ORIE 3500/5500, Fall ’10
HW 10 Solutions
Assignment 10 Solutions
Problem 1
Let
X
be the time in jail and
Y
be the result of the ﬁrst draw. From the
information given,
P
(
Y
= 0) =
P
(
Y
= 1) =
P
(
Y
= 3) = 1
/
3
.
Now, we have
E
(
X
) =
E
(
E
(
X

Y
))
=
E
(
X

Y
= 0)
P
(
Y
= 0) +
E
(
X

Y
= 1)
P
(
Y
= 1)
+
E
(
X

Y
= 3)
P
(
Y
= 3)
=
1
3
(
E
(
X

Y
= 0) +
E
(
X

Y
= 1) +
E
(
X

Y
= 3)
)
.
But we know that
E
(
X

Y
= 0) = 0
,
E
(
X

Y
= 1) = 1 +
E
(
X
)
,
E
(
X

Y
= 3) = 3 +
E
(
X
)
.
The ﬁrst equation is clear: if the prisoner draws the number zero, he will be
released immediately. The second equation is true because if the prisoner draws
the number 1, he will stay in jail for one year and then will be at the exact same
situation as before he made the ﬁrst draw. His expected additional time until
freedom will be just
E
(
X
). The third equation follows from similar reasoning.
Hence we obtain
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 '10
 FROHMADER
 Probability theory, 3k, 1 k, 4 k, 4 K, 1 3K

Click to edit the document details