 Some Notation
 Outcomes and Events
 Odds
 Expectation
 Fairness and Bias
 House Edge
 Dice Outcomes
 Determining the Nbr of Ways to Roll Any Particular Nbr
 The Sum Rule
 Extending the Sum Rule
 Independent Events and the Product Rule
 Conditional Probability
 Another Look at Independence
 The Odds Against Making a Point
 Probability of Winning a Pass Line Bet
 Tree For Pass Line Bet
 Geometric Progressions
 Two Views of the DON’T PASS Bet
 DON’T PASS — View 1 — Calculating The Probability to Win
 DON’T PASS — View 2 — Calculating The Probability to Win
 Making Free Odds Bets Does Not Change the Expectation
 Combining Pass Line Bets With Free Odds Bets
 Some Example Cases of Pass Line Bets Combined With Free Odds Bets
 View #1 of Combining DON’T PASS Bets With Free Odds Bets
 The Side Bets in Craps
 Place Bets and Buy Bets
 Lay Bets
 Place To Lose Bets
 Big Six
 Big Eight
 The Horn Bet
 The Field Bet
 The Proposition Bets
 Any Craps
 CrapsEleven ( C & E )
 Craps 2
 Craps 12
 Big Red [ also called Any 7 ]
 Hard 4
 Hard 10
 Hard 6
 Hard 8
 Items The Shooter’s Baby Might Need
 Additional Exercises
SOME NOTATION
Symbol  Meaning  Example 

→  “implies or if … then  “x is a horse” → “x has 4 legs” 
↔  “equivalence” or “if and only if”  x + 1 = 5 ↔ x = 4 
 x   the absolute value of x  x = – 6 →  x  = + 6 
XY or X * Y  X times Y  2 * 3 = 6 
x ** y  raise x to the y power  4 ** 3 = 4*4*4 = 64 
≈  is approximately equal to  7.999999 ≈ 8 
k !  k factorial = 1 * 2 * 3 * … * k  4! = 1*2*3*4 = 24 
( special case : 0! is defined to be 1 ) 
If x is any number that can be rolled with a pair of dice then
W( x ) = number of Ways to roll the number x with a pair of dice
P( x ) = Probability of rolling the number x
If E is an event then P( E ) = the probability that E occurs.
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An event is a collection of outcomes associated with some activity. The probability of an event is a measure of how likely the event is to occur. The higher the probability of the event, the more likely the event is to occur.
In many practical situations the outcomes that comprise an event are elementary outcomes. These are outcomes that cannot be further subdivided into simpler outcomes. In these cases, the probability of the event can be found by dividing the number of outcomes favorable to the event by the total number of outcomes possible. ( We are assuming that the number of outcomes possible is finite. )
If E is an event whose outcomes are elementary outcomes, then
P( E ) = probability of E = ( nbr of outcomes favorable to E ) / ( total nbr of outcomes possible )
Example
Suppose you shuffle a deck of cards and randomly draw one card.
Let E be the event of drawing an ace. ( A deck has 52 cards, including 4 aces. )
Then
 P( E ) = ( nbr of outcomes favoring ace ) / ( total nbr of outcomes possible )
 = 4 / 52 = 1 / 1
To minimize the chance of making mistakes, you should ensure that the outcomes
you choose to look at are elementary outcomes. The next example illustrates this.
Example
Flip two coins simultaneously.
Let’s say that the result is a match if both tosses yield heads or both tosses yield tails.
We will say that there is a mismatch if one coin is heads and the other is tails.
What is the probability of a mismatch ?
Here is an INCORRECT solution.
There are 3 outcomes possible:
 a) Both coins are heads.
 b) Both coins are tails.
 c) One coin is heads and the other is tails
Hence P( mismatch ) = # outcomes favoring mismatch / total nbr of outcomes = 1 / 3 [ This result is wrong ]
The mistake in the above solution is due to the fact that (c) is not an elementary outcome — there are two ways to get a mismatch. Thus, the (a), (b), and (c) outcomes are not equally likely.
Here is a correct solution.
Imagine that one of the two coins is painted red on its head and tail sides and the other is painted green on both sides. Then the elementary outcomes are given by this table
Red Coin  Green Coin 

heads  heads 
heads  tails 
tails  heads 
tails  tails 
P( mismatch ) = 2 / 4 = 1 / 2 [ correct solution ]
Example
There are 38 numbers on an American roulette wheel. The numbers zero and doublezero are both green. Half of the other numbers are red, and the rest are black. If you bet on black, what is your probability of winning ?
