## Exercises for Tuesday 24 October

Many of these borrowed from Mike Titelbaum.

For the following problems, assume that "a confirms b" (symbolized as aCb) is true for a probability distribution p iff p(b|a) > p(b). Assume that "a disconfirms b" (symbolized as aDb) is true for p iff p(b|a) < p(b).

When giving examples, you may find it helpful to work with a sample space that is a 12-sided die, where all twelve sides are initially regarded as equally likely.

1. Give an example where aCb, bCd, but aDd.

Answer: Suppose the sample space is our fair 12-sided die. Let a be the hypothesis that the die came up one of 1,2,3,4,5. Let b be the hypothesis that the die came up one of 3,4,5,6,7,8. Let d be the hypothesis that the die came up one if 5,6,7,8,9,10. Then p(b) = 1/2, and p(b|a) = 3/5 which is higher. Also p(d) = 1/2, and p(d|b) = 4/6 which is higher. But p(d|a) = only 1/5 which is lower than 1/2.

1. Give an example where aCb, aCd, but aD(b or d).

Answer: as in the previous problem, let our sample space be a fair 12-sided die. Let a be the hypothesis that the die came up one of 1,2,3,4,5,6,7. Let b be the hypothesis that the die came up one of 3,4,5,6,7,8,9,12. Let d be the hypothesis that the die came up one of 2,3,4,5,6,10,11,12. Then p(b) = 8/12, and p(b|a) = 5/7 which is higher. Also p(d) = 8/12, and p(d|a) = 5/7. But p(b or d) = 11/12, and p(b or d|a) = 6/7 which is lower.

1. Give an example where aCd, bCd, but p(d|ba) = p(d|b), so "b and a" doesn't confirm d any more than b does alone. (This is described as b "screening off" a's confirmatory effect on d.) Assume that b does not entail a.

Answer: as in the previous problems, let our sample space be a fair 12-sided die. Let a be the hypothesis that the die came up one of 1,3,5. Let b be the hypothesis that the die came up one of 3,4,5,6. Let d be the hypothesis that the die came up one of 1,2,3,4. Then p(d) = 4/12. p(d|a) = 2/3, which is higher than 4/12. p(d|b) = 1/2, which is also higher than 4/12. p(d|ab) = 1/2, which is the same as p(d|b).

1. Give an example where a rational agent conditionalizes on new evidence E, yet her credence in a proposition H1 that's consistent with E decreases. That is, H1 does not entail not-E, but p(H1|E) < p(H1).

Answer: using our fair 12-sided die, let H1 be the hyppthesis that the die came up one of 1,2,3,4,5,6,7. Let E be the hypothesis that the die came up either 7 or 8. Then p(H1|E) = 1/2, which is less than p(H1) = 7/12.

1. Prove that when an agent conditionalizes on new evidence E, her credence in a proposition H2 that entails E cannot decrease.

Answer: we assume that p(E) is not 0, so p(H2|E) is defined as p(H2 & E)/p(E). But if H2 entails E, then H2 & E is logically equivalent to H2, thus the previous expression is equivalent to p(H2)/p(E). Now p(E) ≤ 1, so p(H2)/p(E) will be ≥ p(H2).

1. You're about to roll five fair six-sided dice (as in the game Yahtzee). How confident should you be that there will be no pairs (where there's a pair if there are two or more dice with the same result)?

Answer: p(No pairs) = p(die 2 differs from die 1 & die 3 differs from dies 1 and 2 & ...) = p(die 2 differs from die 1) * p(die 3 differs from dies 1 and 2 & ... | die 2 differs from die 1) = p(die 2 differs from die 1) * p(die 3 differs from dies 1 and 2 | die 2 differs from die 1) * p(die 4 differs from dies 1 and 2 and 3 & ... | die 2 differs from die 1 & die 3 differs from dies 1 and 2) = p(die 2 differs from die 1) * p(die 3 differs from dies 1 and 2 | die 2 differs from die 1) * p(die 4 differs from dies 1 and 2 and 3 | die 2 differs from die 1 & die 3 differs from dies 1 and 2) * p(die 5 differs from dies 1 and 2 and 3 and 4 | die 2 differs from die 1 & die 3 differs from dies 1 and 2 & die 4 differs from dies 1 and 2 and 3) = 5/6 * 4/6 * 3/6 * 2/6 = 20/216 = 5/54

1. Continuing the previous problem, how confident should you be that there will be no triples?

Answer: p(No triples) = p(No pairs) + p(There is one pair but no triples) + p(There are two pairs but no triples). (In the original answer, I neglected this third factor.)

