Bayes’ Rule: The Monty Hall Problem Josh Katzenstein Kerry Braxton-Andrew Problem From: _theorem...
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Bayes’ Rule: The Monty Hall Problem Josh Katzenstein Kerry Braxton-Andrew Problem From: http://en.wikipedia.org/wiki/Bayes'_theorem Image from: http://www.grand-illusions.com/monty2.htm
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Transcript of Bayes’ Rule: The Monty Hall Problem Josh Katzenstein Kerry Braxton-Andrew Problem From: _theorem...
Bayes’ Rule: The Monty Hall Problem
Josh KatzensteinKerry Braxton-Andrew
Problem From: http://en.wikipedia.org/wiki/Bayes'_theorem
Image from: http://www.grand-illusions.com/monty2.htm
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: www.istockphoto.com, www.flickr.com and www.commissionersam.com
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(PR|G) = Probability of prize in red if host picks green
• P(G|PR) = Probability of host picking green if prize is in red
• P(PR) = Probability of prize in red
• P(G) = Probability of host picking green
)(
)(*)|()|(
GP
PPPGPGPP RR
R =
The Setup
• The probability of the car being in the red door P(PR) 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
)(
)(*)|()|(
GP
PPPGPGPP RR
R =
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|PR)=P(B|PR)=1/2
– If the car is behind the green door, then the host would have to pick the blue door: P(G|PG)=0
– If the car is behind the blue door, then the host would have to pick the green door: P(G|PB)=1
• P(G) = 1/3 * 1 + 1/3 * 0 + 1/3 * 1/2 = 1/2
The Odds
3
2
2131*1
)(
)(*)|()|(
0
2131*0
)(
)(*)|()|(
3
1
2131*21
)(
)(*)|()|(
===
===
===
GPPPPGPGPP
GPPPPGPGPP
GPPPPGPGPP
BBB
GGG
RRR
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.comBut 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
11
10
30
113
1*1
)(
)(*)|()|(
0
30
113
1*0
)(
)(*)|()|(
11
1
30
113
1*
10
1
)(
)(*)|()|(
===
===
===
GPPPPGPGPP
GPPPPGPGPP
GPPPPGPGPP
BBB
GGG
RRR
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!