The number of outcomes favoring black is 36 / 2 = 18. The total number of (elementary) outcomes is 38. So, P( win ) = P( black ) = 18 / 38 = 0.47368….
Thus, we’d expect black to win about 47% of the time.
ODDS
Instead of discussing the probability of an event, it is sometimes more convenient to talk about the odds in favor of or the odds against an event. Probabilities and odds are simply two different ways of measuring the same thing, namely, how likely it is for the event in question to occur.
The odds in favor of and the odds against are reciprocally related. For example, if the odds against something are 3 to 2, then the odds in favor of it are 2 to 3. The odds against an event is simply the ratio of unfavorable to favorable outcomes for that event. Here is how odds against are defined:
Let E be any event [ composed of elementary outcomes for some activity ].
Let a be the number of outcomes in favor of E. [ we assume a > 0 ] Let b be the number of outcomes not in favor of E.
Let p be the probability of E.
Let g be the odds against E.
Then g is defined to be the number of unfavorable outcomes
divided by the number of favorable outcomes.
That is, g is defined to be the ratio of b to a ( i.e. g = b / a ).
Since the probability of E is defined to be the number of favorable outcomes
divided by the total number of outcomes possible,
the probability of E is p = a / ( a + b ).
This observation provides us with an easy way to convert odds to probability and viceversa :
Simply notice that 1/p = (a + b) / a = a/a + b/a = 1 + b/a = 1 + g.
Tip: You should memorize the formula 1 / p = 1 + g
Example
What are the odds against drawing an ace from a shuffled deck of cards ?
 Here p = 4 / 52 = 1 / 13 → 1 / p = 13 / 1.
 So 1 / p = 1 + g → 13 = 1 + g → g = 12 [ which equals 12 / 1 ]
Hence, the odds against drawing an ace are twelve to one.
( The odds in favor of drawing an ace are one to twelve. )
Alternative Solution:
 odds against = #unfavorable outcomes / #favorable outcomes = 48 / 4 = 12 = 12 / 1.
 So, the odds against are 12 to 1.
Example
Suppose that 95% of the people who contract a certain disease will eventually recover from it. What are the odds in favor of recovery ?
Let p = probability to recover.
Then p = 95 / 100, so 1 / p = 1 + g → 100 / 95 = 1 + g.
Thus g = ( 100 / 95 ) – 1 = 5 / 95 = 1 / 19.
Since the odds against are 1 / 19, the odds in favor are 19 to 1.
“Odds” Word Usage
The term “Odds” has at least three uses:
 An “odds bet” ( i.e. “free odds bet” ) is a type of craps bet ( described in Part I )
available to line bettors after the shooter establishes a point.  The “odds in favor of” or the “odds against” an event are ways to measure
how likely it is for that event to occur.  The “payoff odds” show the ratio of dollars to be won per dollar bet.
For example, if the payoff odds are 7 to 5, then winners who bet an
amount B are paid ( 7 / 5 ) * B.
EXPECTATION
In statistics courses, students study random variables and their distributions. For any type of random variable [ in engineering, medicine, gambling, or whatever ], the expected value of the random variable can be found by looking at all the possible values the variable can take on, multiplying each value by the probability that the variable takes on that value, and then adding all of those products. You can think of the expected value of a random variable as the “average” value of the random variable.
In casino games, the most important random variable is the amount of money you win for a particular wager.
The expected value of this random variable is called your mathematical expectation [ expectation for short ].
We could use two random variables ( Amount Won and Amount Lost ) to track our gambling results, but instead we will simplify many calculations and avoid the need for an Amount Lost variable by adopting the convention that negative winnings represent losses. For example, if we lose $12 then we say that we won negative twelve dollars.
FAIRNESS AND BIAS
A game is said to be fair if every player’s expectation is zero. If any player’s expectation is positive, then the game is biased in that player’s. Casino games are usually biased in favor of the house; each of the other players has a negative expectation. As long as there is no cheating by the house or by the other players, the house will be in a good position to make big profits; and, the player will lose in the long run.
In any gambling game where you place a bet, there is a simple rule relating odds to bias. The rule says that the amount you stand to win should be at least as large as the product of the amount you bet and the odds against winning. Let’s examine this rule.
 Let B = amount you bet
 Let W = amount you stand to win
 Let g = odds against winning.
The bias rule has two parts.
Bias Rule :
 a) W = Bg → game is fair
 b) W > Bg → game is biased in your favor.
In other words, in order for a game to be either unbiased or biased in your favor, you need to have
W ≥ Bg .
