This is a famous problem from probability, based on the old game show Let's Make a Deal and its host, Monty Hall. There are many, many explanations of this problem on the internet. Feel free to read them, but write your answer in your own words. There are three doors, A, B, and C. Before the show, Monty puts a fancy car behind one of the doors. We assume he chooses the door randomly, and that each door has an equal probability of being chosen. He puts goats behind the other two doors. A contestent (hoping to win a car) chooses one of the three doors-let's say door A. Before opening door A to see whether to contestent won the car, Monty opens one of the other two doors, and asks the contestent if she would like to switch her choice. If the car is actually behind door B, Monty will open door C (showing the contestant a goat). If they car is actually behind door C, Monty will open door B (showing the contestant a goat). If the car is actually behind door A, Monty randomly chooses between B and C (with equal probability) and opens that door (showing the contestant a goat). Here's the question: say Monty opens door B. Should the contestant switch her choice to door C, or stick with door A? Or does it not make a difference? What is the probability she wins a car if she switches to door C? Hint: We know Pr(car behind A)=Pr(car behind B)=Pr(car behind C)=3, and (given that the contestent chose A) we know Pr(Monty opens B|car behind C)=1 and Pr(Monty opens B car behind A)=1. What is Pr(car behind C|Monty opens B)? Say you've been watching Let's Make a Deal carefully for some time, and you've found that contrary to what I wrote above-Monty actually puts the car behind door A about half the time, and behind doors B and C about a quarter of the time each. If you are scheduled to be a contestent on the show, and you want a new car, what is your best strategy? Which door should you choose first? Under what circumstances, if any, should you switch your choice?