CMPSCI 291b: Reasoning About Uncertainty

Solutions to Practice Final Exam

David Mix Barrington

11 May 2008

  Q1: 10 points
  Q2: 10 points
  Q3: 10 points
  Q4: 10 points
  Q5: 10 points
  Q6: 10 points
  Q7: 10 points
  Q8: 20 points
  Q9: 10 points
  Q10: 20+10 points

  Total: 120 points

• Question 1 (10): True or false with justification: If I choose two cards from a standard 52-card deck without replacement, the probability that both are spades is exactly 1/16.

  FALSE. The correct probability is the number of pairs of spades, (13 choose 2), divided by the number of pairs of cards, (52 choose 2). This is 13*12/1*2 over 52*51/1*2, which is 13*12/52*51 which evaluates to 1/17, not 1/16. Another way to view this is that the first card has a 1/4 chance to be a spade, but when the second card is dealt there are only 12 spades left in the deck of 51 cards, so the second card has a 12/51 = 4/17 chance to be a spade if the first is a spade.

• Question 2 (10): True or false with justification: Suppose we deal one card to each of three players A, B, and C and place a fourth card face-up on the table. The probability that at least one player has a card of the same rank as the card on the table (for example, both sevens) is strictly less than 3/13.

  TRUE. The chance that A's card matches rank with the one on the table is at most 1/13 (actually 1/17 as we saw on a previous exam). The chance for B and C to match the card on the table is the same. The total chance of these three events is at most their sum, by the Union Bound, and it is less than their sum because there is a nonzero probability of more than one of the events happening. The sum is at most 3/13 (actually 3/17), so the probability is strictly less than 3/13.

• Question 3 (10): True or false with justification: If I throw three fair, independent, six-sided dice (i.e., I "throw 3D6"), then the probability that the total of the numbers on the three dice is 11 or more is exactly 1/2.

  TRUE. The expected value of the sum of the three numbers is 3*(3.5) = 10.5 -- while there is no reason in general that a random variable should be equally likely to be above or below its expected value, this is true in this case. The easiest way to see this is a symmetry argument. For every possible throw (i,j,k), there is another throw (7-i,7-j,7-k), and the first throw is 11 or more exactly if the second isn't. So exactly half of the possible throws are 11 or more.

  If you don't see this argument, you can count up how many of the 6*6*6 = 216 throws of 3D6 come to 11 or more. There is one way to get 18 (666), three ways to get 17 (566, 656, 665), six ways to get 16 (466, 646, 664, 556, 565, 655), ten ways to get 15 (three orders of 366, six of 456, and 555), 15 ways to get 14 (six orders of 356, three each of 266, 455, and 446), 21 ways to get 13 (six each of 256 and 346, three each of 166, 355, and 445), 25 ways to get 12 (six each of 156, 246, and 345, three each of 255 and 336, and 444), and finally 27 ways to get 11 (six each of 146, 236, and 245, three each of 335, 344, and 155). This totals 108, exactly half the cases.

• Question 4 (10): True or false with justification: If I throw 3D6 as in Question 3, the probability that two dice have the same number and the third has a different number is exactly (6*6*6 - 6*5*4 - 6) divided by 6*6*6. (Here "*" is ordinary multiplication.)

  TRUE. We could calculate this number directly as (6 choose 2) times 2 times 3, because we can pick the two numbers, pick which occurs twice, then pick which of the three dice has the unique number. This gives 15*2*3 = 90, and 6*6*6 - 6*5*4 - 6 = 216 - 120 - 6 = 90.

  But the formula given suggests an easier and also correct argument. The three dice have 6*6*6 possible values. There are 6*5*4 ways for all three to be different, and just 6 ways for all three to be the same. The case we want is what remains of the 6*6*6 when the other two cases are subtracted, so the given formula is correct.

• Question 5 (10): True or false with justification: Suppose I am playing a game of ordinary ("American") odds-and-evens, where I and my opponent choose to put out either one or two fingers, and my payoff is +1 if the total number of fingers is even and -1 if it is odd. Suppose I know that my opponent plans to put out one finger with probability 1/3 and two fingers with probability 2/3. Then my optimal strategy is also to put out one finger with probability 1/3 and two fingers with probability 2/3.

  FALSE. The suggested strategy has a (1/3)(1/3) + (2/3)(2/3) = 5/9 probability of winning. But if I always put out two fingers, my winning probability is 2/3 > 5/9, so this is a better strategy than the suggested one against this particular strategy.

  It is true that putting out one or two with equal probability is an "optimal strategy" for this game, in that I can guarantee an expected winning probability of 1/2 (and hence expected value 0) against any strategy by my opponent. But if I know what strategy my opponent is using, I can do better.

• Question 6 (10): True or false with justification: If 80% of all suspects are guilty, and 80% of all guilty people are suspects, I may conclude that the number of guilty people is equal to the number of suspects.

  TRUE. We know that Pr(G|S) = Pr(G∩S)/Pr(S) and Pr(S|G) = Pr(S∩G)/Pr(G) are each 0.8, which is only possible if Pr(S) and Pr(G) each equal (5/4) Pr(G∩S) and thus equal each other.

