Every Variant Of Monty Hall That I Know

本站添加: 2026-06-01

[00:00:00]Pretty much everyone knows the Monty Hall problem these days. A quick refresher if you don't remember, you have three doors in front of you with what lies behind them randomized.

[00:00:09]>> [music] >> Behind one of them is a car and behind the other two are goats. Assuming you want the car, the game show host Monty will then open a door after your initial selection. As the show's host, he obviously knows where everything is and purposefully chooses to reveal a goat to make things more interesting. Monty then offers you the choice to switch doors.

[00:00:31]Imagine all goats and cars for this video are indistinguishable unless otherwise noted and of course, feel free to pause the video to try any puzzles yourself. Now, if you've never encountered this problem, the initial strategy appears to be unchanged since it looks like it's still a 50/50 chance of where the car will be. So, randomly switching to the other door wouldn't improve your odds. Unfortunately for the unacquainted, there does exist a better tactic. By switching doors, your chances of winning actually increases to two in three. This is because your initial selection has a one in three chance of being correct and always switching in this case means you have a two in three chance of switching onto the correct door. This can be seen a little bit more clearly in the variant with an arbitrary number of doors. Suppose the show finds a budget surplus and decides to spend it on 97 more doors and 97 more goats.

[00:01:24]With 99 goats and one car, the same rules apply. So, when Monty opens 98 doors to reveal 98 goats, switching should always become even more optimal.

[00:01:35]And when applying the same logic from beforehand, this strategy checks out.

[00:01:39]You had a 99 out of a 100 chance of initially picking a goat, so always switching yields a 99% chance of winning the game.

[00:01:49]If this isn't making sense, perhaps another way to see the game is through the lens of bundles. Suppose Monty is forced to downsize a bit and now only has five doors instead of 100. With only one car to go about still, he presents a new twist. You get to select three of the doors, and then he will reveal two goats from your bundle of three, and one goat from his bundle of two.

[00:02:13]Do you now stay or switch?

[00:02:16]You should actually choose to stay. The way you conceptualize this is that your bundle had a three in five chance of containing the car prior to Monty opening any doors, while the leftover bundle only has a probability of two in five to contain the car initially. If you apply this logic to the original Monty Hall problem, your chosen bundle becomes one door, while Monty gets to keep two doors. So, logically, his bundle is more likely to contain the car, and you should switch into it when more information is revealed. Of course, all versions discussed so far rely on the concept that Monty knows he will always open a door with a goat.

[00:02:53]In this third variant, suppose we're back to the original statement, but Monty didn't get enough sleep last night.

[00:02:59]>> [music] >> He kindly informs you ahead of time that he can't seem to remember what is behind any of the doors. But, to not penalize you unfairly, the rules of the game still apply. Unfortunately for you, the door he randomly chooses still ends up containing a goat, and you do not win the game by default. What should your strategy be here?

[00:03:18]If you said your strategy should mirror Monty's, i.e., you just choose randomly, you'd actually be correct.

[00:03:24]In this particular situation, staying and switching both win with probability 1/2. [music] The reason why is because Monty's filtering of the doors is no longer an informed process. So, intuitively, you don't actually get any information just because he revealed a goat. The logic about bundle picking and sizes also doesn't help since the unpicked bundle started with a 2/3 probability of being picked, but had a 50% chance of revealing another goat, yielding a 1/3 probability of picking a car, which is equivalent to your bundle's chance of having a car, which is just 1/3 from the start.

[00:04:02]Now, suppose Monty feels that a little bit of randomness is actually what the show needed. With the same base setup, this time Monty brings out a coin before he opens any doors. Once you've made your initial selection, he then randomly tosses it. If it's heads, he'll play his usual game and always reveal a goat. If it's tails, he'll have one of his staff tell him in a headset a random door to open and will always listen to them, even if it has a car. He won't tell you the result of the coin flip, but does tell you it's a fair coin and that he'll always have a slight delay to make his move, so you can't distinguish between the outcome of his flip and his actions.

[00:04:38]Suppose as usual you don't win on a triviality and end up seeing a goat revealed.

[00:04:43]What is your strategy now?

