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### Odds of Winning in Dice Game of Craps

```Date: 02/20/2007 at 08:47:58
From: Ekal
Subject: Independent repeated trials

The dice game Craps is played as follows.  The player throws two dice,
and 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).

Calculate the probability that the player wins.  Isn't the process
infinite?  Then how to find the answer?

```

```
Date: 02/20/2007 at 12:12:12
From: Doctor Anthony
Subject: Re: Independent repeated trials

Hi Ekal -

We start by finding the probability of winning at the first roll,
that is obtaining 7 or 11:

Pr(7)
1 6       6/36 = 1/6   probability of getting 7
2 5
3 4      (use fractions rather than decimals)
4 3
5 2
6 1

Pr(11)
5 6       2/36 = 1/18   probability of getting 11
6 5

So the probability of winning on first roll = 1/6 + 1/18 = 2/9

Now we find the probability of losing at the first roll (i.e. getting
2, 3, 12):

Pr(2)           Pr(3)            Pr(12)
1 1 =  1/36     2 1  = 2/36      6 6  = 1/36
1 2

The probability of losing on first roll = 1/36 + 2/36 + 1/36 = 1/9

Then the probability of making a point will be  1 - 2/9 - 1/9 = 2/3

The individual probabilities for the numbers 4, 5, 6, 8, 9, 10 are
shown below.

Pr(4)(3/36 = 1/12)   Pr(5)(4/36 = 1/9)   Pr(6)( = 5/36)
3 1                  4 1                 5 1
2 2                  3 2                 4 2
1 3                  2 3                 3 3
1 4                 2 4
1 5

Pr(8)( = 5/36)       Pr(9)(4/36 = 1/9)   Pr(10) (3/36 = 1/12)
6 2                  6 3                 6 4
5 3                  5 4                 5 5
4 4                  4 5                 4 6
3 5                  3 6
2 6

P(4u5u6u8u9u10) = 3/36 + 4/36 + 5/36 + 5/36 + 4/36 + 3/36

=  24/36

=  2/3   (as we have already shown above)

This is the probability of making a point.

We now consider the individual probabilities depending on which
'point' has been obtained.  (E.G., probability of getting 4 as the
'point' = 1/12)

I shall work through this problem in detail and the result can then
be applied to the various 'point' probabilities.

The probability of gaining 4 as the 'point'   = 1/12

"         getting 7                  = 1/6

"  game continuing  = 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) + ..... to infinity

= (1/12)[1 + (3/4) + (3/4)^2 + ..... to infinity]

= (1/12)[1/(1 - 3/4)]  = (1/12)(4) =  1/3

So the total probability of gaining the 'point' 4 and then winning

= (1/12) (1/3)

This will be the same probability if the initial 'point' was 10.

If the 'point' was 5 or 9 the initial probability of 5 or 9 is
1/9 and the probability of continuing thereafter is 1 - 1/9 - 1/6 =
13/18 and by the same argument as above the probability of winning is

(1/9)[1/(1 - 13/18)] = (1/9)(18/5) = 2/5 and the total probability
is (1/9)(2/5)

If the 'point' is 6 or 8 the probability of continuing is 1 - 5/36 -
1/6 = 25/36  and the probability of winning is

(5/36)[1/(1 - 25/36)]  = (5/36)(36/11) = 5/11 and the total
probability is (5/36)(5/11)

We can now find the overall total probability of winning

= 2/9 + 2[(1/12)(1/3) + (1/9)(2/5) + (5/36)(5/11)]

= 244/495  =  0.493

which as you would expect will be less than 1/2 so that over the
long run you will lose to the bank.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability
High School Probability

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