You have a dice that has 70% chance of rolling a 1 and 30% chance of rolling a 2. You win if you roll two 1's and lose if you roll two 2's (doesn't have to be consecutive). What is the chance of winning?
I know there are 3 winning combinations (1 1, 1 2 1 and 2 1 1) out of 6. The other 3 losing combinations being (2 2, 2 1 2 and 1 2 2).
But I'm not sure how to account for the 70% chance of getting 1 and 30% chance of getting 2. Would appreciate some help, thank you!
Best Answer
You could use conditional probabilities (depending on the first result):
If the first roll is a $1$, the next two rolls must not be two $2$'s, and if the first roll is a $2$, the next two rolls must be two $1$'s, so the probability to win is: $$p = 0.7\cdot(1-0.3^2) + 0.3\cdot0.7^2 = 0.7\cdot0.91+0.3 \cdot0.49 = 0.784$$