[00:04:46]Intuitively, you should realize that there is a non-zero probability that the information you received is a signal of something, so it should always be slightly better than 50/50. Your door still starts with a 1/3 chance of containing the car, while Monty's two-door bundle starts with a 2/3 chance. The difference now is that Monty's goat reveal is not guaranteed to be meaningful. If the coin lands heads, then he's playing the normal game and will reveal a goat with probability one.

[00:05:13]If the coin lands tails, then he's opening randomly. So, if the car is in his two-door bundle, he only reveals a goat with probability 1/2. So, we discount Monty's original 2/3 bundle probability by 3/4, the chance that this bundle would survive the goat reveal, since you didn't win on a triviality at the beginning, for a probability of 2/3 * 3/4, which is 1/2.

[00:05:37]Meanwhile, if your original door has a car, Monty will always reveal a goat, no matter what the outcome of the coin flip is, because both the other doors are goats. So, your door keeps the original weight of 1/3. At this point, the remaining two explanations are that your door has the car with weight 1/3, or that Monty's bundle has the car and survived the goat reveal with weight 1/2. The total surviving weight is 1/3 + 1/2, which is 5/6, while choosing to always switch as a strategy puts 1/2 in the numerator, making the overall probability of winning by always switching as 1/2 / 5/6, or just 3/5. A slight edge, but present nonetheless.

[00:06:20]With all these variants and giving away cars, Monty's show has now gone broke.

[00:06:25]He's decided to try a new scheme instead with his last three goats and cars.

[00:06:29]Hoping for a 50/50, Monty places two cars behind a door, a car and a goat behind another door, and two goats behind the last door with these exact combinations. Each door has two compartments behind it. So, opening the main door won't necessarily show what's behind each compartment. Once you choose a main door, Monty randomly opens a sub door without knowing what's within the compartments, happening to reveal a car.

[00:06:55]Finally, given this, he asks you, "What is the probability that the other compartment behind your chosen door also has a car?"

[00:07:04]Surely with no prior on what is within each compartment and the fact that the door with two goats must be ruled out, it must be a 50/50 between two cars and one car and one goat, right?

[00:07:16]No.

[00:07:17]This is because the two remaining doors were not equally likely to have produced the thing you just saw. If your chosen door had two cars behind it, then Monty had two possible compartments he could open, and both of which would reveal a car. But, if your chosen door had one car and one goat, then only one of the two compartments would reveal a car.

[00:07:36]Therefore, the probability that the other compartment has a car is actually 2/3, not 1/2. Note this only works because Monty explicitly said the layout would be car, car, car, goat, and goat, goat. Funnily enough, if the opening of the compartment was not random and instead deliberate, the answer would actually change to 1/2. Since both the two car door and the one car, one goat door become equally capable of producing an intentional car reveal.

[00:08:05]This problem is also isomorphic to the Bertrand's box paradox, if you're curious for more.

[00:08:11]Having taken two more cars, Monty is down to his last, and his goats are a little less indistinguishable.

[00:08:18]Without getting into too many of the details, let's just say they're labeled as healthy Harry and sickly Steve, where both you and Monty can tell who is who.

[00:08:27]Monty, deciding to go back to the classics, returns to our familiar three-door setup while knowing where everything is. Once again, being fair, he tells you that if possible, he would prefer to reveal sickly Steve over healthy Harry.

[00:08:42]If Monty reveals a goat after your initial selection, what would your strategy and odds be?

[00:08:48]Going case by case, it's actually quite simple. If healthy Harry is revealed, that means that sickly Steve has to be behind your initial door. Otherwise, he would have been revealed. So, always switching has a 100% chance of yielding a car.

[00:09:03]If sickly Steve is revealed, either your original door had the car and Monty chose sickly Steve because he prefers revealing him.

[00:09:11]Or your original door had healthy Harry and Monty revealed sickly Steve because he was the only goat available.

[00:09:20]And thus, with Monty out of cars and myself out of examples, I must humbly end this video of mine.

[00:09:27]If you've made it to the end and enjoyed your time here, feel free to leave a like and comment. If you want more, please subscribe so you don't miss out on any of my my future videos.

[00:09:36]Otherwise, that's all for me this time.

[00:09:39]Thanks for watching.

[00:09:42]Hope to see you next time.