Transcript To see the Craps probability solution go here.

Craps Probability Solution
Or:
More than you ever wanted to
know about conditional probability!
Ways to Win
• There are three ways to win the game:
– Roll a 7 on the comeout roll
– Roll an 11 on the comeout
– Roll a point (4, 5, 6, 8, 9, 10) on the comeout
and roll it again before a 7 comes up
• These three ways are mutually exclusive,
so find each probability and add to get the
overall probability of winning.
– This is the “disjoint or” case
Winning on the Comeout Roll
• There are 6 ways to roll a 7 out of 36 possible
outcomes, so P(7) = 6/36
• There are 2 ways to roll an 11, so P(11) = 2/36
• So P(win on comeout) = 6/36 + 2/36 = 8/36
Make a Point and Win
• Consider the probabilities involved in rolling
a 4 on the comeout and then rolling another
4 to win.
• Since this is an “independent and” problem,
we multiply the appropriate probabilities:
– P(4 AND 4 before 7) = P(4) x P(4 before 7)
• P(4) is easy: there are 3 ways to roll a 4, so
P(4) = 3/36
• Now find P(4 before 7) and multiply by 3/36
– That will give P(win by making a point of 4)
– We will then do the same for the other 5 points
P(4 before 7)
•
There are three game-relevant outcomes on
each roll:
1. Roll another 4 and win: P(4) = 3/36
2. Roll a 7 and lose: P(7) = 6/36
3. Roll any other number: P(not 4 or 7) = 27/36
•
•
•
1 – (3/36 + 6/36) = 27/36
We are finding P(win), so assume we roll
anything but a 7.
That means we win on the first roll (after the
comeout) OR the second roll OR the third roll…
etc.
– Since these are disjoint events, we add probabilities
P(4 before 7) Continued
• The math we just described looks like this:
3 27 3 27 27 3


...
36 36 36 36 36 36
• But that is just an infinite series with a first
term of 3/36 and a common ratio of 27/36
• Recall from your study of Sequences and
Series that there is a formula for adding
a
such a series: S  1  r
• So the sum (and the probability we want) is
1
n
3
3
3
36
 36 
27
9
9
1
36
36
A Simpler Approach
• Notice that after we established the point of 4,
we no longer cared about any outcome other
than 4 or 7
• The basic definition of probability is the number
of ways to succeed divided by the number of
possible outcomes (ways to succeed plus ways
to fail)
• If we ignore the non-4’s and non-7’s, then there
are three ways to succeed (roll a 4) and 6 ways
to fail (roll a 7), so P(4) = 3/(3+6) = 3/9
– That’s the same result we got with so much effort!
P(4 before 7) Concluded
• So we have found that P(4 before 7) = 3/9
• Recall that we know P(4) = 3/36 and we
were trying to evaluate:
– P(4 AND 4 before 7) = P(4) x P(4 before 7)
• So the probability of rolling a 4 and then
making another 4 before rolling a 7 is
3 3 1

36 9 36
• Now we need the same calculation for the
other 5 possible points
Probabilities of Other Points
• P(5) x P(5 before 7) = 4/36 x 4/10
– There are 4 ways to roll a 5 and 6 ways to roll
a7
• P(6) x P(6 before 7) = 5/36 x 5/11
• Then we can argue by symmetry that:
– P(8) x (8 before 7) = 5/36 x 5/11
– P(9) x (9 before 7) = 4/36 x 4/10
– P(10) x (10 before 7) = 3/36 x 3/9
Putting It All Together
• Now let’s add the probabilities of the
following 8 disjoint events in Craps:
– Roll a 7
– Roll an 11
– Roll a 4 and another 4
– Roll a 5 and another 5
– Roll a 6 and another 6
– Roll a 8 and another 8
– Roll a 9 and another 9
– Roll a 10 and another 10
6/36
2/36
3/36 x 3/9
4/36 x 4/10
5/36 x 5/11
5/36 x 5/11
4/36 x 4/10
3/36 x 3/9
• Add all these terms to get P(win) = .493
The Only Fair Bet in the House!
• So your probability of winning at Craps is 49.3% and the house has a
50.7% chance.
– The 1.4% edge they enjoy is all they need to get rich because of the huge
volume of play
• There’s one more bet (that they DON’T advertise) that actually pays
mathematically true odds!
• After the comeout roll and you have established a point (assume you
did not win or lose on the comeout), the odds against making the point
come from the probabilities we just found.
– For a 4 or 10, P(win) = 3/9 or 1/3, so odds are 2:1 against you
– For a 5 or 9, P(win) = 4/10 or 2/5, so odds are 3:2 against you
– For a 6 or 10, P(win) = 5/11, so odds are 6:5 against you
• They will now allow you to make a side bet (called “taking odds”) that
pays properly if you win:
– For a 4 or 10, bet an extra $1 (or multiple); they pay $2 if you win (and give
you back your $1)
– For a 5 or 9, bet an extra $2; they pay $3 if you win (and give back)
– For a 6 or 8, bet an extra $5; they pay $6 if you win (and give back)