Here’s why:
 “Unbiased” means everybody’s expectation is zero
 “Biased in your favor” means your expectation is positive
Let
 p = probability of winning
 E = expectation = Wp + ( – B ) ( 1 – p ) = Wp – B + Bp
 game not biased against you → E ≥ 0
E ≥ 0
 → Wp – B + Bp ≥ 0
 → Wp ≥ B – Bp = B ( 1 – p )
 → W ≥ B * ( 1 – p ) / p
 → W ≥ B ( ( 1 / p ) – 1 )
But 1 / p = 1 + g
So, E ≥ 0 → W ≥ B ( 1 + g – 1 ) = Bg
So you know the probability of winning then you can see where you stand by multiplying the size of your bet by the odds against you and then seeing how that compares to the potential winnings.
HOUSE EDGE
When you play a casino game, the House Edge ( also called the House Advantage ) is simply the negative of your expectation divided by your bet.
We usually multiply this result by 100% to express it as a percentage.
Note: For any number x, x = x * 1 = x * 100%
 So, multiplying any number by 100% does not change its value.
For example, if the expectation is 0.18 dollars and your bet is two dollars then the house edge is E / B = – (0.18 ) / 2 = 0.18 / 2 = 0.09 = 0.09 * 100% = 9 %
Using symbols we have:
 W = amount you stand to win
 B = amount you bet
 p = probability of winning
 E = expectation = expected value of W = Wp + ( B )( 1 – p )
 H = house edge = ( – E ) / B
If the expectation is a negative number, which is almost always the case, then the formula for H can be rewritten in terms of the absolute value of the expectation :
 H= E / B =  E  / B =  E / B  =  E / B  * 100%
The house edge is a useful tool for comparing one bet against another. The bet having the lower house edge is a better value for the player.
Since expectation is almost always negative, and since a negative win is really a loss, the house edge is answering this question : In the long run, what percentage of your bet should you expect to lose ?
The house edge measures the average amount lost for each dollar bet. For example, if the house edge is 5% then, in the long run, you can expect to lose about one nickel for each dollar that you bet.
DICE OUTCOMES
What is the probability of rolling a 4 with a pair of balanced dice ?
In casino games, dice are usually white; but it is easier to see what is going on if we imagine that one die is painted red and the other green.
Each die comes to rest on one of its six faces. The number opposite this face is added to the corresponding number on the other die to determine the number rolled. The following table shows a systematic listing of the 36 possible outcomes [ some lines are omitted for brevity ]. Saying that the dice are balanced means that we assume each of these 36 elementary outcomes is equally likely to occur.

To find out how many outcomes result in a 4, we could make a complete list ( as outlined above ) and then count the lines where the sum equals 4. There are 3 such lines :
1  3  
2  2  
3  1 
Since there are 3 lines where the outcome is 4, the number of outcomes favoring 4 is 3.
Hence, P( 4 ) = probability of rolling a four = 3 / 36 = 1 / 12.
Example
Find the probability of rolling any particular number with a pair of balanced dice. Using the method employed in earlier example, we could easily find the results summarized in the following table:
Number  # Ways To Roll  Probability 

1  0  0 
2  1  1 / 36 
3  2  2 / 36 
4  3  3 / 36 
5  4  4 / 36 
6  5  5 / 36 
7  6  6 / 36 
8  5  5 / 36 
9  4  4 / 36 
10  3  3 / 36 
11  2  2 / 36 
12  1  1 / 36 
x  0  0 / 36 where x is any integer greater than 12 
DETERMINING THE NBR OF WAYS TO ROLL ANY PARTICULAR NBR
When you go to a casino you probably donâ€™t want to be loaded down with a bunch of pencils, paper, calculators, or other paraphernalia. Most of the craps calculations you need are simple enough that they can be done mentally if you know the right tricks.
In craps games we sometimes need to know how many ways a particular number can be rolled with the dice. In example 8 the first two columns of the table show us that there are 6 ways to roll a 7, 3 ways to roll a 10, etc. But letâ€™s explore a simple method for figuring this out mentally.
Let x be any number that can be rolled with a pair of dice, and
let W( x ) = the number of ways to roll x.
From the table in example 8 we have
x :  2  3  4  5  6  7  8  9  10  11  12 
W(x) :  1  2  3  4  5  6  5  4  3  2  1 
Note the symmetry of the W( x ) values about the point where W( x ) = 6.
Also note that if x ≤ 7, then W( x ) is simply x minus one.
Using these observations, we see that we can mentally compute W( x ) as follows:
step 1
Is x less than or equal to 7 ?
If so, then W( x ) = x – 1, and we are done.
