Download Problem 1 - Art of Problem Solving

Problem 1: Two cows play a game where each has one playing piece, they begin by having the
two pieces on opposite vertices of an octahedron, and the two cows take turns
moving their piece to an adjacent vertex. The winner is the first player who moves its
piece to the vertex originally occupied by its opponent’s piece. Because cows are not
the most intelligent of creatures, they move their pieces randomly. What is the
probability that the first cow to move eventually wins?
Solution 1:
Call the first cow to move A, and the second cow B.
Let X be the probability that A will (eventually) win when it is his turn to move and the
two cows are both on their original vertices. Let Y be the probability that A will
(eventually) win when it is his turn to move and both cows are on one of the middle
vertices. Let Z be the probability that A will (eventually) win when he is on his original
vertex and B is on one of the middle vertices. And let W be the probability that that A
will (eventually) win when he is on one of the middle vertices and B is on his original
vertex.
From these designations we get:
X = Y, because from their original vertices they must each transition to one of the
middle vertices
Y = ¼ + 1/16 X + ¼ Y + 1/8 Z + 1/8 W
Because A can either win immediately with probability 1/4, or both cows
return to their original vertices with probability 1/16, or both stay on the
middle vertices with probability ¼, or have A return to his original vertex and
B stay in the middle with probability 1/8, or B return to his original vertex and
A stay in the middle with probability 1/8
Z=½Y+¼W
Because A must return to one of the middle vertices while B could either stay
in the middle, probability ½, or return to his original vertex, probability ¼.
W=¼+½Y+¼Z
Because, from one of the middle vertices, A can either win immediately with
probability ¼, or stay on one of the middle vertices with probability ½, or
return to his original vertex with probability 1/4.
This gives us a system of equations and X is the value we seek.
Solving gives us X = 14/25, Y = 14/25, Z = 11/25, and W = 16/25.
Problem 2: A fair coin is tossed repeatedly. Find the probability of obtaining five consecutive
heads before two consecutive tails.
Solution:
Let
be the desired probability,
be probability given first flip is heads,
be
probability given first flip is tails.
by adding cases where first flip is heads or tails.
because either the second flip is tails (failure) or heads (begins a new
run with heads). So
.
Now. if the first flip is heads, then either:
- Four heads follow (success) with probability
.
- Up to three heads follow, but then tails. In order to have a chance at success,
the flip after tails must be heads. If it is, then a new run begins with heads, which
is neutral. (That means success is independent of these flips.)
The combined probability of TH, HTH, HHTH, HHHTH is
.
Then
Multiplying by
.
,
.
Practice Problems
1. Two cowboys participate in a duel. They take turns shooting at one another until one of
them is hit. Cowboy 1 has an accuracy of 1/4 and cowboy 2 has an accuracy of 1/3. What
is the probability that cowboy 1 will win the duel?
2. Two cows play a game where each has one playing piece. They begin by having the two
pieces on opposite vertices of a square, and the two cows take turns moving their piece
to an adjacent vertex. The winner is the first player who moves its piece to the vertex
originally occupied by its opponent’s piece. Assuming the cows move their pieces
randomly, what is the probability that the first cow to move eventually wins?
3. Again two cows play a game where each has one playing piece. They begin by having
the two pieces on opposite vertices of a hexagon, and the two cows take turns moving
their piece to an adjacent vertex. The winner is the first player who moves its piece to the
vertex originally occupied by its opponent’s piece. Assuming the cows move their pieces
randomly, what is the probability that the first cow to move eventually wins?
4. Erica, Nate, and Jason play a game in which they take turns flipping a fair coin. Erica flips
first, followed by Nate, followed by Jason, and so on. The first person whose coin flip
matches their predecessors flip wins the game. What is the probability that Nate will win?
5. You throw a coin multiple times. What's the average amount of throws required to
obtain 2 heads in a row?
6. Two players, A and B, start with 8 and 4 coins, respectively. They flip a fair coin, and
based on the results of each flip, one player pays the other one coin. The game ends
when one player goes broke. What is each player's probability of winning, and how
long, on average, does the game last?
7. Suppose you are on a number line, and you start at 1. You repeatedly roll a die. If it
lands on 1 or 2, go down one integer. If it isn't a 1 or a 2, go up one integer. What is
the probability that you land at zero at some point?
8. Find the probability that if a fair coin is tossed infinitely and the results are recorded in
a sequence, "HH" will occur before "THTH"
Answers:
1. 1/2
2. 2/3
3. 4/7
4. 4/7
5. 4
6. Player A = 2/3, Player B = 1/3, Expected length of game = ???
7. ½
8. 3/4