Let [tex]X[/tex] be a random variable denoting the amount of winnings in any given round of the game. There is some probability [tex]p[/tex] of winning where [tex]P(X=99,500)=p[/tex], while [tex]P(X=-500)=1-p[/tex], assuming these are the only two possible outcomes. So we have
[tex]\mathbb E[X]=\displaystyle\sum_xxP(X=x)=99,500p-500(1-p)=100[/tex]
[tex]100,000p-500=100\implies p=\dfrac3{500}=0.006[/tex]
