EE 278
Handout #4
Statistical Signal Processing
Monday, July 13, 2009
Homework #1 Solutions
1. (15 points)
(a) The sample space consists of triplets of the form (curtain with gold behind it,
curtain chosen by player, curtain that Monty opens). If we denote the curtains
by
A
,
B
, and
C
, then we can write our sample space as
Ω =
{
(
A, A, B
)
,
(
A, A, C
)
,
(
A, B, C
)
,
(
A, C, B
)
,
(
B, B, A
)
,
(
B, B, C
)
,
(
B, A, C
)
,
(
B, C, A
)
,
(
C, C, A
)
,
(
C, C, B
)
,
(
C, B, A
)
,
(
C, A, B
)
}
.
(b) As dicussed in lecture notes 1, for a discrete sample space, the probability measure
is completely specified by the probabilities of the single outcome events. For this
problem we can calculate the probabilities as follows:
P(
A, A, B
) = P
{
Monty opens
B

gold behind
A
, player’s first choice is
A
} ·
P
{
gold behind
A
, player’s first choice is
A
}
=
1
2
P
{
gold behind
A
} ·
P
{
player’s first choice is
A
}
=
1
2
×
1
3
×
1
3
=
1
18
.
Note that Monty Hall’s choice of curtain to open is
not
independent of where
the gold is placed and what the player’s initial choice was. That is why we used
conditional probability in the above derivation. Using the same approach we find
P(
A, A, C
) = P(
B, B, A
) = P(
B, B, C
) = P(
C, C, A
) = P(
C, C, B
) =
1
18
.
This covers all the cases where the gold placement and the initial choice of con
testant coincide. For the other cases,
P
{
(
A, B, C
)
}
= P
{
Monty opens C

Gold behind
A
, player’s first choice is
B
} ×
P
{
Gold is behind
A
, player’s first choice is
B
}
= 1
×
P
{
Gold is behind
A
} ×
P
{
player’s first choice is
B
}
= 1
×
1
3
×
1
3
=
1
9
.
Using similar arguments,
P(
A, C, B
) = P(
B, A, C
) = P(
B, C, A
) = P(
C, A, B
) = P(
C, B, A
) =
1
9
and we now have a complete description of the probability space for this random
experiment. To find the probability of winning if the player decides to switch, we
1