Proving The Monty Hall Problem

This is a method to prove to yourself the results of the Monty Hall Problem. There is something very difficult in trying to describe why you get the result you do. Doing something physical like this can help foster understanding.

[MoodyRaiNz]

Throughout the game, these are the possible combinations:-

A. CAR GOAT GOAT
B. GOAT CAR GOAT
C. GOAT GOAT CAR


Suppose you chose door 1.
In Case A, switching will not win a car.
In Case B, switching will win a car.
In Case C, switching will win a car.

As seen above, Switching gives 2/3 (66%) chance to win the car, and that my friend is the solution to the Monty Hall problem ;D

[gardyloocomics]

this is how i like to think of it.
a) when you initially choose a card, your odds of having the car is 1/3.
b) the dealer keeps two cards, the odds of the dealer having the car is 2/3.
c) the dealer reveals one of his cards he knows to be a goat. then he asks you if you want to switch.
d) you should switch because the odds that you have the car is still 1/3. the odds that the dealer has the car is still 2/3 because he will only ever reveal a goat.
hope that helps!

[createataco]

2/3 chance you will choose a goat, so when he reveals another goat that means you get 2/3 chance to get the car.

If someone just explained this to me first time I would have got it.

[willh]

To the people that don't get it.  Always Switching is the same as picking the other 2 doors.  If he let you have the other 2 doors after 1st pick would you take  them?

[CobraABC]

Nice video. It really demonstrates that your first pick only offers you a 1/3 chance. Therefore it is always better to choose from the other two (when Monty has made it an easy choice).

[Do_not_assume]

This is another easier explanation which continues from what I learnt from this video: There's an 2/3 (66%) chance that you've selected one of the goats from the beginning. Then Monty reveals one of the goats and eliminates it. Because of the 66% probability from the beginning you have more likely to have selected the other goat, whereas the car is only 33% chance, thus it makes sense to switch. (There's a high probability that you have selected the goat and the other goat has been removed, it makes sense to swtich)

[Sdawkminn]

I like to think of it as a choice between 2 doors, but Monty adds an extra goat to the beginning to increase the odds of you making that choice when you currently have a goat. After that has been achieved, he takes the extra goat away because it has served its purpose.

So really, the game is a choice between 2 doors, but is rigged to have your starting door conceal a goat 2/3 of the time.

Another way is to imagine a different game with the same outcome: Monty gives you a 3-sided die with numbers 1, 2, and 3 on it. He has a piece of paper in his pocket with one of those numbers on it. You roll the die and then Monty asks if you think the side you rolled matches the number in his pocket. If you answer correctly, you win.

[spuriusscapula]

The single detail that is necessary for this to work is that the host (or in your case, the dealer) knows what lies behind each door/card, and knowingly chooses the wrong one to reveal. And that increases the possibility of the switch being correct because they purposefully avoided revealing the other one.

[nd_irish_fan]

Rules: There is only one red card, and you win if you end up with the red card. The easiest way to think about this type of problem is you must first pick a single card at random. Now before revealing anything about any of the three cards, you are given the option to (1) Keep the single card you selected, or (2) Trade your single card selected for BOTH of the two cards you didn't select.

The best choice is obvious, and most everyone will immediately jump at the chance to exchange one card for the other two because they can "see" they have just doubled their chances of winning by having two chances of winning as opposed to only one. Anytime we "switch", we are in effect trading one card for two, and revealing one of the two cards doesn't change the fact that we are getting "two for the price of one" when we switch.

[Tambleguide]

For all you people who say 50/50 you are wrong. If you pick a goat you will have a 66 percent chance of getting a goat. If you pick a goat Monty will be forced to show you the other goat and if you switch you will get a car. This means by picking a goat and switching to get a car you will always win the car if you pick a goat and switch, and the chances of you picking a goat is 66 percent. Like so everyone can see please.

[Bigbuddyandblue]

Your first pick is always most likely wrong, so always switch.

[The_Gamer_Couple]

Having proof of this is great, now the truth of this is solidified for me, thanks for making this video

[sergioramio]

the reason it's not 5050 is because the host ALWAYS shows you a goat and never a car, but no one explains that.

If the host could flip a door and show a car, then it'd be a fair 33 percent

[videoshomepage]

Got a guy at work who insists wrongly, it's 50/50. He just doesn't get it.

[Walkkyh]

When you pick a goat and then switch you'll get the car and when you pick the car and then switch you'll get the goat and since there are two goats and a one car in the beginning you will more likely pick a goat than a car, simple as that

[Swordfishstick]

And just for full disclosure, I myself struggled mightily with this when it was first presented to me, and eventually made a little computer algorithm to simulate the game, and to "prove" that switching didn't make a difference. Had it run for a few hundred trials, and to my astonishment, the switch proved to, indeed, double your chances of winning. Only after that did I put enough effort into understanding why this is, and finally "got" it. Now that I understand it, it seems obvious.

[KirkBocek]

Yes, fancy explanations are probably not needed. And yet we are still from time to time blessed with pearls from those of mighty intellect who *insist* the second choice is 50-50.

[nospamnc]

I shall make it simple.
At your first selection, there is 1 chance out of 3 to select the winning door, when you'd better keep your choice
but there is 2 chances out of 3 to select a loosing door, when you'd better switch your choice.
As you don't know the result of your first choice, odds of winning is larger (=2/3) by shifting.

[joshrob]

The host opening one of the remaining doors reveals information about the likelihood of the prize being behind the remaining door, while giving you zero new information about the likelihood that the prize is behind your door (because the host could never open the door you picked). The problem is called a paradox not because both probabilities are correct, but because people overwhelmingly believe the probability is 1/2 when that is wrong. It's a paradox of behavior, not math.

[paddygilbert]

It's quite simple to explain.

There are 3 doors, 2 incorrect (goats), 1 correct (car!). You choose one of them. One of the incorrect doors is revealed to be incorrect. You can then switch if you want to. Switching will only work if at first you chose one of the wrong doors. Seeing there are 2/3 wrong doors at the beginning, switching will work 2/3 of the time.