Suppose you are at a casino and want to know which game has the
highest probability of winning. It's actually the dice game Craps, which
is played as follows.
The player throws two dice. If the sum is 7 or 11, then he wins. If the sum is 2, 3 or 12,
then he loses. If the sum is anything else, then he continues
throwing until he either throws that number again (in which case he
wins) or he throws a 7 (in which case he loses).
To find the probability of winning, first finding the
probability of winning on the first roll by getting a 7 or 11.
The number of ways to get a sum of 7 on two dice is
(1,6), (6,1), (2,5), (5,2), (3,4), (4,3) which is 6.
The number of possible outcomes when rolling two dice is 36.
Therefore the probability of getting 7 is 1/6.
There are 2 ways in which get get a sum of 11, (6,5) and (5,6).
Therefore the probability of getting 11 is 2/36 = 1/18.
The probability of winning on first roll = 1/6 + 1/18 = 2/9
Next we find the probability of losing on the first roll, which is getting the sum
of 2,3 or 12 on two dice.
There is only 1 way to get a sum of 2 on a pair of dice, (1,1) so the probability
of getting a 2 is 1/36.
There are 2 ways to get a sum of 3 on a pair of dice (1,2), (2,1), so the
probability of getting a 3 is 1/18.
There is only 1 way to get a sum of 12 on a pair of dice (6,6), so the probabiliy
of getting a 12 is 1/36.
Therefore, the probability of losing on the first roll is 1/36 + 1/36 + 1/18 = 1/9.
The probability of rolling a sum of 4,5,6,8,9 or 10 which enables you to continue
rolling until you either win or lose is 1  2/9  1/9 = 2/3.
You can also list the probability of obtaining each sum and adding, but there
is no need to do this. We simply subtract the probabilities of obtaining all
other possible sums (2,3,7,11,12) from 1.
Now focus on the individual probabilities depending on which sum has
been obtained which enables you to keep rolling until you win or lose.
The probability of getting 4 as is = 1/12 since there are 3 ways (1,3), (3,1)
(2,2) of getting a 4 and 36 possible outcomes.
The probability of getting 7 = 1/6
So the probability that the game continues = 1  1/12  1/6 = 3/4
So the probability of winning in this case is
(1/12) + (3/4)(1/12) + (3/4)^2(1/12) + .....
= (1/12)[1 + (3/4) + (3/4)^2 + ..... to infinity] (factor out 1/12)
= (1/12)[1/(1  3/4)] = (1/12)(4) = 1/3
Therefore the probability of getting a 4 and then eventually winning is
= (1/12) (1/3) = 1/36.
Notice that this is the same as the probability of winning when rolling
a 10 first because getting a 10 is equally likely as getting a 4 (3 possible
outcomes, (4,6), (6,4), (5,5).
Similarly the probability of getting a 5 or 9 is the same and is
1/9 since there are 4 possible ways to get a 5 and 4 possible ways to get
a 9 ( (1,4), (4,1), (2,3), (3,2) and (4,5), (5,4), (3,6), (6,3), respectively))
The probability of continuing after is 1  1/9  1/6 = 13/18.
Using the same reasoning as above the probability of winning is
(1/9)[1/(1  13/18)] = (1/9)(18/5) = 2/5.
The probability of getting a 5 or 9 and eventually winning is
is (1/9)(2/5) = 2/45
The probability of getting a sum of 6 or 8
is 5/36 ( (1,5), (5,1), (2,4), (4,2), (3,3) and (2,6), (6,2), (3,5), (5,3),
(4,4) respectively )
The probability of continuing after is 1  5/36  1/6 = 25/36
The probability of winning is
(5/36)[1/(1  25/36)] = (5/36)(36/11) = 5/11
The probability of getting a 6 or 8 and eventually winning is
(5/36)(5/11)= 25/396
Since we went through all the possibilities, the total probability of winning
= 2/9 + 2[1/36 + 2/45 + 25/396]
(probability of winning on 1st roll + probability of winning rolling 4 or 10
first + probability of winning rolling 5 or 9 first + probability of winning
rolling 6 or 8 first)
= 0.493
So even the game with the best chance of winning, you will lose slightly more often
than not.
