Math 501–1, Spring 2006 University of Utah

Solutions to Problem Assignment #5
Math 501–1, Spring 2006
University of Utah
Problems:
1. A gambling book recommends the following “winnin strategy” for the game of roulette:
It recommends that the gambler bet $1 on red. If red appears (this has probability
18/38), then the gambler should take her $1 profit and quit. Else, she should make
additional $1 bets on red on each of the next two spins of the roulette wheel, and then
quit. Let X denote the gambler’s winnings when she quits.
(a) Find P{X > 0}.
Solution. Let Ri denote the event, “red on the ith trial.” Because P(Ri) = 18/38 and
P(Rc
i ) = 20/38,
P{X = 1} = P(R1) + P(Rc1
\ R2 \ R3) =
18
38
+

20
38 ·
18
38 ·
18
38

= 0.591¯7.
On the other hand,
P{X = −1} = P(Rc1
\ R2 \ Rc3
) + P(Rc1
\ Rc2\ R3)
=

20
38 ·
18
38 ·
20
38

+

20
38 ·
20
38 ·
18
38

0.2624,
P{X = −3} = P(Rc1
\ Rc2
\ Rc3
) =

20
38
3
0.1458.
In particular, P{X > 0} = P{X = 1} = 0.591¯7.
(b) Are you convinced that the strategy is indeed a “winning” strategy? Explain your
answer.
Solution. One may think at first that this is a winning strategy because P{win} > 0.5.
But it is the expectation that counts, and we will see next that EX < 0. This
means that the strategy is a losing strategy.
(c) Compute EX.
Solution. We have
EX = 0.591¯7 − 0.2624 − (3 × 0.1458) −0.108.
Therefore, every time we play we expect to lose about 10 cents.
2. One of the numbers 1 through 10 is chosen at random. You are to try and guess
the number chosen by asking questions with “yes-no” answers. Compute the expected
number of questions that you need to ask in each of the following two cases:
(a) Your ith question is to be, “Is it i?”, for i = 1, . . . , 10.
Solution. Let Q denote the number of questions asked. It follows that P{Q = i} = 1/10.
Therefore,
E(Q) =
1
10
+
2
10
+ · · · +
10
10
= 5.5
(b) With each question you try to eliminate one-half of the remaining numbers, as
nearly as possible.
Solution. Consider the following strategy: Divide what you have in halves and ask, “is it
greater than or equal to x?” where x is the middle of the numbers. For instance,
your first question is, “is it greater than or equal to 5.5?”. [All other strategies of
this type are similar. But you need to be consistent.] Let X denote the random
number, so that P{X = i} = 1/10 for i = 1, . . . , 10. If X = 1, 2, 3, 6, 7, 8, then
Q = 3. If X = 4, 5, 9, 10 then Q = 4 (check!). Therefore, P{Q = 3} = P{X =
1} + P{X = 2} + P{X = 3} + P{X = 6} + P{X = 7} + P{X = 8} = 0.6, and
P{Q = 4} = 0.4. Thus, E(Q) = (3 × 0.6) + (4 × 0.4) = 3.4.
3. A person tosses a fair coin until a tail appears for the first time. If “tail” appears
on the nth flip, then the person wins 2n dollars. Let X denote the person’s winnings.
Show that EX = 1. This is known as the “St.-Petersbourg paradox.”
Solution. For all n 1,
P{X = n} = P(H1 \ · · · \ Hn−1 \ Tn) =
1
2n .
Else, P{X = n} = 0. Therefore,
EX =
1X
n=1

2n ×
1
2n

= 1.
4. A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If
they are the same color, then you win $1.10; if they are different colors, then you win
−$1.00 (i.e., you lose one dollar). Compute:
(a) the expected value of the amount you win;
Solution. Let Ri denote the event, “red on ith draw.” Then, P(R1 \ R2) = (5/10)(4/9)
and P(Rc1
\ Rc2
) = (5/10)(4/9). Let X denote the win, where X < 0 means loss.
Then, P{X = 1.10} = 2×(5/10)(4/9) = 4/9, and P{X = −1} = 5/9. Therefore,
EX = (1.1 × 4
9 ) − 5
9 = −0.0¯6.
(b) the variance of the amount you win.
Solution. Evidently, E[X2] = (1.1× 4
9 )+ 5
9 = 1.0¯4. Therefore, Var(X) = 1.0¯4 −(−0.0¯6)2 =
1.04.
Theoretical Problems:
1. Let X be such that P{X = 1} = p and P{X = −1} = 1 − p. Find a constant c 6= 1
such that E[cX] = 1.
Solution. First, for all real c,
E[cX] = cp + c−1(1 − p).
Set this equal to one to find that cp + (1 − p)/c = 1. I.e., pc2 − c + (1 − p) = 0.
This is a quadratic equation (in c) and has two roots:
c =
1 ±
p
1 − 4p(1 − p)
2p
.
But 1 − 4p(1 − p) = 1 − 4p + 4p2 = (1 − 2p)2. Therefore,
c =
1 ± (1 − 2p)
2p
= 1 and
1 − p
p
.
So the answer is c = (1 − p)/p.
2. Prove that if X is a Poisson random variable with parameter , then for all n 1,
E[Xn] = E[(X + 1)n−1]. (1)
Use this to compute EX, E[X2], VarX, and E[X3].
Solution. We calculate directly:
E[Xn] =
1X
k=0
knP{X = k}
=
1X
k=0
kn e− k
k!
=
1X
k=1
kn e− k
k!
=
1X
k=1
kn−1 e− k−1
(k − 1)!
=
1X
j=0
(j + 1)n−1 e− j
j!
= E[(X + 1)n−1].
Therefore,
EX = E[(1 + X)0] = ,
E[X2] = E[(1 + X)] = (1 + ) = + 2,
VarX = E[X2] − (EX)2 = ,
E[X3] = E[(1 + X)2] =

E

1 + 2X + X2
=

1 + 2 + + 2
= + 3 2 + 3.

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