<p>3 doors. 1 has a car, the other 2 are goats</p>
<p>My god the solution seems so bizarre but after trying to swallow the reasoning it seems..logicalish.</p>
<p>Monty</a> Hall problem - Wikipedia, the free encyclopedia</p>
<p>It’s basic conditional probability.</p>
<p>You could think of it like this. You probably (2/3) picked the goat first time. After the other goat is eliminated, switching is clearly better.</p>
<p>21.</p>
<p>That is all.</p>
<p>It sounded confusing to me, until my Algebra III teacher explicated. What helped me was imagining 100 doors and only 1 car and 99 goats. You select your door, then close 98 doors that contained goats. Hopefully that helped.</p>
<p>The trick is that Monty does not pick the door he opens randomly. He always chooses one with a goat.</p>
<p>I’ll expand on the above point. Pretend there are 1,000 doors, and you select one. The host then removes 998 doors and tells you to pick between the remaining door and the one you chose. Think for a second: it’s highly unlikely (1/1000) that the door you selected at first is the correct door, meaning the other remaining door is most likely (999/1000) the correct one. </p>
<p>The same principle applies with three doors, just to a smaller degree. You pick one door which has a 1/3 chance of being the correct one, meaning there’s a 2/3 chance the correct door is one of the other two. And thanks to the host opening one of those doors, you know which of those two would be the right one.</p>
<p>Heh, did you see this on wikipedia’s list of paradoxes page too?</p>
<p>Man… this problem is such a classic. It makes me love it.</p>
<p>You can figure out that switching is better quite easily.</p>
<p>You and your friend have 3 cards. 1 represents the car, the other 2 the goats. One person picks. After you pick, you must switch after your friend removes a goat. </p>
<p>If you pick a goat, your friend removes the other goat and since you have to switch, you end up with the car. If you picked the car, your friend removes one of the goats and since you have to switch, you end up with the other goat. </p>
<p>Picking the car on the first try means you lose, and since there is 1 car in 3 cards, the probability of losing if you switch is 1/3. Likewise, picking the goat and switching means you win, and the probability of picking the goat on the first try is 2 cards out of 3, or 2/3. Hence, switching doors yields 2/3 probability of winning and is the better choice.</p>
<p>Breaking it down piece by piece… it’s really not that difficult a concept to understand.</p>
<p>What if, instead of exactly two goats and one car, each door has either a goat or a car according to a random 2/3 / 1/3 die roll?</p>
<p>In this case, there might be no car at all, two cars, or even three cars.</p>
<p>@OtherWindow yes 
it was the one that stood out with a diagram.</p>
<p>I understand it, just didn’t seem reasonable at first. Guess thats why its so beautiful to ppl like MIT :)</p>
<p>Truly beautiful. Wow. It took me a few minutes to full comprehend it.</p>
<p>As a kid at a school full of idiots, I can’t count the number of times I’ve explained why the probability doesn’t magically shift to 50/50 once one door is removed.</p>
<p>^But there are only two doors! The goat is either behind one, or it isn’t - 50/50!</p>
<p>What if the host is aware of the Monty Hall problem, so he can trick you into switching?</p>
<p>Why do you think we have Deal or no Deal?</p>
<p>^It puts insomniacs to sleep.</p>
<p>Has anyone figured out the formula the Banker uses for the offers?</p>