In this answer there is a covering design that guarantees guessing right two numbers in a lottery that draws $6$ numbers out of $70$, by playing $35$ tickets with $6$ numbers each.
Now I would like to compute the expected number of winning couples of numbers that we can get with those $35$ tickets. After a comment, I clarify that all couples in a played ticket matching a couple in the $6$ extracted numbers win (there is a multiple win). Since the tickets are divided in $5$ groups, each covering all possible couples within $14$ contiguous numbers (totalling to $14 \times 5 = 70$), I thought about the Birthday problem, because numbers are chosen uniformly, therefore we can think of a year with $5$ days when the $6$ numbers can have their birthday.
The Wikipedia page gives under the section "Number of people with a shared birthday" the following formula:
$$n\left(1 - \left( \frac{d-1}{d} \right)^{n-1}\right)$$
And replacing $n=6$ and $d=5$ we get $4.03392$.
Is this correct? It seems too high.
But anyway, then I want to know the expected number of couples, not numbers of numbers in a couple. How can we compute its exact value?
In order for $\{x,y\}$ to be a "winning couple," two things must occur:
$x$ and $y$ are both drawn numbers.
$x$ and $y$ are both in the same "group". The are five groups: $[1,14],[15,28],[29,42],[43,56]$, and $[57,70]$, where $[a,b]:=\{a,a+1,\dots,b\}$.
These two conditions are sufficient as well, because the covering design construction implies that all pairs of numbers in the same group will appear together on some ticket.
There are $5\cdot \binom{14}2$ pairs of numbers which satisfy the second condition alone. For each of these pairs, the probability they both appear among the six drawn numbers is $$ \binom{68}4\Big/\binom{70}6 $$ Therefore, by linearity of expectation, the expected number of winning couples is $$ 5\cdot \binom{14}2\cdot \binom{68}4\Big/\binom{70}6\approx 2.83. $$