Bayes’ Rule: The Monty Hall Problem Josh Katzenstein Kerry Braxton-Andrew Problem From: Image from:

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Presentation transcript:

Bayes’ Rule: The Monty Hall Problem Josh Katzenstein Kerry Braxton-Andrew Problem From: Image from:

The Problem: You are given a choice between three doors: Red, Green and Blue Behind one is a car, behind the other two are goats You pick the red door Photos from: and

The Problem: The host knows where the car is, but is supposed to open a door without the prize The host opens the green door, and a goat is behind it You are given the option to change your choice to blue What is the probability that the car is behind the blue door given that the presenter chose to open green? Photo from: library.thinkquest.com

The Method Use Bayes’ Rule to solve Bayes’ rule says: –Where: P(P R |G) = Probability of prize in red if host picks green P(G|P R ) = Probability of host picking green if prize is in red P(P R ) = Probability of prize in red P(G) = Probability of host picking green

The Setup The probability of the car being in the red door P(P R ) is 1/3 This is the same as the probability of the car being behind any other door Similar relationships can be set up for the other probabilities

The Solution Assuming the host has no bias, the probabilities break down like this –If the car is behind the red door, then the host can pick blue or green to open: P(G|P R )=P(B|P R )=1/2 –If the car is behind the green door, then the host would have to pick the blue door: P(G|P G )=0 –If the car is behind the blue door, then the host would have to pick the green door: P(G|P B )=1 P(G) = 1/3 * 1 + 1/3 * 0 + 1/3 * 1/2 = 1/2

The Odds

So! Where’s the Car? Looking at the odds –Behind the red door still: 1/3 –Behind the green door: 0 –Behind the blue door: 2/3 SO…ODDS SAY TO SWITCH YOUR PICK EVERY TIME Image from: johnfenzel.typepad.com But what if……..

A Bias Say you’ve watched this show before, you see that the host hates MSU, so rarely picks the green door unless he has to (only 10% of the time) What does this do to your odds???

Reevaluating the Odds This changes P(G) P(G)=1/3*1+1/3*0+1/3*1/10=11/30 –The first term is if he has to –The second term is if he can’t (car is behind green) –The third is if he has a choice, but will only pick green one in ten times

New Odds

So! Where’s the Car Now? The same place! However, rather than the odds being 2/3 to 1/3 they are now 10/11 to 1/11. So what would it have meant if the host had picked blue? What were the odds then? That’s an exercise for you!