# Fake Coins Puzzle

### Question

There are 12 coins before you, one of which is fake. The fake coin is either heavier than or lighter than the other 11 coins. The legitimate coins all weigh the same. You have a balance scale you can use to weigh the coins. You can use the scale at most three times. How do you find the fake coin?

Number the coins 1-12. Separate the coins into a pile of 1,2,3,4, a pile of 5,6,7,8, and a pile of 9,10,11,12. Let "f" denote the fake coin. Let the predicate "H" denote heavy and the predicate "L" denote light.

Weigh 1: 1,2,3,4 x 5,6,7,8. There are three possible outcomes:
Outcome I: 1,2,3,4 = 5,6,7,8. f is among 9,10,11,12.
Weigh 1.2: 1,9 x 10,11, where 1 is clearly not f. There are three possible outcomes:
Outcome I: 1,9 = 10,11. f is 12. DONE.
Outcome II: 1,9 > 10,11. Either H(9) or L(10) or L(11).
Weigh 1.2.3: 10 x 11. There are three possible outcomes:
Outcome I: 10 = 11. Then f is 9. DONE.
Outcome II: 10 > 11. Then f is 11. DONE.
Outcome III: 10 < 11. Then f is 10. DONE.
Outcome III: 1,9 < 10,11. Either L(9) or H(10) or H(11).
Weigh 1.2.3: 10 x 11. There are three possible outcomes:
Outcome I: 10 = 11. Then f is 9. DONE.
Outcome II: 10 > 11. Then f is 10. DONE.
Outcome III: 10 < 11. Then f is 11. DONE.
Outcome II: 1,2,3,4 > 5,6,7,8. f is either H(1), H(2), H(3), H(4), L(5), L(6), L(7), or L(8).
Weigh 1.2: 4,5,6,7 x 8,9,10,11. There are three possible outcomes:
Outcome I: 4,5,6,7 = 8,9,10,11. Then f is either H(1) or H(2) or H(3).
Weigh 1.2.3: 1 x 2. There are three possible outcomes:
Outcome I: 1 = 2. Then f is 3. DONE.
Outcome II: 1 > 2. Then f is 1. DONE.
Outcome III: 1 < 2. Then f is 2. DONE.
Outcome II: 4,5,6,7 > 8,9,10,11. Then f is either H(4) or L(8).
Weigh 1.2.3: 1 x 4. There are two possible outcomes:
Outcome I: 1 = 4. Then f is 8. DONE.
Outcome II: 1 < 4. Then f is 4. DONE.
Outcome III: 4,5,6,7 < 8,9,10,11. Then f is either L(5) or L(6) or L(7).
Weigh 1.2.3: 5 x 6. There are three possible outcomes:
Outcome I: 5 = 6. Then f is 7. DONE.
Outcome II: 5 < 6. Then f is 5. DONE.
Outcome III: 5 > 6. Then f is 6. DONE.
Outcome III: 1,2,3,4 < 5,6,7,8. f is either L(1), L(2), L(3), L(4), H(5), H(6), H(7), or H(8).
Weigh 1.2: 10,11,12,1 x 2,3,4,5. There are three possible outcomes:
Outcome I: 10,11,12,1 = 2,3,4,5. Then f is either H(6) or H(7) or H(8).
Weigh 1.2.3: 6 x 7. There are three possible outcomes:
Outcome I: 6 = 7. Then f is 8. DONE.
Outcome II: 6 > 7. Then f is 6. DONE.
Outcome III; 6 < 7. Then f is 7. DONE.
Outcome II: 10,11,12,1 > 2,3,4,5. Then f is either L(2) or L(3) or L(4).
Weigh 1.2.3: 2 x 3. There are three possible outcomes:
Outcome I: 2 = 3. Then f is 4. DONE.
Outcome II: 2 > 3. Then f is 3. DONE.
Outcome III: 2 < 3. Then f is 2. DONE.
Outcome III: 10,11,12,1 < 2,3,4,5. Then f is either L(1) or H(5).
Weigh 1.2.3: 1 x 4. There are two possible outcomes:
Outcome I: 1 = 4. Then f is 5. DONE.
Outcome II: 1 < 4. Then f is 1. DONE.

# Two Stones Puzzle

## Puzzle

Frank owes the merchant Jack a large sum, which Frank is unable to pay. Jack offers Frank a deal: “Convince your daughter Sally to marry me, and I’ll drop the debt.” Frank asks Sally if she’ll marry Jack, but Sally is uninterested. When Jack relates this to Frank, he responds: “I still may be able to help you out. I’ll drop the debt if you convince Sally to agree to the following deal. We all three go to the nearby river. I’ll grab two stones from the river bank, one white and one black. I’ll place the stones in a bag, then hand the bag to Sally. Sally will then pull a single stone from the bag. If Sally draws a black stone, then we’ll be married. If Sally draws white, then we won’t be married. In either case, I’ll drop the debt.” Jack tells Sally the new offer. Sally agrees given the new chances to save her father. However, while at the river Sally sees Jack put two black stones in the bag instead of one black and one white. Sally doesn’t want to ruin the deal by calling Jack out, but she also doesn’t want to marry Jack. What does Sally do?