Otherwise, goto step 2.
step 2
Since x > 7, find out how much greater [ by computing x – 7 ] and
goto step 3
step 3
What number is the same distance below 7 as the distance found at step 2 ?
Computing W at this number will give us the same value as W( x ).
Here is an example of using the process.
How many ways can you roll a 9 ? [ i.e. what is W( 9 ) ? ]
step 1
Q) Is 9 less than or equal to 7 ?
A) No. Goto step 2.
step 2
Q) How much above 7 is 9 ?
A) 2. Goto step 3.
step 3
Q) What number is 2 units below 7 ?
A) 5.
Therefore, W( 9 ) = W( 5 ) = 5 – 1 = 4
Better Method :
In March of 2006 I received an email from Mr. Dan Lubin, a dice dealer who uses this simpler way to compute W ( x ):
If x <= 7 then W( x ) = x – 1; otherwise, W( x ) = 13 – x
Where can you apply this ?
Suppose you make a pass line bet and the craps shooter establishes a point of 9. You will then probably want to make a free odds bet. If you donâ€™t remember what the payoff odds are for a point of 9, then you can easily figure it out mentally; and, Mr. Lubin’s formula is one of the tools you might use to do the calculations.
THE SUM RULE
Unless we say otherwise, we will always use the word â€œorâ€ to mean “and/or” [ i.e â€œone or the other or bothâ€]. When events consist of elementary outcomes and we want to combine the events using ANDs and ORs, the following â€œsumâ€ rule is often useful:
sum rule: For events A and B, P( A or B ) = P( A ) + P( B ) – P( A and B )
This makes sense intuitively because if we count up all the outcomes favoring A and then add to that all the outcomes favoring B, we will have counted the outcomes favoring both A and B twice. So subtracting those should give us the number of outcomes favorable to â€œA or Bâ€. Then dividing each item by the total number of outcomes possible gives us the probabilities shown in the sum rule.
Example
When a pair of dice is rolled once, what is the probability of getting 7 or 9 ?
Let S be the event of rolling a 7.
Let N be the event of rolling a 9.
We want to find P( S or N ).
Since it’s impossible to get both a 7 and a 9 on the same roll, P( S and N ) = 0.
Thus,
P( S or N ) = P( S ) + P( N ) – P( S and N )
= ( 6 / 36 ) + ( 4 / 36 ) – 0
= 10 / 36 = 5 / 18
So, the answer is 5 / 18.
EXTENDING THE SUM RULE
Two events are said to be mutually exclusive if they can’t both happen at the same time. For example, if we roll a pair of dice once, then rolling a 7 and rolling a 9 are mutually exclusive events because we can’t roll a 7 and a 9 at the same time. Each roll yields just one outcome.
If A and B are mutually exclusive events then P( A and B ) = 0, and
in this case, the sum rule says :
P( A or B ) = P( A ) + P( B )
If the 3 events A, B, and C are pairwise mutually exclusive then
P( A or B or C ) = P( A ) + P( B ) + P( C )
This idea extends to any finite number of events.
Given n events A_{1}, A_{2}, …, A_{n} that are pairwise mutually exclusive, we have this rule:
P( A_{1} or A_{2} or … or A_{n} ) = P( A_{1} ) + P( A_{2} ) + … + P( A_{n} ).
Example
What is the probability that the come out role in a craps game will result in a point being established by the shooter ?
We want to find the probability that the come out roll is one of these 6 numbers: 4 5 6 8 9 10.
Let C4 be the event that the come out is a 4
Let C5 be the event that the come out is a 5
Let C6 be the event that the come out is a 6
Let C8 be the event that the come out is a 8
Let C9 be the event that the come out is a 9
Let C10 be the event that the come out is a 10
Our problem is to find P( C4 or C5 or C6 or C8 or C9 or C10 ); and, since the six events are pairwise mutually exclusive, the answer is
P( C4 ) + P( C5 ) + P( C6 ) + P( C8 ) + P( C9 ) + P( C10 ).
Note that the last 3 terms in the above sum have values matching the first 3:
P( C4 ) = W( 4 ) / 36 = W( 10 ) / 36 = P( C10 )
P( C5 ) = W( 5 ) / 36 = W( 9 ) / 36 = P( C9 )
P( C6 ) = W( 6 ) / 36 = W( 8 ) / 36 = P( C8 )
So the answer is 2 * [ P( C4 ) + P( C5 ) + P( C6 ) ] = 2 * [ ( 3 / 36 ) + ( 4 / 36 ) + ( 5 / 36 ) ] = 2 * ( 1 / 3 ) = 2 / 3
The shooter has a 2 / 3 probability of establishing a point.
