March 2026
Here is such a Cobble deck: \[[A, B, E], [A, C, F], [A, D, G], [B, C, G], [B, D, F], [C, D, E], [E, F, G].\] You should check that it satisfies the four properties from the definition given the problem statement. There are many other ways this could be done, but they all differ from this one by permuting the letters and permuting the cards.
To see how you could have constructed the deck, read through the solution to Question 2 including the note at the end.
Suppose \(A, B, C,\) and \(D\) are four symbols satisfying Property (iv). Then Properties (iii) and (iv) together tell us that each pair of these four symbols appears on a distinct card. This is because if two pairs appeared on the same card, we would have three of the four symbols appearing on the same card, contradicting Property (iv).
There are six distinct pairs you can create from the four symbols \(A, B, C,\) and \(D\), and each needs to be on a distinct card. So, the six cards must look like \[\begin{align*} K_1 = [A,B,\ldots] &\quad K_2 = [A,C,\ldots] \quad K_3 =[A,D,\ldots]\\ K_4 = [B,C,\ldots] &\quad K_5 = [B,D,\ldots] \quad K_6 = [C,D,\ldots].\end{align*}\] We have argued that there must be at least six cards in a Cobble deck, so to complete the question we must show that you cannot add symbols to these six cards to create a Cobble deck.
So far, this collection of cards satisfies Properties (i), (iii), and (iv), but not (ii). Currently, \(K_1\) and \(K_6\) do not have a symbol in common, \(K_2\) and \(K_5\) do not have a symbol in common, and \(K_3\) and \(K_4\) do not have a symbol in common.
Let’s first look at \(K_1\) and \(K_6\). We cannot add \(C\) or \(D\) to \(K_1\), since then \(A, B, C,\) and \(D\) would no longer satisfy Property (iv). Similarly, we cannot add \(A\) or \(B\) to \(K_6\). Therefore, we need to introduce a new symbol to both \(K_1\) and \(K_6\). Call this new symbol \(E\). The six cards must therefore look like this: \[\begin{align*} K_1 = [A,B,E,\ldots] &\quad K_2 = [A,C,\ldots] \quad K_3 =[A,D,\ldots]\\ K_4 = [B,C,\ldots] &\quad K_5 = [B,D,\ldots] \quad K_6 = [C,D,E,\ldots].\end{align*}\] Let’s now focus on \(K_2\) and \(K_5\). By a similar argument as we made for \(K_1\) and \(K_6\), we cannot add \(A, B, C,\) or \(D\) to either \(K_2\) or \(K_5\) to make the two cards contain a unique symbol in common. Furthermore, we cannot add \(E\) to both \(K_2\) and \(K_5\) because then \(K_2\) and \(K_1\) would share more than one symbol (and so would \(K_5\) and \(K_6\), \(K_2\) and \(K_6\), and \(K_1\) and \(K_5\)).
We are therefore forced to introduce another symbol \(F\) to both \(K_2\) and \(K_5\).
A similar argument can be made for \(K_3\) and \(K_4\), whose conclusion is that yet another symbol \(G\) must be introduced to both cards.
Therefore, assuming these six cards form a Cobble deck, they must look like this: \[\begin{align*} K_1 = [A,B,E,\ldots] &\quad K_2 = [A,C,F,\ldots] \quad K_3 =[A,D,G,\ldots]\\ K_4 = [B,C,G,\ldots] &\quad K_5 = [B,D,F,\ldots] \quad K_6 = [C,D,E,\ldots].\end{align*}\] These six cards now satisfy Properties (i), (ii), and (iv). However, (iii) is not satisfied since \(E\) and \(F\) (for example) do not appear together on any card.
The goal now is to force \(E\) and \(F\) to appear together on a unique card. Let’s try to do this for each card one by one.
\(K_1\): Introducing an \(F\) to \(K_1\) would mean \(K_1\) and \(K_2\) share more than one symbol, so we cannot do this.
\(K_2\): Adding \(E\) to \(K_2\) means \(K_1\) and \(K_2\) share two symbols, an unacceptable state of affairs.
\(K_3\): Putting both \(E\) and \(F\) on \(K_3\) would force \(K_2\) and \(K_3\) to share \(A\) and \(F\), which rules this possibility out.
\(K_4\): Like the previous case, adding \(E\) and \(F\) to \(K_4\) implies \(K_4\) and \(K_5\) share two symbols.
\(K_5\): Placing \(E\) on \(K_5\) is also not possible since \(K_5\) and \(K_6\) would then share two symbols.
\(K_6\): Adding an \(F\) to \(K_6\) forces \(K_5\) and \(K_6\) to have two symbols in common, which is simply not an option.