After I came up with a solution, I looked at others. The standard solution (if there is one), seems to be: Sally draws a stone and quickly drops it into the river before anyone can tell what color it is. Sally then observes the group could determine the color of the stone she dropped by looking at the remaining stone in the bag.

That’s fine; it might work. But I’m not a fan of this solution, as it relies on dropping the stone before anyone can see its color.

I prefer my solution: Sally claims she is superstitious, and that drawing a black stone to start a marriage courts bad luck. Sally proposes that rather than she and the merchant marrying if she draws a black stone, they marry if she draws a white stone. Had the merchant been fair, there would be no reason for him to prefer one color over another, so he should have no grounds for rejecting Sally’s proposal.

# Trust Logic, Not Tortoises

Carroll's note What Achilles Said to the Tortoise holds many lessons, many of which related to putative justification - or lack of justification - for basic logical inferences. Recently, Romina Padro, pulling from coursework and discussions with Kripke, has argued one more lesson should be added to the list, namely, that under certain conditions adopting basic logical inferences is impossible. I've a few thoughts on this new lesson, in particular how it might play with the old lessons. Check it out a recent draft here!

# There’s ∃x about Mary

PUZZLE:
Larry is married but Nick isn’t. Larry is looking at Mary, and Mary is looking at Nick.

QUESTION:
Is someone married looking at someone not married?

CHOICES:

This puzzle was posed to me by a student (thanks Richard!) after class one morning. A cursory google suggests 80% choose incorrectly. I am skeptical; I’ve found no empirical evidence supporting this claim (I’d be interested if anyone else has). Be warned, searching for the puzzle will likely turn up solutions, so if you’d like to solve it, best settle here and reflect for a bit. I’ll wait. Once you are finished, check your answer by clicking it above. Afterwards, scroll down for a solution and some discussion.

SOLUTION:
The hallmark of a logical solution to a puzzle is that once presented with the solution, nearly everyone agrees it is correct. The Wason Selection Test is an example. Many choose incorrectly. Bentham reports a psychologist once confessed to him that nearly everyone accepts the standard solution as correct once it is explained . The puzzle under discussion strikes me as logical in this sense. I’m curious if you agree.

I’ve translated the puzzle into a standard classical first-order language, where the predicate and relations symbols are obvious. It is straightforward to prove the solution (I’m using Hardegree’s natural deduction system in Symbolic Logic: A First Course). In symbols:

1.      Ml & ~Mn                                            Premise
2.      Llm & Lmn                                          Premise
3.      SHOW∃x∃y(Mx & ~My & Lxy)          ID
4.                   ~∃x∃y(Mx & ~My & Lxy)      AID
5.                    ∀x∀y(~Mx v My v ~Lxy)      4, Substitution
6.                    ~Ml v Mm v ~Llm                  5, Universal Instantiation
7.                     Mm                                         1,2,6 DE
8.                     ~Mm v Mn v ~Lmn               5, Universal Instantiation
9.                     ~Mm                                      1,2,6, DE
10.                     !                                            7,9 Contradiction

I’ve assumed as premises the information provided in the puzzle. On the SHOW line you’ll find a symbolization of the follow-up question. On line 4, I assume the negation of the SHOW line. On line 5, I substitute negated existential quantifiers for universal quantifiers trailed by negation, which is then distributed via De Morgan application. Since both variables in line 5 are under universal scope, I instantiate without restriction; in particular, to the constants denoting Larry and Mary. The result is line 6, which says Larry is unmarried, Mary is, and Larry is not looking at Mary. However, on line 1 we assumed Larry was married. Similarly, on line 2 we assumed Larry was looking at Mary. Hence, by disjunctive syllogisms, we infer Mary must be married on line 7. Instantiating line 5 once more, this time with Mary unmarried, Nick married, and Mary not looking at Nick, results by similar reasoning in Mary not being married on line 9. Since we already inferred that Mary was married, we now find ourselves in a contradiction. Hence, we infer someone is looking at someone not married.

Another way to think about the solution is to observe (or suppose what is plausible) that Mary is either married or not. If married, then since Mary is looking at Nick who is unmarried, someone is looking at someone who is unmarried. If unmarried, then since married Larry is looking at Mary, someone married is looking at someone unmarried. Either way, the answer is "Yes."

DISCUSSION:
Several of my students, when initially posed with the puzzle, claim there is “Not enough information to answer”. I take this to suggest students have trouble thinking to make certain plausible suppositions. To be clear, I do not think students have trouble making plausible suppositions and reasoning from them in general. They do, after all, deliberate about the future. Rather, it just doesn’t naturally occur to them to make even plausible suppositions in certain contexts, such as the context of puzzles and the contexts of proofs.

Works Cited
 Bentham, J. (2008). Logic and Reasoning: Do the Facts Matter? Studia Logica. 88:67-84