$4\times 4$ matrix game, covering $9$ of $16$ squares for even money bet. Good bet or not? In a hypothetical game, person A offers a challenge to person B saying that A gets a $4\times4$ playing board ($2$ dimensional matrix) and gets to roll a special $16$ sided fair die such that each of the $16$ numbers on it are equally likely to appear each roll.  A will keep randomly rolling the die until he gets $9$ distinct random numbers (from $1$ to $16$ let's say), covering only the corresponding position (square) on the playing board with a marker each time he gets one of those $9$ numbers.  The numbering of the boards $16$ squares is irrelevant but for simplicity you can imagine it is numbered $1$ thru $4$ across the top row, $5$ thru $8$ across the $2$nd row and so on.
To have a winning board, A must have at least $4$ of them contiguously and legitimately marked.  That is, across any row, down any column, or either diagonal.  Thus there are only $10$ ways to win per game, and A can only win once per game.  Obviously the board has to start with no squares covered.  At the end of the game, there will be exactly $9$ squares covered but of course if a win is determined before all $9$ rolls, the game can end early as a win.  For example, if a row is completely filled even after only $7$ rolls.
So the question is, if A tells B he must pay A even money for wins and A loses any wagered money for losses, who is getting the better odds?  A or B?  How much better? 
 A: There are a total of $C(16,9)$ ways to cover the board with 9 markers. Now lets count how many 'winning' possibilities (for $A$)there are, that means in how many ways we can cover the board such that there are 4 in a row. (or column or diagonal).
Lets say we have four in a row or column or diagonal. There are - as you said - exactly 10 ways of doing so. Then we still have to distribute the remaining 5 markes to the remaining 12 fields which all produce a win situation.
So we have a total of $10 \cdot C(12,5)-D$ ways of winning, where $D$ is the number of arrangements where we have two or more rows/columns/diagonals covered (I did not find an elegant way of caluclating how many of those there are yet. But I think you can quite easily write a script that does that for you.). Therefore the probability of $A$ winning the game is
$$ p = \frac{10 \cdot C(12,5)-D}{C(16,9)}$$
(Where $C(n,k) = \begin{pmatrix} n \\ k \end{pmatrix}$ the binomial coefficient.)
If $B$ has to pay whenever $A$ loses but $A$ does not have to pay anything, then the games is not 'fair'. For the game to be fair we have to make the expected price 0. (For both). This depends on the exact rules, wheter $A$ has to pay in order to participate or $A$ only has to pay if he loses etc.
EDIT: I just wrote a small matlab script that calculates all possibilities of distributing 9 markers to 16 fields, and counts how many of them have at least one row/column/diagonal of 4 markers, the result is 6688 of 11440. So the probability is $6688/11440 \simeq 58.4616 \%$
wins = 0;
total = 0;
for n = 0:2^16-1;%go throu all numbers from 0 to 2^16-1
    m = n;
    A = zeros(4,4);
    for j=1:16; %convert them to binary and store as 4x4 field
        A(j) = mod(m,2);
        m = (m-A(j))/2;
    end     
    if sum(sum(A)) == 9 %check whether there are 9 markers on the board
        total = total+1;
        s=sum(sum(A)==4)+sum(sum(A')==4)+(sum(diag(A))==4)+(sum(diag(fliplr(A)))==4);
        if s>=1 %if there is at least one row/col/diag with sum=4 count as win
            wins = wins+1;
        end
    end
end
[wins,total]

A: I wrote a computer program to answer my own question but I was hoping for a complete mathematical answer to help verify it.  When I simulate $1$ billion decisions, it tells me A has about a $58.47$% chance of winning which is a decent "edge" over B so A has the better odds.  When I asked wolframalpha to solve for D in flawr's equation, it is telling me D $\approx$ $1232$ (rounded up) but I would like to know how to get that number mathematically (not using a computer program).  It seems for a problem like this it is a pain to subtract out all the double and triple counts where the board has $2$ or $3$ completed line segments of $4$ adjacent markers in combinations of row(s), column(s), and/or either diagonal.
Update: I also get $6688$ when counting up the number of winners.  The program was actually fairly easy for me to write.  Also, I can change 1 line of code to get the multicounted "wrong" answer of $7920$ but that is useful information because $7920$ - $6688$ = $1232$ which is the value of D we need to subtract out.  Those are the cases where more than $1$ line segment of $4$ markers is found in one game but that only counts as $1$ win, not $2$ or even $3$.
Here is something else to consider. We got a solution to this problem because $16 \choose 9$ is a reasonably small number and can be simulated on a computer with instant runtime.  However, what if we extended this problem so that was not true.  For example, what if we use an $8$x$8$ board instead (like a checkerboard or a chessboard) and draw $32$ random numbers instead to mark $32$ of the $64$ squares?  $64 \choose 32$ is about $1.83$ quintillion (which is $1.83 * 10^{18}$).  That might take a very long time to simulate on a computer.  So how would one go about solving that problem?  That is, on an $8$x$8$ board, covering $32$ random squares such that at least one row, column, and/or diagonal is completely covered.  I wonder what the probability of that would be. I assume that is one of those problems where you just take a small random sample of marker arrangements until you get a few winners and then assume that is a reasonable approximation to the correct probability.  If the probability is something like $50$%, then in theory, only getting $2$ decisions (out of $1.83$ quintillion) could give an excellent approximation but of course it would be much better to get maybe $1$ million decisions.
Also I tried "uncounting" the duplicates and triplicates like when you fill $2$ rows, $2$ columns, $2$ diagonals, $1$ row and $1$ column... but am not yet able to "find" all $1,232$ "multicounts".  This is not an easy counting problem "on paper".  At least not yet for me.
$Update:$ I tried the $8$x$8$ simulation ($1$ million decisions) and was surprised to see only about a $4.3$% probability of covering at least one row, column, and/or diagonal using $32$ randomly placed markers on the board of $64$ squares.
