Monty Hall Problem: Probability Calculation for Switching Doors

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SUMMARY

The Monty Hall Problem illustrates that switching doors increases the probability of winning a car from 1/3 to 2/3. By applying the General Case of Bayes' Rule, one can calculate the probability P(C2|M3), which represents the likelihood of winning the car if the contestant switches to Door 2 after the host reveals Door 3. The analysis confirms that the host's actions provide critical information that alters the initial probabilities, making switching the optimal strategy.

PREREQUISITES
  • Understanding of Bayes' Rule and its general case
  • Familiarity with conditional probability
  • Basic knowledge of probability theory
  • Ability to manipulate mathematical variables and equations
NEXT STEPS
  • Study the General Case of Bayes' Rule in detail
  • Practice calculating conditional probabilities with different scenarios
  • Explore variations of the Monty Hall Problem and their implications
  • Learn about probability distributions and their applications in decision-making
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Homework Statement



Suppose you’re on a game show and you’re given the choice of three doors. Behind one is a car, behind the others are goats. You pick a door, say number 1, and the host, who knows what’s behind the doors, opens another door, say number 3 which has a goat. He says to you, “Do you want to pick door number 2?” Is it to your advantage to switch your choice of doors?”

Use the General Case of Bayes’ Rule to demonstrate mathematically the 2/3 probability of getting the car by switching. Assume for sure that the contestant picked Door 1, that the game show host reveals Door 3 (M3), and you are going to let the car vary between the three doors as Ci where i = {1, 2, 3}. You can now calculate P(C2|M3), that is, the probability of winning the car given that the host reveals Door 3 and you switch to Door 2.

Homework Equations


General Case of Bayes Rule
upload_2016-10-3_18-19-4.png


The Attempt at a Solution


upload_2016-10-3_18-16-18.png

I started by changing the variables to match the problem. But I am not sure where to go from here.

I know that P(M3|C2) means "Probability of Door 3, given Door 2". So for those two I figure that it breaks down to a 1/2 probability. But I feel kind of stuck on which values are equal to what.
 

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First you need to correct your formula under section 3. You have a sum over ##i##, but ##i## does not appear in the summand. Also, I suggest using ##j## instead, as this platform always auto-corrects ##i## to upper case.

##P(M_3|C_j)## is the probability that Monty Hall (the host) will open door 3, given that the car is behind door ##j##, and that you have picked door 1. Your original choice of door 1 is taken as a constant rather than a random event, in order to avoid double-barrelled conditionalities. We can achieve that by deferring the numbering of doors until after your original choice is made, and then numbering them so that yours is number 1.

You need to bear in mind that Monty will neither open a door you have picked, nor a door with a car behind it. Given that, what is ##P(M_3|C_j)## for each of ##j=1,2,3##, and what is each ##C_j##?
 

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