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Fundamental Gambling Theory

by Jerry "Jet" Whittaker
April 17, 2006

It is belief that persistence leads to success sometimes. It is never true for gambling and games of chance in general. Actually, in gambling persistence leads to inevitable bankruptcy. It is a game of probability. Theory of probability is the underlying theory for all games of chance. In short and without difficult math it is explained below. One of the important steps needed to make when considering the probability of two or more events occurring. It Is to decide whether they are independent or related events.

Mutually Exclusive vs. Independent

It is not uncommon for people to confuse the concepts of mutually exclusive events and independent events.

Mutually exclusive event

If event A happens, then event B cannot, or vice-versa. The two events "it was full moon on Monday" and "it was not a full moon on Monday" are mutually exclusive events. When calculating the probabilities for exclusive events you add the probabilities.

Independent events

The outcome of event A has no effect on the outcome of event B. Such as "It was a full moon on Monday" and "I met accident at work ". When calculating the probabilities for independent events multiply the probabilities. In other word it means what is the chance of both events happening bearing in mind that the two were unrelated.

If A and B events are mutually exclusive, they cannot be independent. If A and B events are independent, they cannot be mutually exclusive.

What happens if we want to throw 1 and 6 in any order?  This now means that we do not mind if the first die is either 1 or 6. But with the first die, if 1 falls uppermost, clearly it rules out the possibility of 6 being uppermost, so the two Outcomes, 1 and 6, are exclusive. One result directly affects the other. In this case, the probability of throwing 1 or 6 with the first die is the sum of the two probabilities, 1/6 + 1/6 = 1/3.
The probability of the second die being favorable is still 1/6 as the second die can only be one specific number, a 6 if the first die is 1, and vice versa.
Therefore the probability of throwing 1 and 6 in any order with two dice is 1/3 x 1/6 = 1/18. Note that we multiplied the last two probabilities, as they were independent of each other.

The probability of throwing a double three with two dice is the result of throwing three with the first die and three with the second die. The total possibilities are, one from six outcomes for the first event and one from six outcomes for the second, Therefore (1/6) * (1/6) = 1/36th or 2.77%.








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