There is no path forward without introducing another card, and therefore there is no Cobble deck with six cards.
Note: We can remedy the situation by introducing a seventh card \([E,F,G]\), which results in the Cobble deck from the solution to Question 1.
We’ll first prove a helpful result (which we will call a lemma, a word used in mathematics for an intermediary result used to prove a more desirable result).
Lemma: For any two cards in a Cobble deck, there is a symbol that is not on either card.
Proof. This proof will be a little sneaky. Our strategy will be to assume there are two cards that contain all the symbols that appear in the deck, and arrive at a contradiction.
Assume that \(K_1\) and \(K_2\) contain all the symbols in the Cobble deck. By Property (iv), there are four symbols so that no three of them appear on the same card. Therefore, two of them must appear on \(K_1\) and two must appear on \(K_2\). Let \(A\) and \(B\) be the two symbols appearing on \(K_1\), and let \(C\) and \(D\) be the two symbols appearing on \(K_2\). Note that none of \(A\), \(B\), \(C\), or \(D\) appear on both \(K_1\) and \(K_2\) since then three of them would appear on one of the cards.
Let \(L_1\) be the unique card containing \(A\) and \(C\), and let \(L_2\) be the unique card containing \(B\) and \(D\). Note that \(L_1\), \(L_2\), \(K_1\), and \(K_2\) must be four distinct cards.
Let \(E\) be the unique symbol appearing on both \(L_1\) and \(L_2\). Note that \(E\) is distinct from \(A\) and \(B\) since \(A\) is not on \(L_2\) and \(B\) is not on \(L_1\) (can you see why?). The goal is to show that \(E\) cannot appear on \(K_1\) or \(K_2\), which will contradict our original assumption and complete the proof.
If \(E\) is on \(K_1\), then \(E\) and \(A\) are both on \(L_1\) and \(K_1\), which cannot happen. If \(E\) is on \(K_2\), then \(E\) and \(B\) are both on \(L_2\) and \(K_2\), which is impossible. Therefore, \(E\) is not on \(K_1\) or \(K_2\), completing the proof. ◻
Excellent! Let’s now use the Lemma to give a solution to the original question. Let \(K_1\) and \(K_2\) be two distinct cards. Let \(P\) be a symbol not contained in either \(K_1\) and \(K_2\), which much exist by the Lemma. Suppose \(K_1 = [A_1,A_2,\ldots,A_n]\). For each \(i\) between \(1\) and \(n\), let \(L_i\) be the unique card containing \(A_i\) and \(P\). Let \(B_i\) be the unique symbol contained in both \(L_i\) and \(K_2\).
We want to show that \(K_2 = [B_1,B_2,\ldots,B_n]\). The first thing is to show that each of the \(B_i\) are distinct. Suppose not, and suppose that \(B_i = B_j\) for some \(i \neq j\). Then \(B_i\) would be on both \(L_i\) and \(L_j\). Since \(P\) is the unique symbol on both \(L_i\) and \(L_j\), this would mean that \(B_i = P\). However, \(P\) is not on \(K_2\), a contradiction. Therefore, the symbols \(B_1,B_2,\ldots,B_n\) are all distinct.
We must now show that there are no other symbols on \(K_2\). To that end, suppose \(B\) is a symbol on \(K_2\), and let \(L\) be the unique card containing \(B\) and \(P\). Now, \(L\) and \(K_1\) share a symbol. Since \(K_1 = [A_1,A_2,\ldots,A_n]\), it must be the case that there is a unique \(t\) between \(1\) and \(n\) so that \(A_t\) is on both \(L\) and \(K_1\). Since \(L\) contains \(A_t\) and \(P\), it must be that \(L = L_t\). Since \(B_t\) is the unique symbol on both \(L_t\) and \(K_2\), and since \(B\) is contained on both \(L\) and \(K_2\), we are forced to conclude that \(B = B_t\).
We now have that \(K_2 = [B_1,B_2,\ldots,B_n]\). Therefore, if a card contains \(n\) symbols, then every other card must also contain \(n\) symbols.
One of the (many) amazing things about Cobble decks is that roughly, you can change the roles of cards and symbols and prove similar statements. For example, the result we will prove in this question is the counterpart to the one in the previous question.
To begin this question, we will first prove the counterpart of Property (iv), which we will call Property (v).
Property (v): In a Cobble deck, there are four cards so that no three of them contain the same symbol.
Proof. Let \(A, B, C,\) and \(D\) be the four symbols satisfying Property (iv) for the Cobble deck. Let \(K_1\), \(K_2\), \(K_3\), and \(K_4\) be the unique cards that contain \(A\) and \(B\), \(B\) and \(C\), \(C\) and \(D\), and \(D\) and \(A\) respectively. The claim is that these four cards satisfy the property that no three of them contain the same symbol.
