Two (6-sided) dice roll probability table. Single die roll probability tables….Probability of rolling more than a certain number (e.g. roll more than a 5).

While contemplating a gambling problem posed by Chevalier de Mere in 1654, Blaise Pascal and Pierre de Fermat laid the fundamental groundwork of probability theory, and are thereby accredited the fathers of probability.

At the heart of this definition are three conditions, called the axioms of probability theory.

• Axiom 1: The probability of an event is a real number greater than or equal to 0.
• Axiom 2: The probability that at least one of all the possible outcomes of a process (such as rolling a die) will occur is 1.

Disjoint events are either-or events. For example, consider flipping a coin. The two events–flipping heads or flipping tails–have no outcomes in common. If there are no outcomes in common, the two events are disjoint.

The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on.

a) Consider the complement problem, there is a 5/6 probability of not rolling a six for any given die, and since the four dice are independent, the probability of not rolling a six is (5/6)4 = 54/64 = 625/1296. The probability of rolling at least one six is therefore 1 − 625/1296 = 671/1296 ≈ . 517.

1/6 x 1/6 x 1/6 = 1/216 = 100/216% = 0.463% (to 3 s.f.) 3 times in a row must equate to 1/6 x 1/6 x 1/6. JD33: The probability (to 1 s.f) is, indeed, 0.005 but you need to multiply that by a hundred to convert it to a percentage, giving 0.5% (to 1 s.f.)

Assuming that the dice is fair and that the rolls are independent, which is what one would expect, the probabilities multiply. So the probability is indeed (1/20)3=1/8000.

Probabilities for the two dice

98.46%