( We will use this fact later in the section named Combining Pass Line Bets With Free Odds Bets. )
INDEPENDENT EVENTS AND THE PRODUCT RULE
Two events can be either independent or dependent. Being independent means that the occurrence or nonoccurrence of one event does not change the probability of occurrence of the other event. Mutually exclusive events are never independent; they are always dependent.
For example, when you roll a pair of dice once, one outcome is to get a 7 and another is to get a 9. Each of these events does depend on the other. If, for example, we know that you rolled a 7, then we know that you did not roll a 9. That is, if P( 7 ) = 1, then P( 9 ) must be zero rather than 4 / 36. Thus, the events â€œrolling a 7â€ and â€œrolling a 9â€ are dependent.
For another example, suppose you flip a coin twice. The outcome ( heads or tails ) on the second flip is independent of the outcome on the first flip. In particular, getting heads on the 1st flip and getting tails on the 2nd flip are independent but not mutually exclusive events.
Independence is an important concept because of this fact:
Two events are independent if and only if the probability that they both occur equals the product of their probabilities.
That is, for events A and B [ comprising elementary outcomes ],
A is independent of B ↔ P( A and B ) = P( A ) * P( B )
If we know that two events are independent then we can find the probability that they both occur by multiplying their probabilities.
Example
When rolling dice repeatedly, what is the probability of rolling a 7 twice in a row ?
Our intuition tells us that the dice donâ€™t remember the outcome of any roll. So getting a 7 on the 1st roll should have no effect on the probability of what we get on the 2nd roll. Getting 7 on the first roll and getting 7 on the second roll are independent events.
 Let S1 be the event of rolling a 7 on the 1st roll.
 Let S2 be the event of rolling a 7 on the 2nd roll.
Since S1 and S2 are independent,
 P( S1 and S2 ) = P( S1 ) * P( S2 ) = ( 1 / 6 ) * ( 1 / 6 ) = 1 / 36.
The answer is 1 / 36.
To help us compute the odds against making various points in a craps game we need to know a little bit about conditional probability.
CONDITIONAL PROBABILITY
The probability of an event can change if we are given new information that we didnâ€™t know before. For example, suppose that we again consider the probability of drawing an ace from a deck of cards. Suppose we happened to notice that the card at the bottom of the deck is the four of spades, and we know we will not choose this card. Now the probability of drawing an ace is no longer 4 / 52. The probability of drawing an ace, given that the bottom card can be ignored is 4 / 51. The new information has slightly reduced the odds against drawing an ace.
There is a formula that is sometimes useful for computing conditional probability.
In symbols it looks like this: P ( A  B ) = P( A and B ) / P( B )
Here A and B are events, and we read A  B as A given B.
The formula says that the probability of event A, given that event B has happened ( or will happen ) can be found by first finding the probability of A and B and then dividing that by the probability of B.
P( A  B ) = P( A and B ) / P( B ) [ independence doesn’t matter here ]
Example
Suppose we roll a pair of dice repeatedly until we get either 7 or 9. What is the probability that the 9 will appear before the 7 ?
( For example, if a craps shooterâ€™s point is 9, what is the probability that he will win his pass line bet ? )
 Let N = the event of rolling a 9
 Let E = the event of rolling either 7 or 9.
The event rolling 9 before 7 means rolling 9, given that we rolled either 7 or 9.
Thus our problem is to find the probability of N given E.
P( E ) = P( 7 or 9 ) = ( 6 / 36 ) + ( 4 / 36 ) = 5 / 18.
Since getting a 9 and getting a 7 or 9 can both happen only by getting a 9,
P( N and E ) = P( 9 ) = 4 / 36 = 1 / 9.
P( N  E ) = P( N and E ) / P( E )
= ( 1 / 9 ) / ( 5 / 18 )
= ( 1 / 9 ) * ( 18 / 5 ) = 2 / 5.
The answer is 2 / 5.
Exercise
In a craps game what are the odds against rolling a 9 before a 7 ?
Let p = probability of rolling 9 before 7.
From example earlier, p = 2 / 5.
Using 1/p = 1 + g → 5 / 2 = 1 + g → g = 3 / 2
Thus, for a pass line bettor whose point is 9, the odds against winning are three to two.
ANOTHER LOOK AT INDEPENDENCE
Earlier we said that two events are independent if the occurrence or nonoccurrence of one does not effect the probability of the other. Using conditional probability we can restate this more precisely.