The unique symbol common to \(K_1\) and \(K_2\) is \(B\), the unique symbol common to \(K_2\) and \(K_3\) is \(C\), the unique symbol common to \(K_3\) and \(K_4\) is \(D\), and the unique symbol common to \(K_4\) and \(K_1\) is \(A\).
Since \(K_2\) contains \(C\), \(K_1\) contains \(A\), and the unique symbol common to \(K_1\) and \(K_2\) is \(B\), \(K_1\) does not contain \(C\) and \(K_2\) does not contain \(A\). Similarly,
\(K_2\) does not contain \(D\) and \(K_3\) does not contain \(B\),
\(K_3\) does not contain \(A\) and \(K_4\) does not contain \(C\), and
\(K_4\) does not contain \(B\) and \(K_1\) does not contain \(D\).
Since \(K_3\) does not contain the unique symbol common to \(K_1\) and \(K_2\) (which is \(B\)), there is no symbol that appears on \(K_1\), \(K_2\), and \(K_3\). Since \(K_4\) also does not contain \(B\), there is no symbol appearing on the three cards \(K_1\), \(K_2\), and \(K_4\).
Similarly, \(K_1\) and \(K_2\) do not contain the unique symbol common to \(K_3\) and \(K_4\) (which is \(D\)), so there is no symbol that appears on all three cards \(K_1, K_3, K_4\), as well as \(K_2, K_3, K_4\). This checks every possibility of three cards from the set \(K_1, K_2, K_3, K_4\), and these four cards satisfy that no three of them contain the same symbol. ◻
Now, Properties (ii) and (iii) are counterparts of each other, and Properties (iv) and (v) are counterparts of each other. To run the argument from the previous question, we need a counterpart to the lemma from the previous question. While reading the proof, notice that the proof is the same as that of the Lemma in the previous question, but with the roles of the symbols and cards switched.
Lemma: For any two symbols in a Cobble deck, there is a card that does not contain either of them.
Proof. Let \(A\) and \(B\) be two symbols, and assume (towards a contradiction) that every card contains \(A\) or \(B\) (or both). By Property (v), there are four cards, no three of which contain the same symbol. Then two of these cards must contain \(A\) and not \(B\), and the other two must contain \(B\) and not \(A\). Let \(K_1\) and \(K_2\) be the cards containing \(A\), and let \(K_3\) and \(K_4\) be the cards containing \(B\).
Let \(C\) be the unique symbol appearing on \(K_1\) and \(K_3\), and let \(D\) be the unique symbol on \(K_2\) and \(K_4\). Let \(K\) be the unique card that contains \(C\) and \(D\). To complete the proof of the lemma, it is enough to show that \(K\) does not contain \(A\) or \(B\).
If \(K\) contains \(A\), then \(K\) and \(K_1\) contain both \(A\) and \(C\), so they must be the same card. Also, \(K\) and \(K_2\) both contain \(A\) and \(D\), so they must also be the same card. But \(K_1\) and \(K_2\) are distinct cards, so this cannot happen and \(K\) cannot contain \(A\). A similar argument shows that \(K\) cannot contain \(B\), completing the proof of the lemma. ◻
Great, now everything is in place to run the same argument as in the previous question, but with the roles of the symbols and cards exchanged.
Let \(A\) and \(B\) be two distinct symbols. Let \(K\) be a card not containing \(A\) or \(B\), whose existence is guaranteed by the Lemma. Let \(K_1,K_2,\ldots,K_n\) be the cards containing \(A\). For each integer \(i\) satisfying \(1 \leq i \leq n\), let \(A_i\) be the unique symbol on \(K_i\) and \(K\), and let \(L_i\) be the unique card containing \(A_i\) and \(B\).
It remains to show that \(L_1,L_2,\ldots,L_n\) is the set of cards containing \(B\). First we must show that if \(i \neq j\), \(L_i\) and \(L_j\) are distinct cards. If they are the same card, then \(A_i\) and \(A_j\) must be the same symbol. This would imply \(K_i\) and \(K_j\) are the same card, since they both contain \(A_i\) and \(A\) (which are two distinct symbols since \(K\) contains \(A_i\) but not \(A\)). However, we assumed that if \(i \neq j\), \(K_i\) and \(K_j\) are distinct cards. Therefore, \(L_i\) and \(L_j\) must be distinct.
Now, to show \(L_1,L_2,\ldots,L_n\) is the complete set of cards containing \(B\), suppose \(L\) is some card containing \(B\). We want to show that \(L\) is the same card as \(L_i\) for some \(i\).