The first factor was calculated in the previous problem to be 5/54.

We can calculate the second factor as follows: p(1,2 match but all the others are different) + p(1,3 match but all the others are different) + p(1,4 match but all the others are different) + p(1,5 match but all the others are different) + p(2,3 match but all the others are different) + p(2,4 match but all the others are different) + p(2,5 match but all the others are different) + p(3,4 match but all the others are different) + p(3,5 match but all the others are different) + p(4,5 match but all the others are different).

All of those factors will be symmetrical, so this = 10*p(1 and 2 match but all the others are different).

So we have 10*p(1,2 match but all the others are different). The key idea here involves realizing that p(AB|C) = p(B|C) * p(A|BC). Now, that p(1,2 match but all the others are different) = p(1,2 match) * p(3 differs from 1,2 & 4 differs from 1,2 and 3 & 5 differs from 1,2 and 3 and 4 | 1,2 match) = (here we apply the key idea) p(1,2 match) * p(3 differs from 1,2 | 1,2 match) * p(4 differs from 1,2 and 3 & 5 differs from 1,2 and 3 and 4 | 1,2 match but 3 differs) = p(1,2 match) * p(3 differs from 1,2 | 1,2 match) * p(4 differs from 1,2 and 3 | 1,2 match but 3 differs) * p(5 differs from 1,2 and 3 and 4 | 1,2 match but 3 differs from 1,2 & 4 differs from 1,2 and 3) = 1/6 * 5/6 * 4/6 * 3/6 = 5/108.

We can calculate the third factor as follows: p(1,2 match and 3,4 match but differ from 1,2 and 5 differs from the rest) + p(1,3 match and 2,4 match but differ from 1,2 and 5 differs from the rest) + p(1,4 match and 2,3 match but differ from 1,2 and 5 differs from the rest) + ... corresponding options where it's 1 which differs from the rest, or 2, or 3, or 4.

That's 15 possibilities in all, which are symmetrical, so this = 15*p(1,2 match and 3,4 match but differ from 1,2 and 5 differs from the rest).

So we have 15*p(1,2 match and 3,4 match but differ from 1,2 and 5 differs from the rest). That p(1,2 match and 3,4 match but differ from 1,2 and 5 differs from the rest) = p(1 and 2 match) * p(3,4 match but differ from 1,2 and 5 differs from the rest | 1 and 2 match) = p(1 and 2 match) * p(3,4 match but differ from 1,2 | 1 and 2 match) * p(5 differs from the rest | 1,2 match and 3,4 match but differ from 1,2) = 1/6 * p(3 differs from 1,2 | 1 and 2 match) * p(4 matches 3 | 1,2 match and 3 differs from 1,2) * 4/6 = 1/6 * 5/6 * 1/6 * 4/6 = 5/324.

Plugging these numbers back into p(No pairs) + 10 * ___ + 15* ___, we get 5/54 + 25/54 + 75/324 = 255/324.

1. Your friend Heather is addicted to dice games. To help control her addiction, she flips a fair coin each day to decide how much to bet that day. If the coin comes up heads, she bets nothing (but still goes ahead and plays dice games anyway, for fun). If the coin comes up tails, she bets that she'll get at least a triple when tossing five dice. (By "at least a triple," I mean that at least three dice will show the same result. We'll ignore other kinds of rolls that score high in dice games like Yahtzee, such as "straights.") When walking through the department today, you saw Heather emerging from the dice game with a smile on her face. Clearly either Heather bet and won, or she didn't bet and when she rolled the dice anyway they came up bad (no triple). You consider the outcome of the dice tossing to be independent of whether Heather bet.
1. Using a language with the atomic propositions B for "Heather bet" and D for "the dice came up with (at least) a triple", what did you learn when you saw Heather smiling?

