Monty Hall Problem Simulation

The Monty Hall Problem is explained in my previous video. In this video I will simulate the Monty Hall problem 1 000 000 times. I'm curious to see if the simulation gives the same result as the theoretical explanation of the Monty Hall problem.

If you want to simulate the Monty Hall problem yourself and don't want to code, I suggest you use the python code in the link.

Python code from github:

[Tinseltopia]

The simulation works out, the math works out, but I just don't get it. I cannot fathom this problem, my ape brain developed on the African Savannah, just isn't enough for this problem. I have read countess posts explaining why it is better to switch and can see logically, via this simulation... But it doesn't make sense

[benwang]

The way it makes sense to understand this problem is to scale out the number of doors but apply the same rules. If there are 1 million doors and a car behind 1 door, when you select a door and the host drops a blank door and asks if you want to reselect, you are probably going to say yes and keep taking him up on the offer. The question is if you reselect and drop a door, at what point do you feel confident enough to stick to your original choice? It's very easy to visualize why reselecting is better from that standpoint because the likelihood of choosing the wrong door is so high. The main problem here on why people don't get this problem is that they grossly overestimate their odds of picking the right door in a 1/3 vs a 1/30000 and therefore don't understand the statistics portion. There is a significant chance you pick the right door in a 1/3 VS 1/30000 and you only get one chance so statistics is kind of BS in that case, but just like how flipping a coin 5 million times will come out to roughly 50/50, the same applies to this problem. Being given new information by dropping the door is a new problem but I also agree in a 1/3 VS 1/2 really surprising me that it results in 100% (33 VS 67%) delta between the win and loss between the 2 VS something like 33% VS 50%.

[CodeOverReality]

This is reason i have trust issues... with reality itself

[suikun]

I'm not convinced yet. I wanna see a 24 hours equivalent simulation for a more reliable result.😉

[vivaciencia]

I would like to see a field test with people. There is two cabins in a park somewhere, the test can only be taken by 2 people groups. One of the cabins present the monty hall problem exactly as it is, the results are measured. In the second cabin the person is presented just by the second part of the test, with only the information that the person in the first cabin knows for sure (not the probability) too, there are two doors and behind one of them there is a prize, pick one. The results of the second cabin are measured too. I would like very much to know the comparison between the two results.

[alimetlak]

after revealing a door by the host sticking to the first pick is a second new choice between 2 doors the same as switching since switching too is a second new choice between 2 doors so its 50 50

[TristanSimondsen]

There must be hundreds of Monty Hall Problem on YT, yet when you post a comment disputing the 2/3-always-switch-claim, one of these dudes will reply WITHIN HOURS to cyber bully and name call you to shame. If that's not a "cult", I don't know what is. And I'm talking about Matt5250, Araquis, Klaus7443, kt22027, KarlHeizSpock and Hank254. There maybe one or two more on here, but I think they're basically the same group of dudes.

The reason you're getting the a 2/3 no-car probability is because you're coding the simulation with one pick. Player has TWO SEQUENTIAL picks, one after host eliminates a no-car door. Specify in the rules that player has two sequential picks, host ALWAYS eliminates a no-car door, what is player's second pick's probability to NOT GET THE CAR, and you will get the CORRECT ANSWER, which is 1/3.

Because player at no point in time has a probability to ever pick Monty's no-car door.

It's like saying, what if Monty removes the door with the car. Does player still have a 1/3 probability to get the car with the second pick?

[mhpdebunked]

why have you set `The host now opens one of the unchosen doors that contains banana` to be random? Nowhere in the puzzle does it say he is doing this randomly

[erniepike]

So this prolem is ignoring the 1/3 who won on the first round?

[evandegreve]

The music volume is so loud I can barely hear you speak

[rootbeer]

The text is way too small on a mobile. Otherwise I really like your Monty Hall problem simulation.

[kenjulikiera]

This solution omits an important piece of information. Assume three doors: 1, 2, and 3. You pick 1, Monty opens 2 revealing a goat. That leaves doors 1 and 3 unknown. Doors 2 and 3 have a .67 probability of having money. But, importantly, there is a second combination, doors 1 and 2, which also have a .67 probability of having money. At this point you have two combinations, each with a .67 probability. It becomes a choice between two equally probable combinations. Effectively this is the same as a probability of .5 There is no advantage to either staying or switching.

[t-rex]

I tried this. 4/10 when stuck and 7/10 when switched

[knarftrakiul]

Still makes no sense lol it should be 50/50 but just by changing your pick increase ur odds. This is some quantum mechanics "spooky action from a distance" shit going on lol