Let \(C\) be the unique symbol on \(L\) and \(K\). Since \(A\) is not on \(K\), \(C\) and \(A\) are distinct. Therefore, there is a unique integer \(i\) satisfying \(1 \leq i \leq n\) so that \(K_i\) contains \(C\) and \(A\). Since \(C\) is on \(K_i\) and \(K\), it must be that \(C = A_i\). Then \(L\) contains \(A_i\) and \(B\), and so \(L = L_i\).
We have proved that if one symbol appears on exactly \(n\) cards, then so does any other symbol.
To complete the question, we need to show that the number of cards that contain a particular symbol is equal to the number of symbols on a particular card.
To do this, suppose \(A\) is a symbol and let \(K\) be a card not containing \(A\). Suppose \(K = [B_1,B_2,\ldots,B_n]\). We want to show that \(A\) is contained in exactly \(n\) cards.
For each integer \(i\) satisfying \(1 \leq i \leq n\), let \(K_i\) be the unique card containing \(A\) and \(B_i\). Since each of the \(B_i\) are distinct, each of the \(K_i\) are distinct. Conversely, suppose \(L\) is a card containing \(A\). Let \(B_t\) be the unique symbol on \(L\) and \(K\). Then \(L\) contains \(B_t\) and \(A\), and so \(L = K_t\). Therefore, the list of cards \(K_1,K_2,\ldots,K_n\) is precisely the list of cards containing \(A\), and \(A\) is contained in exactly \(n\) cards.
From the previous question, we know that in any Cobble deck, the number of symbols on every card is equal to the number of cards containing any particular symbol. Call this number \(n\). Let’s count how many cards are in such a Cobble deck, in terms of \(n\).
Let \(K = [A_1,A_2,\ldots,A_n]\) be a card. By the previous question, each of the \(A_i\) are on exactly \(n-1\) other cards. This gives \(n(n-1)\) other cards in the deck. We claim that these \(n(n-1)\) cards, along with \(K\), form all the cards in the deck.
To see that all these cards are distinct, let \(L_1\) and \(L_2\) be two of the \(n(n-1)\) cards. By the way they are constructed, neither are equal to \(K\). Let the unique symbol shared by \(L_1\) and \(K\) be \(A_t\), and let the unique symbol shared by \(L_2\) and \(K\) be \(A_s\). If \(A_t= A_s\), then \(L_1\) and \(L_2\) are distinct by the way they were constructed. If \(A_t\neq A_s\), then \(A_t\) is on \(L_1\) and \(A_t\) is not on \(L_2\), so \(L_1\) and \(L_2\) are distinct.
To see that these cards form all the cards in the Cobble deck, suppose \(L\) is some other card. Then \(L\) and \(K\) share a unique symbol \(A_t\). Therefore, \(L\) is one of the \(n-1\) other cards containing \(A_t\), so we have accounted for all the cards in the Cobble deck.
In terms of \(n\) (and adding one for the card \(K\)), the total number of cards in the Cobble deck is \(n(n-1)+1 = n^2 - n + 1\). So, if there was a Cobble deck with \(2026\) cards, there would have to be a positive integer \(n\) satisfying \(n^2 - n + 1 = 2026\). However, there are no integer solutions to this equation (which can be checked by solving the quadratic in \(n\) with the quadratic formula). Therefore, there is no Cobble deck with \(2026\) cards.
Challenge: No solution is given to this, because I do not know of a solution. In fact, no one does. This is an open problem. It is equivalent to the existence of what’s called a finite projective plane of order \(12\).
The solution to Question 5 tells us that the number of cards in a Cobble deck must be an integer of the form \(n^2 -n +1\), where \(n\) is a positive integer. In Questions 1 and 2 we showed that the smallest \(n\) for which there is a Cobble deck is \(n = 3\). Cobble decks are known to exist when \(n = p^k + 1\) where \(p\) is a prime number and \(k\) is a positive integer. It is conjectured that these are the only possible values of \(n\) for which a Cobble deck exists, but very little has been proven about the nonexistence of Cobble decks.
It is known that Cobble decks do not exist for \(n = 7\) and \(n = 11\), but that’s it! The smallest number for which this question is unsettled is \(n = 13\), which corresponds to a Cobble deck with \(13^2 - 13 + 1 = 157\) cards.
The game of Dobble (also known as Spot It) is played with what is almost a Cobble deck. The deck consists of cards, each with \(8\) symbols, and almost with each symbol appearing on \(8\) cards. If a Dobble deck were a Cobble deck, there would be \(8^2 - 8 +1 = 57\) cards in the deck. Inexplicably, there are only \(55\) cards in a Dobble deck. Go figure!