Short Answer: 11/12 Comment The easy way to do this is the figure out the probability for 2 and 3. Add those two together and subtract from one.
Total number of ways to throw a pair of dice. A normal die has 6 sides. Each side is numbered with a 1, 2, 3, 4, 5, 6. If you imagine them with different colors (say red and green) then all together the red die offers you 1/6 of a chance of getting any one number. The green die will offer the same probability. P(all combinations) = P(red die) * P(green die) P(all combinations) = 1/6 * 1/6 P(all combinations) = 1/36
Probability of rolling a two There is only one way to do that. There are thirty six possible rolls. Only 1 of them will give you a two. P(red_1) + P(_1) = 1/36
Probability of rolling a Three This can be done two ways Red Green 2 1 1 2
That means that you have two ways which will bring success. P(3) = 2/36 This can be reduced, but there is no real point for this question.
Probability of throwing a 2 or 3 P(2 or 3) = 1/36 + 2/36 = 3/36 P(2 or 3) = 1/12
Probability of throwing 4 or more The probability of throwing 4 or more is 1 - 1/12 = 11/12 <<<<< Answer
That means if you go to Vegas and you play this game and you want to throw the dice 11 times out of 12 you should throw something that is 4 or more.
