What are the optimal strategies for the "prime-game"? A and B are playing the following game :
A and B choose a number from 1 to 100, not knowing the number chosen by the
opponent.
A wins if the sum of the chosen numbers is prime, otherwise B wins.
Which mixed strategies are optimal for A and B ? Which player wins
with which probability, if both play optimal ?
This optimal strategy can be calculated with the simplex algorithm, but I
did not have access to a program calculating large simplex tableaus.
Is there an easier method to solve such games ?
If I remember right, I found out that player B wins between $\frac{6}{11}$ and $\frac{3}{5}$ units per game , but I do not remember the strategies.
 A: The sum is $\le 200$ hence any composite number has one of the primes $2,3,5,7,11,13$ as a factor.
If for some $0\le n\le 70$, $B$ picks one of the numbers $n+1, n+2, \ldots, n+30$ at random (each with same probability) then the distribution $A+B\bmod 30$ will be uniformly random, hence $A+B$ will have a divisor $\in\{2,3,5\}$ with probability $\frac{22}{30}=\frac{11}{15}$ (i.e. this divisibility fails only if the remainder $\mod 30$ is one of $1,7, 11, 13, 17,19,23,29$). Also, $A+B\ge n+2$ so that a choice of $n\ge 4$ guarantees that $A+B$ is composite with probability at least $\frac{11}{15}$.
Since already this simple strategy gives $B$ an expected winning probability greater than $\frac35$, you seem to remember wrong. 
By the same argument, $A$ can enforce that the smallest divisor of $A+B$ is $>5$ with probability at least $\frac 4{15}$. The only composite numbers with this property up to $200$ are $7^2, 7\cdot 11, 7\cdot 13, 7\cdot 17, 7\cdot19, 7\cdot 23, 11^2,11\cdot 13, 11\cdot 17, 13^2$. It is feasible that a suitable choice of $n$ is able to modtly avoid theseten numbers, thus allowing $A$ to almost reach this upper limit of $\frac4{15}$. 
For a precise result, it seems one really has to try something like the simplex approach you suggested.
