31
$\begingroup$

So the setup of the scenario here is exactly the traditional Monty Hall problem. Except, before the game starts, you decide to cheat and open a door and it happens to be a goat. Before you can peek at the two other doors, the game begins, and you pick a door that's not the one you just peeked into (obviously, because you peeked a goat). Here it seems the probability to win a car is $\frac{1}{2}$, because you improved your original $\frac{1}{3}$ chance of winning a car by cheating.

Then the game host opens the same door you just peeked at before the game started (knowing it was a goat, like the traditional Monty Hall problem). Is it still a $\frac{2}{3}$ chance of winning the car if you switch, even though you already knew the informatiom beforehand? Or does it still remain $\frac{1}{2}$ because you already knew the information before, so you just screwed yourself out of the better $\frac{2}{3}$ probability by cheating?

If you originally picked a door with a goat, then the game host must pick the door you peeked at, because the other door contains the car. In this situation, it doesn't seem like anything changes because you already knew the door the game host opened beforehand by cheating.

If you originally picked a door with a car, the game host can open either other door, because they both have goats. If the game host opens the door you did not peek at and reveals a goat, then you know with $100\%$ probability that the door you originally picked contains the winning car. However, if the game host opens the door you peeked at, then it seems like the same situation in the paragraph I described before this one.

$\endgroup$
  • 7
    $\begingroup$ the door happens to be a goat? that's nontraditional Monty Hall $\endgroup$ – zhw. Jul 26 '17 at 15:34
  • 3
    $\begingroup$ As zhw pointed out, doors are not goats! I believe you mean "Pick the door you KNOW the goat is behind. Other than that, you comment is excellent. $\endgroup$ – user247327 Jul 26 '17 at 19:06
  • 1
    $\begingroup$ Also, when these Monty questions pop up no one ever clarifies if the player in question mostly needs a new door or a car or a goat or a host. $\endgroup$ – mathreadler Jul 27 '17 at 0:26
  • 8
    $\begingroup$ since you cheated and you know where one goat is, then there's a strategy that guarantees you a win. which is pick the door behind which you know there is a goat. $\endgroup$ – A.Ellett Jul 27 '17 at 5:06
  • 1
    $\begingroup$ If Monty will Always show you the door you peeked behind, Gregory is right. If he still follows the rules of the game, Linnell is right. $\endgroup$ – Weckar E. Jul 27 '17 at 13:24
34
$\begingroup$

If you pick the car first, as you say, the host has 50% chance of confirming you have the car, and 50% of giving you no new information.

If you pick the unknown goat first, the host gives you no new information.

Ultimately the host has a $\frac{3}{4}$ chance of revealing the goat you already knew about.

If you're picking between those two randomly, then in $\frac{1}{4}$ of all cases, you pick the car, then the host reveals the second goat, so you win the car. In the other $\frac{3}{4}$ of cases, you're back to the $\frac{2}{3}$ - $\frac{1}{3}$ situation, since in $\frac{2}{3}$ of those $\frac{3}{4}$, you had the goat to begin with, and in the other $\frac{1}{3}$ you had the car.

$$P(\mathrm{car}|\mathrm{unknown\,goat\,revealed}) = 1$$

$$P(\mathrm{car}|\mathrm{known\,goat\,revealed}) = \frac{P(\mathrm{car}\cap\mathrm{known\,goat\,revealed})}{P(\mathrm{known\,goat\,revealed})} = \frac{1/4}{3/4} = \frac{1}{3}$$

You can expect to get the car $\frac{3}{4}$ of the time by switching if the goat you knew about is revealed and staying if not.

Of course, if you pick randomly between the two unpicked doors, you're a fool, because picking the door you know to be a goat forces Monty to show you the other goat and hence brand new car :)

$\endgroup$
  • 2
    $\begingroup$ There are only 3 outcomes, but they aren't equally likely. In the original Monty Hall, there are only 4 outcomes - pick goat A and reveal goat B, pick goat B and reveal goat A, pick car and reveal goat A, or pick car and reveal goat B. But these aren't equally likely - the former two are twice as likely as the latter two. Similarly, the three outcomes here are: pick unknown goat and reveal known goat, pick car and reveal known goat, pick car and reveal unknown goat. Again, the former is twice as likely. Unlike you, Monty isn't picking randomly - picking a goat forces Monty to reveal the other. $\endgroup$ – T. Linnell Jul 26 '17 at 16:00
  • 11
    $\begingroup$ IIRC, in the real game Monty didn't always give the option to switch. So picking the goat door could backfire $\endgroup$ – Ben Aaronson Jul 26 '17 at 17:00
  • 1
    $\begingroup$ @CarstenS The standard problem is about the situation where he does ask you whether or not to switch. If he doesn't... well, that's pretty trivial, and not very interesting: you pick a door at random and you're stuck with it. Either way, your strategy for the initial door pick in the standard version is just to pick one at random (or however you like), because you have no information so there's nothing better you can do. In this situation you have some info before even your initial door pick because you cheated. $\endgroup$ – Ben Aaronson Jul 27 '17 at 14:18
  • 2
    $\begingroup$ @BenAaronson, I am just saying that the standard version at least implicitly assumes that the host will always open one of the other doors, or at least that his choice does not depend on the contestant's choice. Because, if the host would only ever open another door and let the contest switch if the contestant chose the car in the first round, then switching would not be a good strategy. $\endgroup$ – Carsten S Jul 27 '17 at 15:39
  • 1
    $\begingroup$ @CarstenS Ah, I see what you're getting at. Yes, Monty giving the option whether or not to switch would have to be random (or at least fair). I agree that if we didn't know Monty's protocol for choosing whether or not to give the option, then we wouldn't have enough information to make the right choice or judge the probabilities. $\endgroup$ – Ben Aaronson Jul 27 '17 at 17:28
33
$\begingroup$

As a side note, wouldn't it be better to pick the door you knew had a goat? Thus Monty will have to reveal the other goat?

