Craps Probability Solution

1
Craps Probability Solution

• Or
• More than you ever wanted to know about
  conditional probability!

2
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

3
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

4
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

5
P(4 before 7)

• There are three game-relevant outcomes on each
  roll
• Roll another 4 and win P(4) 3/36
• Roll a 7 and lose P(7) 6/36
• 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

6
P(4 before 7) Continued

• The math we just described looks like this
• 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 such a series
• So the sum (and the probability we want) is

7
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-4s and non-7s, then there
  are three ways to succeed (roll a 4) and 6 ways
  to fail (roll a 7), so P(4) 3/(36) 3/9
• Thats the same result we got with so much effort!

8
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
• Now we need the same calculation for the other 5
  possible points

9
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 a
  7
• 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

10
Putting It All Together

• Now lets add the probabilities of the following
  8 disjoint events in Craps
• Roll a 7 6/36
• Roll an 11 2/36
• Roll a 4 and another 4 3/36 x 3/9
• Roll a 5 and another 5 4/36 x 4/10
• Roll a 6 and another 6 5/36 x 5/11
• Roll a 8 and another 8 5/36 x 5/11
• Roll a 9 and another 9 4/36 x 4/10
• Roll a 10 and another 10 3/36 x 3/9
• Add all these terms to get P(win) .493

11
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
• Theres one more bet (that they DONT 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
  21 against you
• For a 5 or 9, P(win) 4/10 or 2/5, so odds are
  32 against you
• For a 6 or 10, P(win) 5/11, so odds are 65
  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)