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Solutions to Assignment 2, STAT230- Fall 2012
Problem 1- (How an inferior player with a superior strategy can gain an advantage.)
Boris is about to play a two-game chess match with an opponent, and wants to find the strategy that
maximizes his winning chances. Each game ends with either a win by one of the players, or a draw. If
the score is tied at the end of the two games, the match goes into sudden-death mode, and the players
continue to play until the first time one of them wins a game (and the match). Boris has two playing
styles, timid and bold, and he can choose one of the two at will in each game, no matter what style he
chose in previous games. With timid play, he draws with probability pd > 0, and he loses with probability
1 − pd . With bold play, he wins with probability pw , and he loses with probability 1 − pw . Boris will
always play bold during sudden death, but may switch style between games 1 and 2.
(a) Find the probability that Boris wins the match for each of the following strategies:
(i) Play bold in both games 1 and 2.
Let W be the event that Boris is the grand winner, Li be the event that Boris loses in the ith match,
Wi be the event that Boris wins in the ith match. Then
P (W ) = P (L1 ∩ W2 ∩ W3 ) + P (W1 ∩ L2 ∩ W3 ) + P (W1 ∩ W2 )
= P (L1 )P (W2 )P (W3 ) + P (W1 )P (L2 )P (W3 ) + P (W1 )P (W2 )
= (1 − pw ) × pw × pw + pw × (1 − pw ) × pw + pw × pw
= 3p2w − 2p3w
(ii) Play timid in both games 1 and 2.
Let W be the event that Boris is the grand winner, Li be the event that Boris loses in the ith match,
Wi be the event that Boris wins in the ith match and Di be the event of draw on the ith match. Then,
P (W ) = P (D1 ∩ D2 ∩ W3 ) = P (D1 )P (D2 )P (W3 ) = p2d pw .
(iii) Play timid whenever he is ahead in the score, and play bold otherwise.
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Let W be the event that Boris is the grand winner, Li be the event that Boris loses in the ith match,
Wi be the event that Boris wins in the ith match and Di be the event of draw on the ith match. Note that
Boris starts by playing bold. Then
P (W ) = P (W1 ∩ D2 ) + P (W1 ∩ L2 ∩ W3 ) + P (L1 ∩ W2 ∩ W3 )
= P (W1 )P (D2 ) + P (W1 )P (L2 )P (W3 ) + P (L1 )P (W2 )P (W3 )
= pw × pd + pw × (1 − pd ) × pw + (1 − pw ) × pw × pw
= p2w (2 − pd − pw ) + pd pw .
(b) Assume that pw < 21 , so Boris is the worse player, regardless of the playing style he adopts. Show
that with the strategy in (iii) above, and depending on the values of pw and pd , Boris may have a better
than a 50-50 chance to win the match. How do you explain this advantage?
As an example, if pw = 0.45 and pd = 0.9, with strategy (iii) we have P (W ) ≈ 0.54. With strategies
(i) and (ii), the corresponding probabilities of a win can be calculated to be approximately 0.43 and 0.36,
respectively. What is happening here is that with strategy (iii), Boris is allowed to select a playing style
after seeing the result of the first game, while his opponent is not. Thus, by being able to dictate the
playing style in each game after receiving partial information about the matchs outcome, Boris gains an
advantage.
Problem 2- Two players take turns removing a ball from a jar that initially contains m white and n
black balls. The first player to remove a white ball wins. Let pm,k be the probability that the starting
player wins when the jar initially contains m white and k black balls. Prove the recursive formula
pm,k = 1 −
k
pm,k−1 .
m+k
What is pm,0 ? Calculate p5,6 .
Let V be the victory event for the first player to start the game. Let W be the event that the 1st draw
is white and B be the event that the 1st draw is black, i.e., B = W̄ . We have
pm,k = P (V ) = P (V |W )P (W ) + P (V |B)P (B).
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Bur, P (V |W ) = 1, P (W ) =
m
m+k
and P (B) =
k
.
m+k
Also,
P (V |B) = 1 − P (V̄ |B) = 1 − P (the second player wins|B).
