not restricting to the subset in which a goat is. – Ronald Becerra Aug 14 '23 at 17:35

1 Answers1

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I do not understand the question, so going by your title and intro, I will provide the intuition behind the solution of the Monty Hall Problem. If this does not answer your question, please comment that down below.

Let us say we are playing the "Lets Make a Deal" Game. I am Monty and you are the player.

We have three doors to choose from: A, B and C.

As you have already explained the rules of the game, let us just jump into the solution, shall we?

Solution

Let us assume that A and C have the goats and B has the car.

                                   A B C

Now, if you pick C, then I am force to show you a goat, that meaning I have to show you door A. If you switch, then you win the car!

If you picked door A, then I will be forced to show you door C; If you switch, you win the car again.

The probability of losing comes when you pick the right door. If you pick B, then I will show you either A or B. After showing you a goat, if you switch, you lose.

However, we just came up with a conclusion: Out of three games, you win two times, and you lose once. That being said:

P(Winning) = $2/3$

P(Losing) = $1/3$

SMK
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