If the probability of event A is not changed by knowing that event B has already occurred ( or will occur ) then A is independent of B. So, independence means that P( A ) = P( A  B ).
If A and B are events with nonzero probabilities then the following 5 statements are equivalent:
 A is independent of B
 B is independent of A
 P( A ) = P( A  B )
 P( B ) = P( B  A )
 P( A and B ) = P( A ) * P( B )
If A and B are not independent then they are dependent.
THE ODDS AGAINST MAKING A POINT
The following table shows the odds against winning a pass line bet when the shooter has established a point
Point  Probability To Win  Odds Against Winning 

4 or 10  1 / 3  2 to 1 
5 or 9  2 / 5  3 to 2 
6 or 8  5 / 11  6 to 5 
If you are going to play craps then you should memorize the odds stated in the above table. But if you should happen to forget those values then the following process can be used to reconstruct them. ( After a little practice you could easily do the key steps of these 3 cases in your head. )
Case 1 — The point is 4
When the point is 4, the probability for the shooter to win a pass line bet is the probability to roll a 4 before a 7.
P( win )  =  P( 4 before 7 ) = P( 4  (4 or 7) ) 
=  P( 4 and (4 or 7) ) / P( 4 or 7 )  
=  P( 4 ) / ( P(4) + P(7) ) 
But P( 4 ) = W( 4 ) / 36
and P( 7 ) = W( 7 ) / 36
Thus P( 4 ) / ( P(4) + P(7) ) = W( 4 ) / ( W(4) + W(7) )
P( win ) = 3 / ( 3 + 6 ) = 1 / 3
1 / p = 1 + g → 3 = 1 + g → g = 2 = 2 / 1
Since W( 4 ) = W( 10 ), the same results apply to a point of 10.
Case 2 — The point is 5 ( or 9 )
P( win ) = W( 5 ) / ( W( 5 ) + W( 7 ) ) = 4 / ( 4 + 6 ) = 2 / 5
1 / p = 1 + g → 5 / 2 = 1 + g → g = 3 / 2
Case 3 — The point is 6 ( or 8 )
P( win ) = W( 6 ) / ( W( 6 ) + W( 7 ) ) = 5 / ( 5 + 6 ) = 5 / 11
1 / p = 1 + g → 11 / 5 = 1 + g → g = 6 / 5
PROBABILITY OF WINNING A PASS LINE BET
What is the probability for the shooter to win a pass line bet in a craps game ? The shooter must either roll a natural or roll a point and then make that point.
P( win ) = P( natural ) + P( win on a point )
P( natural ) = P( 7 or 11 )
 = P( 7 ) + P( 11 )
 = W( 7 ) / 36 + W( 11 ) / 36
 = ( 6 / 36 ) + ( 2 / 36 ) = 8 / 36 = 2 / 9
To help us find P( win on a point ), let’s define some helper events:
 A1 = the event of getting a point of 4 on the come out roll
 A2 = the event of rolling a 4 before a 7 after the come out roll
 B1 = the event of getting a point of 5 on the come out roll
 B2 = the event of rolling a 5 before a 7 after the come out roll
 C1 = the event of getting a point of 6 on the come out roll
 C2 = the event of rolling a 6 before a 7 after the come out roll
 D1 = the event of getting a point of 8 on the come out roll
 D2 = the event of rolling an 8 before a 7 after the come out roll
 E1 = the event of getting a point of 9 on the come out roll
 E2 = the event of rolling a 9 before a 7 after the come out roll
 F1 = the event of getting a point of 10 on the come out roll
 F2 = the event of rolling a 10 before a 7 after the come out roll
We can cut our work in half by recognizing these duplications
 W( 8 ) = W( 6 ) → P( D1 and D2 ) = P( C1 and C2 )
 W( 9 ) = W( 5 ) → P( E1 and E2 ) = P( B1 and B2 )
 W( 10 ) = W( 4 ) → P( F1 and F2 ) = P( A1 and A2 )
 P( win with a point ) = P( A1 and A2 ) + P( B1 and B2 ) + … + P( F1 and F2 )
 = 2 * [ P( A1 and A2 ) + P( B1 and B2 ) + P( C1 and C2 )
The probability of getting a 4 on ANY roll is W( 4 ) / 36, therefore P( A1 ) = 3 / 36.
 P( A2 ) = P( 4 before 7 ) = W( 4 ) / ( W(4) + W(7) ) = 3 / ( 3 + 6 ) = 1 / 3.
