Here is a construction of size $38$. Call the points *cards* and divide the deck into $4$ *suits*, so that $2$ suits have ranks $1,...,13$ and $2$ suits have ranks $1,...,12$. Every hand of $5$ cards must have at least $2$ cards in some suit. So, if we cover all possible pairs in each suit, the result will be a wheel.

The [La Jolla Covering Repository][1] says that [$C(12,5,2)=9$][2] so we can cover all pairs within the short suits with $9$ hands each:

    1  2  4  7  9 
    1  3  4  6 12 
    1  3  7 10 11 
    1  5  6  8 12 
    2  3  5  7  8 
    2  6 10 11 12 
    3  6  7  9 12 
    4  5  8 10 11 
    5  8  9 10 11 

Similarly, it says [$C(13,5,2) \le 10$][3] so we can cover all pairs within the long suits with $10$ hands each:

    1  2  3  4  5
    1  6  7  8  9
    1 10 11 12 13
    2  3  6  7 10
    2  3  8  9 11
    4  5  6  7 11
    4  5  8  9 10
    2  3  4 12 13
    5  6  7 12 13
    1  8  9 12 13

The total is $38$ hands which intersect each hand of $5$ cards in at least $2$ cards. So, $L(50,5,5,2) \le 38$.

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Edit: The best this method can do with the coverings known in the La Jolla Repository is $37$. That can be done by splitting a $50$ card deck into suits of sizes $(17, 11, 11, 11),$ $(13,13,13,11)$, or $(15, 13, 11, 11)$.  

[$C(17,5,2) \le 16$][4].

[$C(15,5,2) \le 13$][5].

[$C(13,5,2) \le 10$][6] as shown above.

[$C(11,5,2) \le 7$][7].


  [1]: http://www.ccrwest.org/cover.html
  [2]: http://www.ccrwest.org/cover/t_pages/t2/k5/C_12_5_2.html
  [3]: http://www.ccrwest.org/cover/t_pages/t2/k5/C_13_5_2.html
  [4]: http://www.ccrwest.org/cover/t_pages/t2/k5/C_17_5_2.html
  [5]: http://www.ccrwest.org/cover/t_pages/t2/k5/C_15_5_2.html
  [6]: http://www.ccrwest.org/cover/t_pages/t2/k5/C_13_5_2.html
  [7]: http://www.ccrwest.org/cover/t_pages/t2/k5/C_11_5_2.html