Conditional probability - Monty Hall problem

An example of the use of conditional probabilities applied to the famous 'Monty Hall problem'.

[ddddsdsdsd]

I'd like to mention that we can go beyond the Monty Hall problem and touche on a fundamental issue in probability and statistics: the comparability of events with different sample spaces or magnitudes.

1. A first event with a magnitude of 3 (three doors)
2. A second event with a magnitude of 2 (two doors)

While a great introduction into probability, the Monty Hall Problem only works if one accepts the comparison of two events of different magnitudes as logical. To dismiss people who cannot agree with this comparison as they do not get it is a problem if you ask me. I see the importance of highlighting both ways of thinking.

In many contexts, comparing probabilities from events with different magnitudes or sample spaces is indeed problematic or even meaningless. For example, to provide another example of fundamentally different events:

Comparing the probability of rolling a 6 on a die (1/6) with the probability of flipping heads on a coin (1/2) doesn't make much sense in isolation.

In more complex scenarios, like comparing stock performance across different markets or time frames, not accounting for differences in magnitude can lead to serious misunderstandings.

Additionally, people think if your initial choice was the car (which happens 1/3 of the time), then switching would be the wrong move. In this case, the host's reveal of a goat door doesn't help you at all. You've already won, and switching would make you lose.

If one accepts the comparison of two events of different magnitudes, the Monty Hall strategy isn't about "always switch" but rather "switch because it's more likely you initially picked a goat." The host's reveal doesn't create new winning chances. The host doesn't change the fact that you probably (2/3 chance) guessed wrong at first.

[dark_knight]

Was searching for this kind of explanation using bayes theorem. Wonderfully explained

[seanzhou]

I am confused by p(a, b)= 1/3 x 1/2. Does this assume A and B are independent? If they are independent, p(a|b)=p(a); no need to go through all the rest of the derivation steps.

[GregTom]

Well I think I can say that I'm now confident in my understanding of Baye's theorem and Bayesian inference.

[dihan]

I am sorry, but I have to point out that your method is totally wrong. I don’t want people to get more confused by seeing this video.
First of all, event B can not be described as “a goat shows up behind door 2”. That is also very tricky and critical insight of the Monty Hall Problem. The most correct description of event B is “the host had to choose a door between door 2 and door 3 to reveal a goat, and the door 2 was chosen”. The difference is that event B is INDEPENDENT of event A!
I want to mention there are other mistakes you made in the later part of this video. I would say, basically you used a wrong method to get a correct answer.

[joehsiao]

I am confused about the part when you say p(B) = 1/2.

[dlschys]

Wow His explanation is very straightforward.
But I am kinda afraid of his approach cuz it cannot go further.
Actually, you should define 3 events.
A_i: Your choice. (i = 1, 2, 3)
B_j: Host's choice (j = 1, 2, 3)
C_k: Where is Car. (k = 1, 2, 3)
2/3 can be derived by P(C_1^c/B_2, A_1), (^c is complement of a evernt)

Now, let our strategy be more complicated.
If door 2 opens you do not switch. If door 3 is opened, you switch.
(Host randomly chooses the remaining door)
The answer is 1/2 so this strategy is worse than the previous one.

[omid_tau]

The probability of p(B) is 1/2 there can only be 2 choices to make and it's independent from your choice youre using that information in your solution to P(A, B) hence p(a)p(b) so why do you over complicate it by then going into great detail about the marginal value for p(B) ? p(a|b) = p(a) because of independence .

[RealMcDudu]

Here's another way to look at it: the contestant only has to choose either to switch a curtain or not. If he chose not to switch, than it's obvious his chances of winning (i.e. getting the car) is 1/3, which was his chance in the beginning (since no matter what curtain you choose, monty could always show you a goat). Now - what's the possibility of winning if he does switch? Well - if he chose the car in the beginning (1/3) then there's 0 chance of winning in case of switching. But if he didn't choose the car (2/3) than there's 1 (=100%) chance of winning because Monty will open the other goat, and then the other curtain has to be the car. So the chance of winning if you don't switch is 1/3, while the chance of winning if you do switch is 2/3 - i.e. 2x more => SWITCH!!!

[AlleyCat]

When this problem was first posted in the 70s, scores of people refused to accept the solution as it is so counterintuitive, even famous mathematician Paul Erdos. However, this solution is correct.

[CarinaZheng]

Well explained!! Your videos have been helping me with my Stat! Great thanks!! :D

[dawyer]

Maybe thinking this Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a pair of doors with sheep, leaving only two pairs of doors, one sheep and one car, for the player to choose. If the player predicts the door of the car, what will the host do? of couse Want the player to change their choice. The host asks you to calculate the odds with misleading you already have once choice before. you swapping your choice, 2/3 will win. On the contrary, the player predicted the door with the sheep. What will the host do to make the player still choose the door with the sheep? you can tell.
Dawyer's door problem, calculate the chance of the host winning.

[furi]

Very well done m9, finally a proper explanation. Get rekt Monty Hall.

[FreeMarketSwine]

I've always hated this question because there are a lot of assumptions behind it that are never addressed, namely that 1) the host will always open a door, 2) he has knowledge of where the car is and is acting adversarially to the contestant (i.e., he will never open door with the car behind it), and 3) he's either not aware of conditional probability or he is and believes the contestant is not.

For example, if the host has the ability to decide whether to open the door but is aware of conditional probability and believes the contestant is also aware of conditional probability, well good luck in that case.

[sirajnakhuda]

i don't really understand where he got the 1/2 at 3:31.Can someone please explain to me in more detail.

[rogerbodey]

Over complicated. Here's a very short proof:

1. Whenever you switch, you get the "opposite" of you original choice. (ie car to goat/ goat to car)
2. 2/3 of the time your initial choice will be a goat. (2 goats/ 1 car)
3.2/3 of the time you will get the car by switching.

[dawyer]

Each sheep is also an independent and unique individual. This game cannot mislead players by treating every sheep as the same thing. There are three options, two of which also stand on their own. If the player can tell them apart, there is no chance of winning like 2/3

[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.

[MyKombucha]

P(B), that a goat is shown regardless of whether or not there is a car behind door A is 1/2. I believe it, but so counterinteruitive...

[calebstacey]

does this apply in the real world or just theoretically