\nthis is a question for Skinny:<\/p>\n
\tbased on the following facts: 1. odds of a seven 1 in 6 2. each roll of he dice is an independent trial – the past or future has no affect<\/p>\n
\tso how can five count be proven mathematically? fact number 2 negates the five count <\/p>\n
\tplease elaborate<\/p>\n
Posted by:<\/strong> Set44 on January 21, 2018, 4:23 am<\/p>\n Posted by:<\/strong> Skinny on January 21, 2018, 5:56 am<\/p>\n \tYou asked a good question. I will explain the answer in two ways. First I will give you another example that has the same criteria you list and show how a similar premise can be proven mathematically despite having the two facts that you listed. Second I will give you a link to an article that shows how Dr. Catlin proved the 5-count mathematically.<\/p>\n \tIn my first example I will discuss tossing a fair coin with heads on one side and tails on the other. I think you will agree that the odds of throwing a head is 1 in 2. Thus a coin toss satisfies both of your facts.<\/p>\n \t1. odds of a head is 1 in 2. Each toss of the coin is an independent trial – the past or future has no effect.<\/p>\n \tNow if I toss the coin twice in a row I can expect one of the four following results: HH, TH, HT, TT<\/p>\n \tIf I were to toss the coin twice in a row 1,000 times I can expect to see two heads (HH) approximately 250 times.<\/p>\n \tAll I have done is used both facts to determine the probability of throwing 2 heads in a row.<\/p>\n \tThe five count does the same thing for craps. It determines the probability of a random roller sevening out before completing the five count. The five count is using the probabilities of tossing 2 fair die by a random roller. But because it is a bit more complicated than my simple coin toss example, Dr. Don Catlin set up a computer simulation to do the calculation.<\/p>\n \tDr. Catlin did a computer simulation of 200 million random shooters placing 6 and 8 for $6 each. One player, called the "Bet-All" player, obviously bet on every shooter who rolled the dice. The 5-Counter only placed the 6 and 8 after the shooter made it through the 5-Count. The 5-Count eliminated approximately 57 percent of the random shooters. During the period of play the 5-count shooter lost about 57% less than the "Bet All" player.<\/p>\n