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Hi ZS, assuming that whether or even otherwise one wins or loses on one scratch ticket (what is that, in any case?) is independent from winning or losing on any other scratch ticket, you treat each event as persistent 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. Prevails the probability of [losing] on 30 scratch tickets one after the other (if that maybe what your problem is asking) must be (2/3)^30 = approximately 5.2 x 10^-6, which is about.0000052, or 52 out of 10 million, which amounts to 1 chance out of 192,307.
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