Fun! Here are a few ideas:

• Hey, D’Artagnan, does the combined troy weight of your two coins exceed the troy weight of my two coins that are identical to yours in volume and are both composed of a metal that weighs less than gold but more than silver?

• Hey, D’Artagnan, if you place your two coins on the table with opposite faces showing, will you have a gold coin showing a face that will match the showing face of the coin I’m about to flip?

• Hey, D’Artagnan, if gold coins are worth more than 10 livres and silver coins are worth less than 10 livres, are your coins worth more than 20 livres?

• Hey, D’Artagnan, if you were to choose one of your coins at random from a bag, would you be able to find a word in English that rhymes with its type of metal?

• Hey, D’Artagnan, would you do at least as well financially if you traded in your coins for two random coins of gold or silver?

But the real challenge is to find questions for which the “one of each” scenario does NOT lead to the “I don’t know” response:

• Hey, D’Artagnan, would your two coins be acceptable to Pierre the merchant, who will only accept two identical coins made from a precious metal that is on his exclusive list (which includes gold)?

]]>D’Artagnan was rewarded with two coins, each of which was either gold or silver. Ask him one question to which he can answer “Yes,” “No,” or “I do not know,” so that you will be able to figure out his coins.

From an information-theoretic point of view, it seems clear that the two versions of the problem are the same, because you want to distinguish between three possibilities and you’re allowed to ask a question that will extract one trit of information. From that perspective, a natural solution is to ask, “Say ‘yes’ if you have two gold coins, ‘no’ if you have two silver coins, and ‘I do not know’ otherwise.” However, this solution seems to violate the spirit of the problem quite badly!

A less flagrantly cheating solution to my version is to ask something like, “Imagine that the coins you got came from a set of six coins, three gold and three silver, and that the rest of the coins were given…” However, this solution also seems unnatural to me, because of the weird counterfactual.

So, is this problem really a linguistics puzzle disguised as an information theory puzzle? If so, is it possible to lay out explicit rules for such puzzles?

]]>Start by weighing 1+3+7 against 2+4+5.

If 1+3+7 is lighter, weigh 1+2 against 3 next.

If 2+4+5 is lighter, weigh 1+4 against 5 next.

If the first weighing balances, then 6, 8, or 9 is the culprit; weigh 2+6 against 8 to find out.

There’s probably a ton of other solutions.

]]>1+2<3, 1+3<4: 1 light

1+2<3, 1+3=4: 2 light

1+2<3, 4<1+3: 3 heavy

1+2=3, 1+3<4: 4 heavy

1+2=3, 1+3=4: impossible

1+2=3, 4<1+3: 4 light

3<1+2, 1+3<4: 3 light

3<1+2, 1+3=4: 2 heavy

3<1+2, 4<1+3: 1 heavy

In the case of four coins marked 1-2-3-4, there are only two such weighings: 1+2 against 3, and 1+3 against 4. So those must be the two weighings; if they don’t work, there is no solution. Let’s see if it works:

1+2<3, 1+3<4: 1 light

1+2<3, 1+3=4: 2 light

1+24: 3 heavy

1+2=3, 1+34: 4 light

1+2>3, 1+33, 1+3=4: 2 heavy

1+2>3, 1+3>4: 1 heavy

So it works out.

No clue yet on the nine coin puzzle.

]]>There are four silver coins marked 1, 2, 3, and 4. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is either lighter or heavier than it should be. Find the fake coin using the balance scale twice.

There are nine silver coins marked 1, 2, 3, 4, 5, 6, 7, 8, and 9. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

]]>5 + 2 = 4 + 3 -> 1

Otherwise: 2 + 3 against 5 will tell you which one is wrong.

I wonder if it is still possible if you only know that one of the weights is incorrect

]]>There are five silver coins marked 1, 2, 3, 4, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

]]>Yes => 2 silver

IDK => 1 silver, 1 gold

No => 2 Gold. ]]>