Dan H. writes:

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... We assume that each [Bingo] board is equally likely, though that can't be true--with over 552 septillion possible boards, only a small fraction ... have ever been printed! ...
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Certainly only a small fraction have ever been printed, but that doesn't exclude the possiblity that they are random. Chances are they aren't, but it wouldn't be that hard to hook up a randomizer to the printer so that each board is the result of the appropriate randomization process.  (Hmmm, you may have suggested a likely way to win at Bingo in the long run...)

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There are at most five equivalence classes of lines: center horizontal(1), off-center horizontal(4), center vertical(1), off-center vertical(4), and diagonal(2).  Is the distribution for diagonals the same as for the center horizontal? 
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If the only question is Find the probability P(X;n) that line X is filled after n draws, then there are only 2 relevant categories for X: those that need 5 matching draws, and those that need only 4.  The number of possible sequences of n draws is D_n = 75! / (75-n)!, and given X, the number of winning sequences among these for one of the 8 lines X needing 5 matches (for n >= 5) is W_5 = 5! * ( n! / (5! (n-5)!) ) = n! / (n-5)!, so the winning probability is the quotient W_5 / D_n =

P_5 = n! (75-n)! /  (75! (n-5)!) = n(n-1)(n-2)(n-3)(n-4) / (75*74*...*(75-n+1)),

and likewise for one of the 4 lines X needing only 4 matches, W_4 / D_n =

P_4 = n(n-1)(n-2)(n-3) / 75*74*...*(75-n+1).

--Dan A.