After player 1 picks a black ball, one can think of the second player picking the first ball out of a box
with m white and k − 1 black balls. Hence, P (the second player wins|B) = pm,k−1 . Thus,
pm,k =
m
k
k
+ (1 − pm,k−1 )
=1−
pm,k−1 .
m+k
m+k
m+k
Clearly, pm,0 = 1. Note that
6
6
5
6
5
4
p5,6 = 1 − p5,5 = 1 −
1 − p5,4 = 1 −
1−
1 − p5,3
11
11
10
11
10
9
5
4
3
6
1−
1−
1 − p5,2
=1−
11
10
9
8
6
5
4
3
2
=1−
1−
1−
1−
1 − p5,1
11
10
9
8
7
6
5
4
3
2
1
=1−
1−
1−
1−
1−
1 − p5,0
11
10
9
8
7
6
6
5
4
3
2
1
=1−
1−
1−
1−
1−
1−
≈ 0.641
11
10
9
8
7
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Problem 3- Each of k jars contains m white and n black balls. A ball is randomly chosen from jar 1
and transferred to jar 2, then a ball is randomly chosen from jar 2 and transferred to jar 3, etc. Finally,
a ball is randomly chosen from jar k. Show that the probability that the last ball is white is the same as
the probability that the first ball is white, i.e., it is
m
.
m+n
Let pl be the probability that the ball picked from the lth jar is white where 1 ≤ l ≤ k. Let Wl be the
event that the ball picked from the lth jar is white, i.e., P (Wl ) = pl . We have
pl = P (Wl ) = P (Wl |Wl−1 )P (Wl−1 ) + P (Wl |W̄l−1 )P (W̄l−1 )
= pl−1 P (Wl |Wl−1 ) + (1 − pl−1 )P (Wl |W̄l−1 ).
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Note that P (Wl |Wl−1 ) =
pl =
m+1
m+n+1
and P (Wl |W̄l−1 ) =
m
.
m+n+1
m+1
m
1
m
pl−1 +
(1 − pl−1 ) =
pl−1 +
.
m+n+1
m+n+1
m+n+1
m+n+1
Having this recursion, we show by induction, that pl =
Then, assuming pl−1 =
pl =
Thus,
m
m+n
for any 1 ≤ l ≤ k. Note that p1 =
m
.
m+n
m
,
m+n
1
m
1
m
m
m
pl−1 +
=
+
=
.
m+n+1
m+n+1
m+n+1m+n m+n+1
m+n
Problem 4-(The prisoners dilemma.)
The release of two out of three prisoners has been announced, but their identity is kept secret. One of
the prisoners considers asking a friendly guard to tell him who is the prisoner other than himself that will
be released, but hesitates based on the following rationale: at the prisoners present state of knowledge,
the probability of being released is 2/3, but after he knows the answer, the probability of being released
will become 1/2, since there will be two prisoners (including himself) whose fate is unknown and exactly
one of the two will be released. What is wrong with this line of reasoning?
There are three possibilities for the inmates to be released, i.e., {1, 2}, {2, 3} and {3, 1}. Therefore the
sample space is S = {{1, 2}, {2, 3}, {3, 1}}. Let prisoner 1 considers asking the guard. We assume that
if prisoner 1 is NOT to be released, then the guard randomly selects prisoner 2 or prisoner 3 and tells
prisoner 1 about it. Let Ri be the event that prisoner i is released and Gi is the event that the guard says
prisoner i is to be released. Then,
P (R1 |G2 ) =
P (R1 and G2 )
.
P (G2 )
Clearly,
1
P (R1 and G2 ) = P ({1, 2}) = .
3
Also,
P (G2 ) = P ({1, 2}and G2 ) + P ({2, 3}and G2 ) =
1 1
+
3 6
where the last step is by the fact that P ({1, 2}and G2 ) = P ({1, 2}) =
1
3
and P ({2, 3}and G2 ) =
5
P (G2 |{2, 3})P ({2, 3}) =
1
2
×
1
3
= 61 . Then,
P (R1 |G2 ) =
1
3
1
3
+
1
6
2
= .
3
Therefore, the chance of release does not change.
Problem 5- (A two-envelopes puzzle.)
You are handed two envelopes, and you know that each contains a positive integer dollar amount and
that the two amounts are different. The values of these two amounts are modeled as constants that are
unknown. Without knowing what the amounts are, you select at random one of the two envelopes, and after
looking at the amount inside, you may switch envelopes if you wish. A friend claims that the following
strategy will increase above 1/2 your probability of ending up with the envelope with the larger amount:
toss a coin repeatedly, let X be equal to 1/2 plus the number of tosses required to obtain heads for the
first time, and switch if the amount in the envelope you selected is less than the value of X. Is your friend
correct?
