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STAT 471  Assignment 2 solutions
1. Calculate the chance of needing a game 7 in a bestofseven series when one team has
a 3
/
5 chance of winning each individual game.
Solution:
Game seven is needed if each team wins exactly 3 of the ﬁrst 6 games.
There are
(
6
3
)
= 20 ways that this can happen: WWWLLL, WLWLWL, LLWLWW,
etc. By independence, each of these ways has the same probability of happening,
for example,
P
(
WWWLLL
) =
P
(
W
)
P
(
W
)
P
(
W
)
P
(
L
)
P
(
L
)
P
(
L
) = (3
/
5)
3
(2
/
5)
3
=
216
/
15625. Therefore
P
(
need game
7) =
(20)(216)
15625
=
.
27648
.
2. Players A and B alternately throw an ordinary pair of dice in that order. Player A
wins if he gets a sum of six, while Player B wins if he gets a sum of seven. They keep
playing until somebody gets the sum he needs. If A starts the game, what is his probability
of winning?
Solution:
There are two ways to solve this problem: a clever way using conditional
probabilities and a brute force way that involves inﬁnite sums. Here are both solutions.
a) Let
A
be the event that player
A
eventually wins, and
B
the event that player
B
eventually wins. Let
A
1 be the event that player
A
wins (rolls a six) on the
ﬁrst
toss, and
B
2 the event that player
B
wins (rolls a seven) on the
second
toss. Let
C
(for ‘continue’) be the event that nobody wins on the ﬁrst two tosses, i.e.,
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 Spring '09
 Various.

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