Answer: (B and D) or (not-B and not-D). Also known as "B iff D".

1. After updating on the information you just articulated, how confident should you be that the dice came up with (at least) a triple?

Answer: p(B iff D) = p(B and D) + p(not-B and not-D), since these options exclude each other. The first is p(D|B)p(B). The second is p(not-D|not-B)p(not-B). Since the outcome of the die toss is independent of whether Heather bet, p(D|B) = p(D) and p(not-D|not-B) = p(not-D). So p(B iff D) = p(D)p(B) + p(not-D)p(not-B). Since p(B) = p(not-B) = 1/2, this is p(D)/2 + p(not-D)/2 = 1/2.

(Added later: what the problem asked was: what is p(D|B iff D)? This will be p(D & (B iff D))/p(B iff D) = p(B & D)/p(B iff D). Since we're assuming B is independent of D, this is p(B)p(D) / p(B iff D) = (1/2)p(D)/(1/2) = p(D). We solved for p(D) in problem 7.)

1. How confident should you be that Heather bet?

(New answer: p(B|B iff D) = p(B & (B iff D))/p(B iff D) = p(B & D)/p(B iff D), which we solved for in problem 8b.)

1. Explain why one of the credences you calculated in (b) and (c) matches the value it had prior to you updating on what you learned when you saw Heather smiling, whereas the other differs.

Answer: ...

1. Give an example where a is probabilistically independent of b, and b is probabilistically independent of d, but a is not probabilistically independent of d.

Answer: an easy solution is to let d = not-a. Then for any choice of probabilistically independent a,d, the other constraints will follow.

1. Give an example where a is probabilistically independent of b, and b is probabilistically independent of d, and a is also probabilistically independent of d, but a is not probabilistically independent of "b and d".

Answer: using a fair 12-sided die again, let a be the hypothesis that the die came up one of 1,2,3,4,5,6. Let b be the hypothesis that the die came up even. Let d be the hypothesis that the die came up 1,3,5,8,10,12. Then p(a|b) = 3/6 = p(a), so a and b are independent. Also p(b|d) = 3/6 = p(b), so b and d are independent. Also p(a|d) = 3/6 = p(a), so a and d are independent. But "b and d" = the die came up one of 8,10,12, and this is incompatible with a. So a and "b and d" are not independent.

1. Give an example where p(a|b) = p(a|d), but ≠ p(a|b or d).

Answer: using a fair 12-sided die again, let a be the hypothesis that the die came up one of 1,2,3,4; b be the hypothesis that the die came up 1,2,3,5,6,9; d be the hypothesis that the die came up 2,3,4,7,8,9. Then p(a|b) = p(a|d) = 3/6. But p(a|b or d) = p({1,2,3,4}|{1,2,3,4,5,6,7,8,9}) = 4/9.

1. Your friend Alan has credences p(a) = 1/3 and p(b) = 1/3 and p(ab) = 1/6. His p(a|b) is also 1/3. Describe a set of bets that Alan's credences sanction as fair, but which guarantee that he will lose money, once the truths about a and b are settled. Hint: use a conditional bet, where a bet on X conditional on Y pays one side money if X and Y are both true, the other side money when not-X and Y are true, but no money changes hands when Y is false.

Answer: Alan's credence p(a|b) = 1/3 sanctions as fair a bet against a conditional on b, where he loses \$12 if a and b, but wins \$6 if not-a and b. His credence p(b) = 1/3 sanctions as fair a bet against b, where he loses \$4 if b, but wins \$2 if not-b. His credence p(ab) = 1/6 sanctions as fair a bet for ab, where he wins \$15 if ab, but loses \$3 if either not-a or not-b.

Now, if not-b, then Alan neither gains nor loses on the first bet, he wins \$2 on the second bet, and he loses \$3 on the third bet, for a net loss of \$1. If b but not a, then Alen wins \$6 on the first bet, loses \$4 on the second bet, and loses \$3 on the third bet, for a net loss of \$1. If b and a, then Alen loses \$12 on the first bet, loses \$4 on the second bet, and wins \$15 on the third bet, again for a net loss of \$1. Poor Alan.