This document discusses the Monty Hall problem, a probability puzzle involving a game show with 3 doors and a prize behind one door. It shows that if a contestant initially picks a door, and is then shown a losing door and given the option to switch to the other unopened door, their chances of winning increase from 1/3 to 2/3 if they switch doors. Conditional probability is used to explain this counterintuitive result. Diagrams and examples are provided to illustrate the application of conditional probability rules to the Monty Hall problem scenario.
21. Each door has a 1 in 3 chance of hiding the grand prize.
Suppose we begin by choosing door #1.
In this case Monty may
open either door #2 or
#3
In both of these cases,
Monty is forced to reveal
the only other.
22. So what happens when you switch?
In this case you were
right the first time.
You lose!
In both of these cases,
you switch to the correct
door.
You win!
23. • Conditional probability gives us a way to
determine how the occurrence of one event
affects the probability of another.
The probability of event A given event B is the
probability of both A and B divided by the
probability of B.
Bp
BAp
BAp
|
24. In the following argument:
• Assume that:
– we originally chose door #1.
– Monty opened door #2.
• Our aim is to compute p(#1 | opened #2) and
p(#3 | opened #2).
25.
2#opened
)2#opened1(#
2#opened|1#
p
p
p
2#opened
2#opened3#
2#opened|3#
p
p
p
(If the prize is behind door #1, Monty can open
either #2 or #3.)
(If the prize is behind door #3, Monty must
open door #2.)
BApBpBAp
Bp
BAp
BAp
|2.
|1.
:Rules
6
1
3
1
2
1
3
1
3
1
1
1#1|#2#opened2#opened1# ppp (By rule 2.)
3#3|#2#opened2#opened3# ppp (By rule 2.)