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: 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

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-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!

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 let’s add the probabilities of the following 8 disjoint events in Craps: –Roll a 76/36 –Roll an 112/36 –Roll a 4 and another 43/36 x 3/9 –Roll a 5 and another 54/36 x 4/10 –Roll a 6 and another 65/36 x 5/11 –Roll a 8 and another 85/36 x 5/11 –Roll a 9 and another 94/36 x 4/10 –Roll a 10 and another 103/36 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 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)