UC Berkeley Department of Electrical Engineering and Computer Sciences. EECS 126: Probability and Random Processes

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1 UC Berkeley Department of Electrical Engineering and Computer Sciences EECS 26: Probability and Random Processes Problem Set Spring 209 Self-Graded Scores Due:.59 PM, Monday, February 4, 209 Submit your self-graded scores via the Google form: Make sure you use your SORTABLE NAME on CalCentral.. Miscellaneous Revie (a Sho that the probability that exactly one of the events A and B occurs is P(A + P(B 2P(A B. (b If A is independent of itself, sho that P(A = 0 or. (a The probability of the event that exactly one of A and B occur is P(A B c + P(A c B = P(A P(A B + P(B P(A B = P(A + P(B 2P(A B. (b P(A A = P(AP(A, so P(A = P(A 2 ; this implies that P(A {0, }. Alternatively, suppose for the sake of contradiction that 0 < P(A <. Then, P(A A = P(A, hich contradicts the supposed independence of A ith itself. Hence, P(A {0, }. 2. Coin Flipping & Symmetry Alice and Bob have 2n+ fair coins (here n Z >0, each coin ith probability of heads equal to /2. Bob tosses n + coins, hile Alice tosses the remaining n coins. Assuming independent coin tosses, sho that the probability that, after all coins have been tossed, Bob ill have gotten more heads than Alice is /2. Hint: This problem has a to-line solution. If e let Ω be the sample space consisting of all possible 2n + tosses, then Ω is a uniform probability space by assumption. Define the events A = {there are more heads in the first n + tosses than the last n tosses}, B = {there are more tails in the first n + tosses than the last n tosses}.

2 By symmetry, P(A = P(B, and e note that A B = since it is impossible for the first n + tosses to have more heads and more tails than the last n tosses, and A B = Ω. So, P(A + P(B = and hence P(A = /2. Alternatively, if the probability that Bob has more heads than Alice in the first n tosses is p, then the probability that Bob has feer heads than Alice in the first n tosses is also p, and the probability that they are tied after n tosses is 2p. So, the probability that Bob ins is p + (/2( 2p = /2 since Bob can either in by having more heads than Alice in the first n tosses, or by having the same number of heads as Alice in the first n tosses and then flipping heads on the last toss. 3. Choosing from Any Jar Makes No Difference Each of k jars contains hite and b black balls. A ball is randomly chosen from jar 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. Sho that the probability that the last ball is hite is the same as the probability that the first ball is hite, i.e., it is /( + b. We derive a recursion for the probability p i that a hite ball is chosen from the ith jar. We have, using the total probability theorem, p i+ = + + b + p i + + b + ( p i = + b + p i + starting ith the initial condition p = /( + b. Thus, e have p 2 = + b + + b + + b + = + b. + b +, More generally, this calculation shos that if p i = /( + b, then p i = /( + b. Thus, e obtain p i = /( + b for all i. 4. Passengers on a Plane There are N passengers in a plane ith N assigned seats (N is a positive integer, but after boarding, the passengers take the seats randomly. Assuming all seating arrangements are equally likely, hat is the probability that no passenger is in their assigned seat? Compute the probability hen N. Hint: Use the inclusion-exclusion principle First, let us calculate the probability that at least one passenger sits in his or her assigned seat using inclusion-exclusion. Let A i, i =,..., N, be the event that passenger i sits in his or her assigned seat. We first add the probabilities of the single events (of hich there are N, and the probability of each event is (N!/N! (indeed there are (N! ays to permute the remaining passengers once a specific passenger is fixed, and N! total permutations, so the probability is (N!/N!; next, e subtract the probabilities of the pairise intersections of events (of hich there are ( N 2, and the probability of each 2

3 event is (N 2!/N! (there are (N 2! ays to permute the passengers other than the fixed to; continuing on, e see that ( N P i= A i = ( N (N j! ( j+ j N! j= = ( j+ j!. j= No, the event that no passenger sits in his or her assigned seat is the complement of the event just discussed: ( N P i= A i = ( j+ N j! = ( j. j! j= Taking the limit as N, the expression converges to j=0 ( j /j!, and using the expression for the poer series of the exponential function, e conclude that the probability converges to exp( Expanding the NBA The NBA is looking to expand to another city. In order to decide hich city ill receive a ne team, the commissioner intervies potential oners from each of the N potential cities (N is a positive integer, one at a time. Unfortunately, the oners ould like to kno immediately after the intervie hether their city ill receive the team or not. The commissioner decides to use the folloing strategy: she ill intervie the first m oners and reject all of them (m {,..., N}. After the mth oner is intervieed, she ill pick the first city that is better than all previous cities. The cities are intervieed in a uniformly random order. What is the probability that the best city is selected? Assume that the commissioner has an objective method of scoring each city and that each city receives a unique score. You should arrive at an exact anser for the probability in terms of a summation. Approximate your anser using n i= i ln n and find the optimal value of m that maximizes the probability that the best city is selected. Let B i, i =,..., N, be the event that the ith city is the best of the N cities, and let A be the event that the best city is picked by the commissioner. Then, j=0 P(A = P(A B i P(B i = N i= P(A B i i= using the la of total probability. Next, P(A B i = 0 for i =,..., m because if the best city is among the first m cities, there is no chance of picking the best city. Also, P(A B i = m/(i for i = m +,..., N because P(A B i is the probability that second-best city among the first i cities is ithin the first m cities. Therefore, P(A = m N i=m+ i. 3

