STAT 455/855 – Midterm Solutions
Fall, 2007
1. (15 marks)
(a) (6 marks) Let
p
1
be the probability that player 1 wins when he rolls first and
let
p
1
be the probability that player 1 wins if he doesn’t roll first. By symmetry,
p
1
doesn’t depend on which of players 2
, . . . ,
6 rolls first.
Conditioning on the
outcome of the first roll, we obtain
p
1
= (1)
1
6
+
p
1
5
6
and
p
1
=
p
1
1
6
+
p
1
4
6
+ (0)
1
6
.
From the second equation we obtain
p
1
=
1
2
p
1
. Plugging this into the first equa
tion, we obtain
p
1
=
1
6
+
5
12
p
1
, or
p
1
= 2
/
7.
(b) (3 marks) The number of times the die is rolled until a winner is declared is a
Geometric random variable with parameter
p
= 1
/
6.
Therefore, the expected
number of times the die is rolled is 6.
(c) (6 marks) Now we condition on 3 possible outcomes for the first roll: a 1 is rolled,
a 2 is rolled, or neither a 1 or 2 is rolled. By symmetry, given that neither a 1 or 2
was rolled first, the conditional probability that player 1 wins is the same whether
player 3,4,5 or 6 rolls second. Let
p
1
be as in part(a),
p
2
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 Fall '09
 GLENTAKAHARA
 Probability theory, Harshad number, Roll, p1, ψi pij

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