Ideas in Geometry/Instructive examples/Monty Hall Lesson

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Monty Hall Problem

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car; behind the other two, goats. You pick on door, say No. 1. The host, who knows what lies behind each of the doors, opens another door, say No. 3, which has a goat. He then asks you, " Would like like to pick door No.2?" Is it in your best interest to switch or stay with the same door.

How Should we go about solving this problem?

- Let's first think about the possible door outcomes:

Right? So we choose door one (*). The host opens a door with a goat (o). So, based of of this, the probability must be 1/2 of the time or 50%. Because of this, it doesn't matter if you stay or switch.


But why not? There are only two choices? - The critical thing to know is that the host does not have a choice in the door he opens. He must always open a door that has a goat behind it after the contestant has made their choice. - As long as the 3 choices are equally likely, there can never be a probability of 1/2 since the host's choice does not have an effect on the outcome.

So what is the best thing to do? Should we stay or should we switch?

- Let's start by thinking of all the possible outcomes behind the doors:

- To figure out the outcomes of staying or switching, let's assume that you always choose door one. The results would be as follows:

- Our outcome when we stay yields a car one out of three times or 1/3 of the time. If we were to switch doors, this would yield a car two out of three times or 2/3 or the time.