Where I derive a constraint for Double Dobble.

In previous post Matt Parker on Dobble, I introduced a video by Matt Parker with interesting stuff on the game Dobble.

One interesting thing is a link to Double Dobble, where a variant is introduced with the following rule:

Each pair of cards share exactly two symbols.

The two solutions provided, for $5$ and $9$ symbols per card, have been derived with some code, but it got me thinking on whether they knew beforehand how many cards these attempts would yield (i.e. $11$ and $37$ cards, respectively, which is the same as the overall number of symbols).

I think I got how to calculate these numbers.

Let’s have $N$ symbols and $N$ blocks in total. This means that there are:


distinct pairs in total; each pair must appear in two cards, so we consider $N(N-1)$ total pairs to distribute over $N$ blocks, i.e. each block holds $N - 1$ pairs.

If a block contains $k$ symbols, then it also contains:


distinct pairs. As we saw, this must be equal to $N - 1$, so:

\[N - 1 = \frac{k(k-1)}{2} \\ N = \frac{k(k-1)}{2} + 1\]

This formula tells us that:

\[k = 5 \iff N = 11 \\ k = 9 \iff N = 37\]