Question 798500: ‘Craps’ is a game where two fair die are rolled and the sum of the uppermost faces is observed. The game starts on the first roll, where a sum of 7 or 11 produces a win; a sum of 2, 3, or 12 produces a loss. If the sum is anything else, the die are rolled until either the ‘anything else’ appears again (a win) or a sum of seven is rolled (a lose). For example, if the first toss produces a sum of four, the die is tossed continuously until either a sum of 4 is observed (win) or a sum of seven (loss). The chance of winning a game of ‘Craps’ is 0.4929, or 49.29%. A gambler is to play ‘Craps’ 10 times. What is the chance that he/she will win six games of the ten? 4 digits after the decimal.
I went (0.4929)^6 x (1-0.4929)^4 which is 0.0009. I only have two tries on this assignment and they take the average of the 2 tries and I'm not too sure if I'm correct or not...
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! You are missing one factor, a coefficient in the expansion of  . I'll explain so you understand why a crazy formula is applied. Once you understand, it will be easy to remember where the crazy formula comes from, and you will know how to calculate without memorizing the crazy formula. (Or so I hope).
In one game the probability of winning is  ,
and the probability of loosing is  .
For one game,the probability of either winning or loosing (anythingthat happens is cool, as long as a game is played) is
 .
For 2 games, the probability of an outcome of any sort happening is
where  is the probability of winning both times,
 is the probability of loosing both times,
and  is the probability of winning once and losing once.
The coefficient 2 comes from the 2 equal terms that you get when you multiply
I made them different colors (red for first game, green for second) so that you could tell them apart, but the products highlighted are equal, so we add them up as , and that is where the coefficient 2 comes from.
For 3 games, you would have
For 4 games, you would have
For 10 games, the probability of an outcome happening is
  =   .
The terms in the middle (all except the first and last one) have coefficients that reflect the number of combinations (products) that result from picking the first term in some of the ten  factors and the second term in the rest of the factors.
The term with  has the coefficient  ,
and  is the probability of winning 6 and losing 4 of the 10 games.
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