Since the events A1 and A2 occur on different rolls of the dice, they are obviously independent events; and so,
 P( A1 and A2 ) = P( A1 ) * P( A2 )
 = ( 1 / 12 ) * ( 1 / 3 )
 = ( 1 / 36 )
Similarly,
 P( B1 ) = 4 / 36 = 1 / 9
 P( B2 ) = P( 5 before 7 ) = 4 / ( 4 + 6 ) = 2 / 5
 P( B1 and B2 ) = ( 1 / 9 ) * ( 2 / 5 ) = 2 / 45
 P( C1 ) = 5 / 36
 P( C2 ) = P( 6 before 7 ) = 5 / ( 5 + 6 ) = 5 / 11
 P( C1 and C2 ) = ( 5 / 36 ) * ( 5 / 11 ) = 25 / 396 )
 P( win with a point ) = 2 * [ ( 1 / 36 ) + ( 2 / 45 ) + ( 25 / 396 ) ]
A common denominator for the right hand side is 1980.
 P( win with a point ) = 2 * [ ( 55 / 1980 ) + ( 88 / 1980 ) + ( 125 / 1980 )
 = 2 * [ 268 / 1980 ]
 P( win ) = P( natural ) + P( win on a point ) = ( 2 / 9 ) + ( 268 / 990 )
 P( win ) = ( 220 + 268 ) / 990 = 488 / 990 = 244 / 495
TREE FOR PASS LINE BET
Some people find it helpful to draw a “tree” when calculating the probability of an event composed of several parts. The following image shows a tree that can help us calculate the probability of winning a pass line bet in craps. Here W stands for “Win”, L stands for “Lose”; and, the fractions in the light blue boxes attached to the branch segments are probabilities.
For each branch that leads to a desired outcome ( W in this case ), you simply multiply the probabilities attached to the segments of that branch, and then add all the products obtained ( as shown in the big green box at the bottom of the tree ).
Exercise
Show that in craps the house edge for a pass line bet is about 1.41%
 Let B = the amount bet ( in dollars ) on the pass line
 Let p = P( win ) = 244 / 495 [ shown in an earlier section ]
 Let E = expectation = B * p + ( – B ) * ( 1 – p ) = Bp – B + Bp
 = 2Bp – B = B * ( 1 + 2p ) → E / B = 1 + 2p
 H =  E / B  =  1 + 2p  =  1 + ( 488 / 495) 
 =  – 7 / 495  = 1.41414…. %
 » 1.41 %
Exercise
What would the payoff odds have to be for a pass line bet in order for the bet to be totally fair with no house edge ?
A fair game is one where every player’s expectation is zero.
Let p = probability to win = 244 / 495 [ shown previously ]
Then E = ( kB ) * p + ( B ) * ( 1 – p ) = kBp – B + Bp
E = 0 → kBp = B – Bp = B * ( 1 – p ) → k = ( 1 – p ) / p = ( 1 / p ) – 1
We need k = 251 / 244 = 1.02868852….
The payoff odds would have to be about 1.03 to 1.
Exercise
Show that about 45% of the winning pass line bets are made on the come out roll. Hint: P( natural  win ) = ?
Let p = P( win ) = 244 / 495
Let natural = The event of rolling 7 or 11
The only way you can win on the come out is to roll a natural.
So, P( win on come out  win ) = P( natural  win )
 P( natural  win ) = P( natural and win ) / P( win )
 = P( natural ) / P( win )
 = ( 8 / 36 ) / ( 244 / 495 )
 = ( 2 / 9 ) * ( 495 / 244 )
 = 55 / 122 ≈ 45.1%
So, given that you won, the probability that you won on the come out is about 45%.
Exercise
Suppose that the rules were changed so that the don’t pass bet would be the exact opposite of the pass line bet. What would be the probability to win a don’t pass bet if rolling a 12 on the come out roll were a win instead of a push ?
 Let p = P( win )
 Let A = the event of rolling a 2, 3, or 12 on the come out
 p = P( A ) + 2 * [ P( 4 ) * P( 7 before 4 ) + P( 5 ) * P( 7 before 5 ) + P( 6 ) * P( 7 before 6 ) ]
 = ( 4 / 36 ) + 2 [ ( 3 / 36 ) * ( 6 / 9 ) + ( 4 / 36 ) * ( 6 / 10 ) + ( 5 / 36 ) * ( 6 / 11 ) ]
 = ( 1 / 9 ) + 2 * ( 1 / 36 ) * [ 2 + ( 24 / 10 ) + ( 30 / 11 ) ]
 = 251 / 495 = 0.5070707….
Since p > 0.5, this game would be biased in the player’s favor.