Yes, he is correct! To see why, let a be the smaller amount of the two envelopes and b be the larger
amount of the two envelopes. We are not aware of a and b, and we only know that a < b and a, b ∈ N.
Let N be the number of times we require to toss a coin to get heads for the first time, i.e., X = 0.5 + N .
Clearly, X is a random number. Let W be the event that we eventually pick the envelope of larger amount.
Also, let A and B be the events that we first select the envelop containing $a and $b, respectively. Clearly,
the following four events make a partition:
C1 = A and {a < X},
C2 = A and {a > X},
C3 = B and {b < X}
and
C4 = B and {b > X}.
Then,
P (W ) = P (W |C1 )P (C1 ) + P (W |C2 )P (C2 ) + P (W |C3 )P (C3 ) + P (W |C4 )P (C4 ).
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It is evident that
P (W |C1 ) = P (W |C4 ) = 1, P (W |C2 ) = P (W |C3 ) = 0.
Therefore,
P (W ) = P (C1 ) + P (C4 ).
On the other hand, note that A and {a < X} are independent of each other. Also, B and {b > X} are
independent of each other. Therefore,
1
P (C1 ) = P (A)P (X > a) = P (N > a − 0.5)
2
and
1
P (C4 ) = P (B)P (X < b) = P (N < b − 0.5).
2
Hence,
P (W ) =
P (N > a − 0.5) + P (N < b − 0.5)
.
2
Note that
P (N < b − 0.5) = P (N < a − 0.5) + P (a − 0.5 < N < b − 0.5).
Thus,
P (W ) =
P (N > a − 0.5) + P (N < a − 0.5) + P (a − 0.5 < N < b − 0.5)
2
=
1 + P (a − 0.5 < N < b − 0.5)
1
>
2
2
where we have used the fact that P (N > a−0.5)+P (N < a−0.5) = 1 and P (a−0.5 < N < b−0.5) > 0.
Problem 6- A researcher has heard that 10% of students have cheated on an examination, and wishes
to confirm this result with a confidential study. The researcher prepares a box containing 100 cards, 75
of which contain Question A and 25 of which contain Question B.
Question A: Have you ever cheated on an examination?
Question B: Do you think the Leafs will win the Stanley Cup this decade?
Each student draws a card from the box, answers the question it contains and returns the card back
to the box without showing it to the researcher. Since only the student knows which question is being
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answered, there is little incentive to lie. A separate study determines that 20% of students think the Leafs
will win this decade.
(a) What is the probability that a student answers yes?
Let Y be the event that a student answers yes and A and B be the events that a student answered
Question A and Question B, respectively. Then,
P (Y ) = P (Y |A)P (A) + P (Y |B)P (B) = 0.1 × 0.75 + 0.2 × 0.25 = 0.125.
(b) What is the probability that a student who answers yes was responding to Question B?
P (B|Y ) =
0.2 × 0.25
P (Y |B)P (B)
=
= 0.4.
P (Y )
0.125
Problem 7- Alice and Bob want to choose between the opera and the movies by tossing a fair coin.
The only available coin is biased (though the bias is not known exactly). How can they use the biased
coin to make a decision so that either option (opera or the movies) is equally likely to be chosen?
Fix n ∈ N. One idea is that let Alice toss the coin n times and count the number of heads she gets.
Let this number be X. Then, let Bob toss the coin n times and count the number of heads he gets. Let
this number be Y . Then, let the decision be opera if X > Y , movies if X < Y and draw if X = Y . By
symmetry P (X > Y ) = P (X < Y ), and therefore, cinema and opera is equally likely to happen, but the
problem is that P (X = Y ) > 0. However, it is easy to show that limn→∞ P (X = Y ) = 0. So, the game
is asymptotically without a draw decision.
A smarter and more efficient solution is that let n = 1 and repeat the game if there is a tie, i.e., X = Y .
Let Ak be the event that this game ends on the k th round. Clearly,
1
P (opera|Ak ) = P (cinema|Ak ) = , k = 1, 2, 3, · · · .
2
Then,
P (opera) =
∞
X
k=1
∞
P (opera|Ak )P (Ak ) =
1X
1
1
P (Ak ) = × 1 = .
2 k=1
2
2