• Question 7 (10): When I return from work, there is a 80% chance that my wife or daughter have already fed the cat. On Tuesday I observe that the cat is meowing when I return. He always meows when he has not been fed, but he also meows 80% of the time when he has been fed. What is my best estimate of the probability that he has been fed on Tuesday, given the new evidence that he is meowing?

  Let F be the event that the cat has been fed, and M the event that he is meowing. My prior odds for F are (0.8)/(1 - 0.8) = 4 and I must update them for the evidence M. I must multiply the odds by the likelihood ratio for M with respect to F, which is Pr(M|F) divided by Pr(M|¬F) or (0.8)/(1.0) = 4/5. My new odds are thus (4/5)(4) = 16/5, and the probability is (16/5) divided by (1 + 16/5) = 16/21, about 76%.

• Question 8 (20): In building a Naive Bayes Classifier spam filter, suppose that we have found ten "bad words" that each occur in 2/3 of spam messages and in 1/3 of non-spam messages, and ten "good words" that each occur in 1/3 of spam messages and 2/3 of non-spam messages. We assume that the occurrence of different words in the same message are independent events. Assume that half my mail is spam. If I get a new message containing exactly 7 of the bad words and exactly 2 of the good words, what (according to the classifier) is the probability that this message is spam?

  Let S be the event of spam and note that my prior odds are 1 (equal odds of spam and non-spam). Each bad-word event e has a likelihood ratio of Pr(e|S)/Pr(e|¬S) = (2/3)/(1/3) = 2, and similarly each good-word event has a likelihood of (1/3)/(2/3) = 1/2. The event of not finding a particular bad word has likelihood Pr(¬e|S)/Pr(¬e|¬S) = (1/3)/(2/3) = 1/2, and similarly the event of not finding a particular good word has likelihood (2/3)/(1/3) = 2. For our twenty events, the total likelihood is:

  2⁷ (for the 7 bad words found) times (1/2)³ (for the 3 bad words not found) times (1/2)² (for the 2 good words found) times 2⁸ (for the 8 good words not found). This multiplies to 2¹⁰ = 1024, so the estimated probability of spam is 1024/(1+1024) = about 99.9%.

• Questions 9 and 10 involve the following Markov decision process. There are two states, A and B, and two actions, X and Y. In state A, action X yields either A or B with 1/2 probability each, and action Y always yields state B. In state B, action X always yields state A and action Y yields A or B with 1/2 probability each. The reward function gives +2 for state A and +4 for state B.

• Question 9 (10): Give matrices representing the Markov chain where we always take action X, and for the chain where we always take action Y. Find the steady state distribution in the case where we always take action X.

  The matrix for X has 0.5 in both places on the top row and 1 followed by 0 on the bottom row. The matrix for Y has 0 followed by 1 on the top row and 0.5 in both places on the bottom row.

  If we say that the steady state distribution for X has state A with probability p and state B with probability 1-p, we can solve for p. We have that after one round taking X from distribution (p 1-p), we are in state A with probability p(0.5) + (1-p)(1) and in state B with probability p(0.5) + (1-p)(0). For the steady state p, then, p = p/2 + 1 - p, which solves to p = 2/3. The steady state (for X) has state A with probability 2/3 and state B with probability 1/3.

• Question 10 (20+10):

  • (a,10) Which action, X or Y, maximizes our expected reward on the following turn, for each of the two possible starting states? What is the expected reward from each state if we take this action?

    If we are in A, the expected payoff from taking X is (0.5)(2) + (0.5)(4) = 3 and the expected payoff from taking Y is 4. If we are in B, the expected payoff from X is 2 and the expected payoff from taking Y is (0.5)(2) + (0.5)(4) = 3. So Y is better in each case and our expected reward is 4 from A and 3 from B.

  • (b,10) Which action on the first turn, X or Y, maximizes our expected reward on the following two turns, for each of the two possible starting states? (Remember that on the second turn we will take the action determined in part (a).) What is the expected reward for those two turns from each state if we take these actions?

    If the first move puts us in A, our expected total reward is 2 for being in A after the first turn plus 4 expected from the second turn (from part (a)) for 6. If the first move puts us in B, our expected total reward is 4 for the first turn and 3 expected from the second turn, for 7.

    From state A, then, X gets us (0.5)(6) + (0.5)(7) = 6.5, while Y gets us 7. From state B, X gets us 6 while Y gets us 6.5, so Y is the better move in each case. Taking Y, we expect total return of 6.5 from A or 7 from B.

  • (c,10XC) Which action from each state maximizes the total expected discounted future reward, with a discount factor γ of 0.5? What is the expected discounted future reward?

    Let u be the expected discounted future return from A and v the expected discounted future return from B. Moving into A, then, is worth 2 for being in A plus uγ = u/2 expected discounted future return. Moving into B is worth 4 for being in A plus v/2 expected future discounted return.

    From A, X is worth (0.5)(2 + u/2) + (0.5)(4 + v/2) = 3 + (u+v)/4. From A, Y is worth 4 + v/2. From B, X is worth 2 + u/2 and Y is worth 3 + (u+v)/4.

    If the optimal value is based on X, then u = 3 + (u+v)/4 and v = 2 + u/2. Solving these two equations gives u = 5.6 and v = 4.8. If the optimal value is based on Y, then u = 4 + v/2 and v = 3 + (u+v)/4, which implies that u = 7.2 and v = 6.4, so the optimal policy always uses Y.

Last modified 12 May 2008