Hi ZS, assuming that whether or not one wins or loses on one scratch ticket (what is that, anyhow?) is independent from winning or losing on additional scratch ticket, you treat each event as wages event. Laws of probability tell us to multiply the various probabilities of independent situation. It appears that the probability of [losing] on any particular scratch ticket must be 2/3. So then the probabilities of [losing] on 30 scratch tickets consecutively (if that maybe what your problem is asking) end up being (2/3)^30 = approximately 0.2 x 10^-6, which is about.0000052, or 52 through 10 …