The Monty Hall Problem (Part 1)

[arrowsmithbros]

The correct answer is that you should switch

Ex. You pick the first door. There is a 1/3 chance that the car is behind the first door. There is a 2/3 chance that it’s somewhere else. Monty opens the 2nd door showing you that there’s nothing behind it. So instead of there being a 2/3 chance of it being behind door 2 AND 3, the 2/3 chance shifts to door 3. There is a 1/3 chance it’s behind door 1 and a 2/3 chance it behind door 3.

[duker]

I refuse to believe Captain Holt couldn't grasp the Monty Hall problem 🤷‍♂️😂

[PeskyNoodle]

Adrian’s my favorite, “if it’s dead dogs you want to see, I know where to get my hands on dozens of em” -Adrian Pimento

[isaiahrogge]

It’s a very well accepted answer. It makes complete sense statistically. They tested it and it works 66.36% of the time which is pretty close to the 66.667%

[AngryNerdBird]

If your initial choice was 1/3, removing one wrong choice doesn't change the fact that your initial choice only had a 1/3 chance of success.

[decentish]

Switching is the correct answer here and can be demonstrated quite simply. Let’s label the three doors as left door, middle door and right door. If you initially pick left door there is a 1/3 chance that you’re correct and a 2/3 chance that you’re wrong. Now we examine the following three scenarios:

1. The car is behind left door, the host randomly picks between middle door and right door to eliminate. Either way, sticking with your choice wins.

2. The car is behind middle door. In this case, the host must eliminate right door. If you stick with your choice in this case you lose. If you switch you win.

3. The car is behind right door. In this case the host must eliminate middle door. Once again therefore, if you stick with your choice you lose and if you switch you win.

Therefore as you can see, in 2/3 possible scenarios switching wins while staying only wins in 1/3. This has been confirmed by both simulations and practical tests thousands of times.

[vikkycb]

Best way to wrap your head around it is to change the problem.
Imagine if there were 10000 doors you pick 1 door and host opens all other doors except yours and 1 more.
Your possibility of picking right door was 1 in 10000.

[sethlaske]

If you think it’s wrong you are welcome to do the math or simulate it yourself.
Let’s say the car is door number 1. You have 3 options. If you pick doors 2 or 3, the host reveals the other. If you pick door 1 the host reveals either of the other 2.
Now let’s say you switch. You win the car 2/3 times. If you stay you win the car 1/3 times

[pancakes]

All I’m saying, is if he knows it’s called the Monty hall problem he should probably know the answer

[erichermann]

The reason they say it's 2/3 to switch is because of a little manipulation on the part of Monty Hall.
If you got it wrong (2/3 odds) he MUST eliminate the other wrong door, this preserves the odds. Meaning if you picked wrong the first time (2/3), then the other door at the end MUST have the prize.

This is the worst explained math trick I have ever had to learn.

[georgehilty]

Its bizzare but true. Mythbusters even had an episode about it!

[Aliksander]

Every time i see this situation i feel like this is math for compulsive gamblers. In reality, if the show is legit and the prize is behind one of any number of doors chosen at random, then EACH door has the same chance and that DOES NOT CHANGE. It doesn't matter how many doors you open, each closed door in a set has an equal possibility of being right if the conditions hold (the car was place at random behind one of X starting doors and you pick ONE door at the start and the car position doesn't change).

[anialator]

Using 100 or 1000 doors helps visualize it, but really, its more like the host is offering you the option to pick every door you didn't choose initially and if any of them win then you win. Because the host knows where the winning door is and will never open it, the final result is identical. if you chose door 1 and instead are offered doors 2-100, its obvious you should switch. The host then opens up doors 2-99 showing all of them are losers, but now you are in the identical situation. If the prize was behind door 2-100, you win, vs if it was behind door 1, because switching is to choose every door you didn't choose initially due to the fact that nothing about the doors the host opens is random.

[johnnorton]

They should make a game show out of this problem. Like you could have a bunch of doors to pick from that are all worth different amounts of money. You pick one, then you pick ones to open until the very end, and once you get it down to two doors, you have the option to switch. They could call it, “Agree or Disagree” or something catchy like that…

[pairot]

Imagine there are 100 doors, only one with a prize. You pick door #1, then the host opens up every door except your door and door #73. Do you still think it's a 50/50?

[abedrahman]

Funny thing is someone like Holt who's supposedly learned stats as a kid should understand the Monty Hall problem

[gonzaloruiz]

If you’re still wondering why it isn’t 50/50, imagine you have 10000 doors. You pick one, and the host opens 9998, leaving 1 unopened. Would you swap then?

[HTownsend]

I got this wrong the first time I heard it. But only because it was explained incompletely. It wasn't explained that the host knew which door the car was behind and would always choose an empty door.

[petermatis]

It is so so so dumb.
But basically, you assume you picked wrong.
This is because you're less likely to hit your mark in a chance of 1/3 than in a 50/50.
Thus switching is your best option.

[vladimirturudija]

the host who heard this math "problem" smiles bcz he knows you chose the right door in first pick