Sample space (for 2 dice)
Let’s consider the ‘event’ A of getting the sum of the dice to add up to 7.
There are 6 combinations that each add up to 7, so we can describe the event by
A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
and ring it on the diagram. The probability of A is 6 chances in 36, which can
be written Pr(A) = 6/36 = 1/6. If we let B be the event of getting the sum on the
dice equal to 11 we have the event B = {(5,6), (6,5)} and Pr(B) = 2/36 = 1/18.
In the dice game craps, in which two dice are thrown on a table, you can win
or lose at the first stage, but for some scores all is not lost and you can go onto a
second stage. You win at the first throw if either the event A or B occurs – this is
called a ‘natural’. The probability of a natural is obtained by adding the individual
probabilities, 6/36 + 2/36 = 8/36. You lose at the first stage if you throw a 2, 3
or a 12 (this is called ‘craps’). A calculation like that above gives the probability of
losing at the first stage as 4/36. If a sum of either 4, 5, 6, 8, 9 or 10 is thrown,
you go onto a second stage and the probability of doing this is 24/36 = 2/3.
In the gaming world of casinos the probabilities are written as odds. In craps,
for every 36 games you play, on average you will win at the first throw 8 times
and not win 28 times so the odds against winning on the first throw are 28 to 8,
which is the same as 3.5 to 1.
marcin
(Marcin)
#1