$\endgroup$
  • 5
    $\begingroup$ It's not a side note. This is the correct answer. $\endgroup$ – J Ashley Jul 26 '17 at 20:09
  • 3
    $\begingroup$ It boosts the probability of knowing which door has the right up to 1. Pretty good! $\endgroup$ – DoctorHeckle Jul 26 '17 at 20:52
  • 11
    $\begingroup$ It's the correct answer to "how do you maximize your probability of getting the car?" but that's not what is being asked here. I don't believe this actually answers the question asked. $\endgroup$ – David Z Jul 26 '17 at 21:51
  • $\begingroup$ @DavidZ: This is of course game theory, so you need to consider strategies. It's a given here that the player is cheating, which isn't one of the two strategies in classical Monty Hall. $\endgroup$ – MSalters Jul 26 '17 at 22:12
  • 3
    $\begingroup$ @MSalters This question really is not that much about game theory, but just a straightforward probability question. $\endgroup$ – JiK Jul 26 '17 at 23:03
4
$\begingroup$

Your chance still improves even if Monty picks the same goat you already saw. That may seem odd, so I'll try to explain it intuitively.

Forget Monty Hall for a moment and think about another situation. Your friend has two coins. One's a regular, unbiased coin, the other is double headed. He tells you to pick one at random. Your chance of having picked the double-headed coin is now 50/50.

Now he flips the coin you picked 10 times, and every time it comes up heads. It's still possible you picked either coin, but now it's clearly more likely you picked the double-headed coin.

In fact, even if he just flipped it once and it came up heads, you're still a bit more likely to have picked the double-headed coin. If we label the coins "S" (single-headed) and "D" (double-headed), the combinations are:

SH
ST
DH1
DH2

So eliminating ST, there's now a 2/3rds chance that you have the double-headed coin.

Going back to Monty Hall, we can map this situation directly. Your choice between the two coins maps to the choice between the two remaining doors, and Monty's choice of which door to reveal maps to the coin flip.

If you picked the goat door, Monty has only one option- picking the door you already saw. This corresponds to picking the double-headed coin, where the flip only has the option to come up heads. If you picked the car door, Monty has two options, just like if you picked the fair coin, it could come up either heads or tails.

So the situation where Monty picks the door you already saw is just like getting heads on the first flip- it means there's a 2/3rds chance you have the goat door. So, just like regular Monty Hall, you should switch.

$\endgroup$
-2
$\begingroup$

No, the probability is now $1/2$ not $2/3$ because for it to be $2/3$ with the switching strategy you need a $2/3$ chance of originally chosing the wrong door, but by cheating you lowered that to $1/2$

$\endgroup$
  • 1
    $\begingroup$ So somehow forgetting the information you learned would give you better odds? That's.... silly $\endgroup$ – Weckar E. Jul 27 '17 at 10:54
  • 1
    $\begingroup$ Here's the thing though: IF he reveals the only you already know to be wrong, yeah you have a 50/50 chance. BUT he could immediately confirm your door to be right half the time, too. So your odds actually rocket to .75 $\endgroup$ – Weckar E. Jul 27 '17 at 13:22
  • 2
    $\begingroup$ @GregoryGrant The strategy isn't to always switch. The strategy is to switch if Monty shows you the goat you peeked at, and to not switch if he shows you the goat you didn't peek at, and this strategy gives a win-rate of $3/4$. There are three possibilities, not all equally likely, when you choose to not pick the goat you peeked at: 1) You picked the unknown goat, Monty shows the known goat (as that is his only choice), $p=0.5*1=0.5$. 2) You picked the car, Monty shows the known goat, $p=0.5*0.5=0.25$. 3) You picked the car, Monty shows the unknown goat, $p=0.5*0.5=0.25$. (tbc.) $\endgroup$ – T. Linnell Jul 27 '17 at 13:55
  • 1
    $\begingroup$ Case 3, you stay. In cases 1&2, all you know is Monty showed the goat you already knew about. So the probability of you having picked the car, given Monty showed you the known goat, is not $0.5$, but is in fact $P(\mathrm{car}\cap\mathrm{known\,goat\,shown})/P(\mathrm{known\,goat\,shown} = 0.25/0.75 = 1/3$, and the probability you had the goat given Monty shows the known goat is $0.5/0.75 = 2/3$. While you're equally likely to pick the goat and the car at first, Monty can only reveal the known goat if you pick the other. If you pick the car, Monty has a choice of which goat to reveal. $\endgroup$ – T. Linnell Jul 27 '17 at 13:55
  • 1
    $\begingroup$ T. Linnell is correct. And keep in mind this question is explicitly asking about the situation where he does show you the door you saw from cheating $\endgroup$ – Ben Aaronson Jul 27 '17 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.