4 No, e turn toards approximation. P(A m N (ln N ln m = m N ln m N. Letting x := m/n, then P(A x ln x, and differentiating ith respect to x suggests that the optimal value is x = /e, so e should reject the first N/e cities. Plugging in this value for x into P(A = x ln x gives the optimal probability as P(A /e. Note: This problem is a famous example from optimal stopping theory and is commonly knon as the secretary problem (a boss is intervieing secretaries instead of a commissioner intervieing city representatives. In fact, one may use a dynamic programming approach to see hy the policy outlined here is in fact the optimal policy. If you are interested, the details of such an approach can be found in Dynamic Programming and the Secretary Problem by Beckmann. 6. Superhero Basketball Superman and Captain America are playing a game of basketball. At the end of the game, Captain America scored n points and Superman scored m points, here n > m are positive integers. Supposing that each basket counts for exactly one point, hat is the probability that after the start of the game (hen they are initially tied, Captain America as alays strictly ahead of Superman? (Assume that all sequences of baskets hich result in the final score of n baskets for Captain America and m baskets for Superman are equally likely. Hint: Think about symmetry. First, try to figure out hich is more likely: there as a tie and Superman scored the first point, or there as a tie and Captain America scored the first point? Let T be the event that Captain America and Superman ere tied at least once after the first point. Let C be the event that Captain America scores the first point, and S be the event that Superman scores the first point. In fact, P(C T = P(S T for the folloing reason: given any sequence of baskets here the first point is scored by Captain America and there is a tie, flip all of the baskets until the first tie. This yields a sequence of baskets here the first point is scored by Superman and there is a tie. Thus, the outcomes in C T are in one-to-one correspondence ith the outcomes in S T. Hoever, since Captain America on the game, P(S T = P(S (if Superman scored the first point, then there must have been a point hen Captain America caught up. So far, e have P(T = P(C T + P(S T = 2P(S T = 2P(S. We can calculate P(S = m/(m + n. Thus, P(T = 2m/(m + n. Finally, the question is asking for P(T c = 2m/(m + n = (n m/(m + n. Note: This is yet another famous problem knon as the ballot problem. 4

5 7. [Bonus] Tournament Probabilistic Proof The bonus question is just for fun. You are not required to submit the bonus question, but do give it a try and rite don your progress. In a tournament ith n players (here n is a positive integer, each player plays against every other player for a total of ( n 2 games (assume that there are no ties. Let k be a positive integer. Is it alays possible to find a tournament such that for any subset A of k players, there is a player ho has beaten everyone in A? For such a tournament, let us say that every k-subset is dominated. For example, Figure depicts the smallest tournament in hich every 2-subset is dominated Figure : A tournament ith 7 vertices such that every pair of players is beaten by a third player. In fact, as long as ( n k ( 2 k n k <, it is possible to find a tournament of n players such that every k-subset is dominated. Prove this fact, and explain hy it implies that for any positive integer k there exists a tournament such that every k-subset is dominated. Let us build a tournament ith n players randomly, by deciding each game ith a independent fair coin flip. Let A j, for j =,..., ( n k, be the event that the jth subset of k players is not dominated by anyone. What is P(A j? There are n k other players, and the probability that a player fails to dominate the jth subset is 2 k, so the probability that none of the n k other players dominates the jth subset is P(A j = ( 2 k n k. Hence: (( n k P j= ( n k A j P(A j = j= ( n ( 2 k n k. k Thus, if ( n k ( 2 k n k <, then P( ( n k j= Ac j > 0, i.e., there is a positive probability that every subset of k players is dominated, but this means there exists some tournament ith n players such that every k-subset is dominated. No, the question is hether e can pick n so that ( n k ( 2 k n k <. The term ( n k gros roughly at the rate n k, and the term ( 2 k n k gros 5

6 roughly at the rate c n for some c (0,, and since exponential decay outpaces polynomial ( groth, e can indeed alays pick n sufficiently large so that n k ( 2 k n k < holds. Can ( e figure out the order of magnitude of n? Using the simpler bounds n k n k and ( 2 k n k exp( (n k2 k, then it suffices to have n k exp( (n k2 k <, and upon taking logarithms and rearranging e get the condition n k2 k ln n k > 0. From this e at least need n > 2 k so ln n > k but then k2 k ln n > k 2 2 k, hich means e need to take n > k 2 2 k. If e let n = Ck 2 2 k for some constant C >, then n k2 k ln n k (C k 2 2 k > 0 for large k, so roughly, a constant multiple of k 2 2 k players suffices. 6