Why does switching doors in the Monty Hall Problem increase odds: 2 doors, 50-50

Question

I have read through around 10 articles and webpages on this problem, and still don’t understand. I’ve run simulations and yes, switching does get you better odds, but why?

In: Mathematics

Answer 1

What worked for me is reducing “3 doors, only one has the prize you want*”, to the two outcomes you care about: P(car) (initially 1/3) and P(goat) (initially 2/3),

In other words, there are two sets of doors Cars={car}, Goats={goat, goat}. You are hoping to select the first set/avoid the second*. It should be clear you initially had a probability of 1/3 selecting a door in the Cars set, and 2/3 for a door in Goats set.

Now, once you make your initial selection and the host pauses to open a door to show you a goat, it’s tempting to think that the problem has become Cars={car}, Goats={goat} (so 50/50 for selecting either).

However, we know the host was never going to open a door in the Cars set (if they were selecting randomly and may have shown you a car, then the previous logic does actually apply and it is 50/50 now **). Essentially, what they have done is identified a door from the initial Goat set. They have, in fact, removed all of the doors in Goat set but one. You now have two doors—one ‘representing’ the Cars set and the other representing the Goats set. And because the host knew what they were doing, the initial probabilities apply. So, your initial guess had a 1/3 chance of being correct, and still does! This means there is still is a 2/3 chance that the car is in ‘one of the other’ doors (which in this case, is only one door), which is why you’re twice as better off switching.

I find the problem becomes a lot more obvious if you make it say, 100 doors, and the host opens all but yours and one other (so always ending w/ one door from Cars and one from Goats). What’s more likely—you got a 1/100 choice right from the start, or it was any of the other 99 doors (remember that host knew to make sure none of the 98 doors were a car).

You can even play with the conditions more (e.g. add more doors w/ or w/o a prize, change the number of doors the host opens, etc.) and work out the difference probabilities once you understand the framing.

* Assuming you want a car, not a goat, which everyone seems to ignore.

** Assuming you don’t get shown a car, in which case you can achieve 100% probability of success.