Monty will do a similar thing once you choose for the first time, but, instead of revealing 1 goat, he will reveal 8. What's the probability of choosing the wrong door the first time?Īnother expansion of the thought that several people have brought up: If you choose one of the two wrong doors the first time, switching will always make you win. If you choose the correct door the first time, switching will always make you lose. In the actual Monty Hall problem, you choose twice, and the probability of the second choice is affected by the first. ![]() In the case you've given, you are only choosing once. No, Syomantak, you're changing the situation in an essential way. But I love his explanation for the simplicity. It is more mathematically rigorous than Sal's demonstration, but I was a mathematics major at university. The notation really took work to put together. It only gives you information about * P(not A)*. Notice that Monte's new information tells you nothing about A. So, substituting, we end up withģ) Since P(B) = 0, Then we can substitute '0' for P(B), which gives us Using the last 2 certainties, we end up with the followingġ) P(not A) = P(B)+P(C). ![]() So, let's look at our certain probabilities:Į. So we have the following certain probability: ![]() Monte will never open the door with the prize, or he will be fired. This leads to the following probabilities:Ĭ Monte Opens 'B' - can be any door, but we can call any door he chooses 'B' You Choose 'A' - can be any door, but we can call any door you choose 'A' This is just a way of stating that the prize can be anywhere.ī. So I tried to break it down by probability of winning the car behind doors A, B or C.Ģ) P(A)+P(B)+P(C) = 1 (or 100%). What makes the Monte Hal Problem interesting in that the host knows, and will always chose the goat.
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