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So the typical set up for Monty Hall problem, there are 3 doors where 2 have goats and 1 has a car. I, the contestant, get to randomly guess a door looking to get the one with the car, after this the host will open a door that will always be a goat. Thus out the two doors that are left, I have to choose to stay with the original door I chose or switch to the other door. As many analysis of this problem have been done, switching my choice gives me a higher probability of winning. This largely has to do with the fact that since the host always reveals a goat the asking of whether to stay or not, is the same as did you guess right or not, and you have $\frac{2}{3}$ of wrong so you should switch

Now it seems, this "strange" result largely has to do with the fact that the host always reveals a goat. But what if alternatively you had this situation

You are given 3 doors, 2 with a goat and 1 with a car. You randomly choose a door (looking to get one with the car). The host will randomly choose to reveal what is behind one of the 2 doors you haven't chosen. Given that he reveals goat, what is the probability of getting car if you chose to stay with choice?

My analysis of this problem goes as follows:

Let $D$ be event the door I guessed has car and Let $G$ reprsent the event that host reveals a goat thus what I want to calculate is $P(D|G)$ with this I have $$P(D|G)=\frac{P(D\cap G)}{P(G)}=\frac{P(G|D)P(D)}{P(G|D)P(D)+P(G|D^{c})P(D^{c})}=\frac{1\left(\frac{1}{3}\right)}{1\left(\frac{1}{3}\right)+\frac{1}{2}\left(\frac{2}{3}\right)}=\frac{1}{2}$$

So it seems it doesn't matter if I choose to switch or not, and this is the result most people come up with when first thinking of problem.

Question: First is my analysis correct for this problem? Second, is it true in general that if you guess out of $n$ doors and host reveals $k$ doors that all have goats, will the probability that car is behind door you choose just $\dfrac{1}{n-k}$?

UPDATE

So I ended up asking my statistics/probability professor about this question and he said the result I got was correct. He explained that the reasoning that the Monty Hall problem inherently causes confusion is because many don't notice the only randomness in the original problem is just in your choice while hosts choice of door is deterministic. Now the problem I asked now has two sets of randomness, your original choice of door and the hosts choice thus the problems are inherently different

Theo Bendit
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Kamster
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3 Answers3

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Your analysis is correct. Suppose that there are $n$ doors, one of which has a car, the other have goats. The host randomly chooses $k$ doors and opens them. I will use your notation, so $D$ is the event you have chosen the car and $G$ is the event the host reveals $k$ goats.

Then we have $$ \mathbb{P}(D|G) = \frac{\mathbb{P}(D \cap G)}{\mathbb{P}(G)} = \frac{\frac{1}{n}}{\frac{n-k}{n}} = \frac{1}{n-k}. $$ This is because

  • the probability $\mathbb{P}(G)$ that the host only reveals goats is $\frac{n-k}{n}$ (as it is the probability that the car is among one of the other $n-k$ doors),
  • the probability $\mathbb{P}(D \cap G)$ that you have chosen the car and the host only reveals goat is $\frac{1}{n}$ as this is the same as the probability $\mathbb{P}(D)$ that you have chosen the car.
user133281
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    That seems about right. I was also wondering if the question to this problem is the same as: let there be a deck of $n$ cards you choose one card out of the deck. The deck is then shuffled, you start taking cards off top of the deck, and you have chosen $k$ cards and none of them are your card. What is probability that next card is your card? – Kamster Aug 28 '14 at 21:51
  • I think that is indeed an equivalent situation. – user133281 Aug 28 '14 at 21:53
  • Ok cool, because I think the question explained this way that makes it intuitively obvious that the result should be $\frac{1}{n-k}$ and I always like to have intuitive explanations when I can – Kamster Aug 28 '14 at 21:53
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1/2 because there are only 2 choices left at that point and one has to be the car.

David
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  • What I essentially tried to do was change the problem so it got the result that most people mistakenly get. I said instead revealing goat always like in original monty hall, he chooses door at random, if he happens to show us a goat, then what is probability that have car if stay. The fact that you can have case that car can be revealed now essentially changes the problem because the door reveal now actually gives us information instead of no information like in original monty hall – Kamster Sep 25 '14 at 06:36
  • I reworded the problem in comments user133281 in terms of a deck of cards which is essentially the same problem that after the fact I saw was a lot easier to see the result I hypothesized it would be more intuitively – Kamster Sep 25 '14 at 06:40
  • Ok I respect your opinion and I'll be sure to take that into account and think through problem more – Kamster Sep 25 '14 at 06:57
  • Let us continue this discussion in chat. – Kamster Sep 25 '14 at 07:13
  • Also if you really want to verify which one is correct, do a simulation of about 1000 runs of the game and proportion of which you win in should be close to actual probability by law of large numbers – Kamster Sep 25 '14 at 07:30
  • Ok I think I see your problem. This probability is finding a conditional probability where the condition is that random door revealed is goat. I am not looking at probability that given revealed car, that probability is obviously equal to 0 – Kamster Sep 25 '14 at 07:46
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Your analysis of the intuition looks good. I think your right that people think his way. But no. No matter the number of goats revealed, your odds of initially guessing right are always 1 in $n$.

It helps to imagine that there are 100 doors. And that the host reveals 98 goats. It is easier to see that you should switch.

amcalde
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  • This does not answer the question. – user133281 Aug 28 '14 at 21:43
  • Wow my first down votes. I tried to clarify. – amcalde Aug 28 '14 at 21:46
  • But if the host, is randomly opening doors and the all happen to be goats, that should give me some information, not just leave with no information like before he revealed them – Kamster Aug 28 '14 at 21:48
  • But you made your choice in advance. That shot is just $1/n$ because there was no information. – amcalde Aug 28 '14 at 21:49
  • The option to switch gives you better odds. It took me some time to wrap my mind around this one, but the 100 door case helped me, that's all. – amcalde Aug 28 '14 at 21:50
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    But the fact that randomly chosen doors reveal goats gives you information about your choice: it becomes more likely that you have chosen the car. Please note that this is an alternative Monty Hall problem. – user133281 Aug 28 '14 at 21:50
  • I get what you are saying, but even in your alternative universe where Monty somehow misses the car, your initial guess will always only be right $1/n$ and the more revealed goats you see, the more you should be convinced to switch. What would happen in your alternative version if Monty accidently showed the car? Would you just start over until he only randomly showed goats? If so this alternative game is the same as the original game. – amcalde Aug 28 '14 at 21:54
  • Well thats is the original reason why I knew this problem must be different, because randomly revealing doors has the chance of actually revealing a car, thus changing probabilities. The fact that there was a chance that host could have revealed car but didn't should give us more confidence in our choice of door (i.e. increase probability that car is behind door we chose) – Kamster Aug 28 '14 at 21:58
  • Your origininal odds: $1$ in $n$. Monty's behaviour after that (random or knowing) cannot change your odds of being right on the first guess. – amcalde Aug 29 '14 at 13:32
  • Imagine, as you do, that if he opens a car then the game restarts. Then your probability of winning if you keep will go up because, every time he reveals a car, you have avoided losing by keeping. Hence, in this setting, it is trivial that your probability of winning if you commit to keep your door must be strictly greater than 1/3 (because you will win 1/3 of the time on your first go, and 1/3 of the time you will get a do-over). And, as it happens, your probability of winning will just be 1/2 (which we already know is the answer from the other answers posted in this thread). – guy Feb 22 '16 at 22:18