(I also point out that the game goes more smoothly when MH knows where the car is; but I’d like to be as clear as possible about when certain epistemic factors do or don’t affect probability outcomes, so I leave this bit about knowledge out of the account. Maybe he's acting more or less randomly to keep things unpredictable, or maybe he tries to steer the show in the direction of good TV.
This will be true whether or not anyone knew in advance that the goat was behind the opened door.
Let's now tackle a classic thought experiment in probability, called the Monte Hall problem.
The Monty Hall problem hit the headlines in 1990, when Craig F. Whitaker of Columbia, Maryland, asked Marilyn vos Savant: ‘Suppose you’re on a game show, and you’re given the choice of three doors: behind one door is a car; behind the others, goats. Nevertheless, for many the 3-door scenario continues to be counterintuitive.Feel free to stop by anytime to help me clarify my thoughts — I can always use the help :)You are now given the opportunity to guess which door conceals the car.
Is it to your advantage to switch your choice of doors?The "correct" answer, this last one, is tricky enough to get when you start with the required assumption. I’m fascinated by how a given individual’s intuition and intellect—including my own—come together with these things.
Here is why:So as you can see, by choosing to switch every time, you have a higher probability (2/3) than choosing not to switch (1/3).Let’s say instead of 3 doors, there were one million doors — 999,999 of which contained a goat, and one which contained the car.Using the above image, let’s run through your outcomes when you stick with the strategy to always change doors during Round 2.In Round 1, the likelihood of picking the correct door is incredibly small — one in a million, in fact. 2?”… I’ll illustrate with your version of the MHP.Imagine another variation. The Monty Hall problem is named for its similarity to the Let's Make a Deal television game show hosted by Monty Hall. Part I: The Monty Hall Problem. We could also calculate getting exactly one A (in two pulls with replacement), which could be in the first or second pull, but not both: which is (1/52)(51/52) + (51/52)(1/52), or about .0377219; notice that this is what we get if we subtract getting A twice, i.e. (1/52)^2 from the above probability of getting A at least once.Now, back to my claims about it not mattering what MH knows or intends.In case he is choosing a door randomly, assuming even odds (e.g., flipping a coin), you get: P(G|R) =((1/2)(2/3))/(2/3) = 1/2.
The views expressed in this article are his own and do not necessarily reflect the views of the University.‘Then tell me,’ said A, ‘the name of one of the others who will be executed. The problem is stated as follows.