The classic gambling Problem of probability

Modern dice are popular in the Middle Ages, and the Renaissance prodigal Sir Meyres a mathematical conundrum:

A dice Toss 4 times, at least once is 6 points, two dice toss 24 times, at least once is double 6 points;

Normal thinking, the probability of gambling in both cases is the same:

A dice: the chance of throwing a 6 point =1/6, throwing four times the chance of 6 points =4x (1/6) =2/3

Two dice: throw 1 times the probability of a double 6 points is 1/36, the chance of throwing 24 times 6 points is =24x (1/36) =2/3

But in fact the second situation of gambling, the number of failures more.

Sir Meyres raised this question to his friend, genius Brenz Pascal (1623-1666 years). Although Pascal no longer thought that mathematics could bring pleasure, he agreed to solve the problem of Mel.

Pascal wrote to his friend, Pierre Depelma, a genius. After several letters, they created a modern form of probability theory.

Basic result--basic events

Sample Space

Several laws:

Law of Addition: P (e∪f) =p (E) +p (F)-P (e∩f)

Special addition rule: when E and F repel (i.e. P (e∩f) =0), P (e∪f) =p (e) +p (F)

Subtraction Law: P (E) =1-p (e)

Multiplication rule: P (e∩f) =p (e| f) P (f)

Special multiplication rule: P (e∩f) =p (E) p (F)

Positive solution:

The E represents the event that 4 throws a single dice at least one 6 point at a time, what is P (e)? This problem is more easily explained by a counter example: E indicates a probability event that 4 dice do not appear at 6 points.

If the Ai represents an event that throws 6 points, it is known that: P (Ai) = (5/6), and 4 throws are independent of each other, so

Now, to solve the other half of the problem, so that F is throwing 24 times two dice at least one double 6 points event. Similarly, the inverse of f is easier to describe, which means that there are no 6-point events.

If BI indicates that the 6-point event does not appear in the second throw, that

The classic gambling Problem of probability