Before we compute the probability of winning a donâ€™t pass bet we should review geometric progressions. If you skipped algebra II in high school, then the next section might be an introduction rather than a review.
GEOMETRIC PROGRESSIONS
The study of geometric progressions has several applications in mathematics. We will use this subject to help us find the probability of winning a DON’T PASS bet in craps.
A geometric progression is a sequence of numbers created by these rules:
First, choose a starting number a and a multipler r .
( Both a and r can be any two nonzero real numbers. )
 The 1st term in the sequence is a [ 1st term = a ]
 The 2nd term is r times the first term [ 2nd term = r * a ]
 The 3rd term is r times the second term [ 3rd term = r * r * a ]
Continue this process as follows:
To get the next term in the sequence, you simply multiply the previous term by r.
For example, choose a = 0.3 and r = 0.1
Then the sequence is 0.3 0.03 0.003 0.0003 0.00003 ….
The special cases in which r equals zero or one are of no practical use,
so we will always assume that r is neither zero nor one.
In fact, for our applications we will always assume that 0 < r < 1.
There is a simple trick for finding the sum of the first n terms of the progression.
Let S(n) = sum of the first n terms
S(n) = a + a*r + a*(r^{2}) + a*(r^{3}) + a*(r^{4}) + … + a*( r^{n – 1})
Now we will multiply both sides by r and subtract the 2nd equation from the first.
Most of the terms will cancel each other out.
 S(n) = a + a*r + a*r2 + a*r3 + a*r4 + … + a*r(n – 1)
 r * S(n) = a*r + a*r^{2} + a*r^{3} + a*r^{4} + … + a*r^{(n1)} + a*r^{n}
subtracting gives S(n) – r * S(n) = a – a*(r^{n})
Then factoring both sides gives ( 1 – r ) * S(n) = a * ( 1 – r^{n} )
Since r < 1, 1 – r does not equal zero; and so, we can divide both sides by ( 1 – r ).
r /= 1 → S(n) = a ( 1 – r^{n} ) / ( 1 – r )
Most applications ( including ours ) are only interested in the case where r is between zero and one, and n is very large. In that case, r^{n} approaches zero as n approaches infinity and we get :
0 < r < 1 → 
 [ Formula for later use ] 
Long ago, many people believed that any time you add up an infinite number of terms the result would have to be infinitely large.
The above result shows that an infinite sum ( which we represented by “Ssubinfinity” ) can, in fact, be finite. Applying the formula to the example that preceded it, we have
0.3 + 0.03 + 0.003 + 0.0003 + ….  =  0.3 / ( 1 – 0.1 )  
=  0.3 / 0.9 = 1 / 3 
This confirms something you probably already knew: 0.3333333333… is exactly one third.
In later sections [ e.g. just before exercise 14 ], we will apply this useful formula for infinite geometric progressions :
S_{ } = a + ar + ar^{2} + ar^{3} + …. = a / ( 1 – r ) when 0 < r < 1 
Two Views of the DON’T PASS Bet
In earlier sections [ e.g. just above TREE FOR PASS LINE BET ] we saw that the probability of winning a pass line bet is exactly 244 / 495. But what is the probability of winning a donâ€™t pass bet ? The answer is not as straight forward as weâ€™d like because we first need to clarify what defines the starting and ending points of a “game”.
Consider the following scenario:
Alice walks up to the craps table and places a bet on the donâ€™t pass line. The shooter also bets the don’t pass line and rolls the dice. The come out roll is a 12. The stickman gives the dice back to the shooter who rolls a second time, and again the come out is 12.Alice takes down her bet and leaves the casino.
How many games did the shooter play while Alice’s bet was on the table ?
Some observers might say that the shooter played two games, and others might say that Alice left before the shooter had even finished his first game.
Observers could disagree about how to count the number of games played.
With pass line bets there are only two outcomes; the shooter either wins or loses his line bet. But for don’t pass bets we have a choice to make. We can either retain the view that only two outcomes are possible or we can decide that a game can also end in a tie.
The possibility of a 12 on the come out roll provides us with two viewpoints:
View 1: A Push Continues The Current Game
When the come out roll is a 12, we say that the outcome of this game is not yet decided. The shooter must continue rolling until the come out is not a 12. Every game results in a Win or a Loss for the shooter –there can be no ties.
View 2: A Push Ends The Current Game
When the come out roll is a 12, we say that the current game has ended in a tie. The shooter did not win and did not lose. The very next roll of the dice will be a come out roll for the next game.
The people who are playing craps don’t need to be concerned about these viewpoints. Is this the 3rd game of craps I’ve bet on tonight or is it the 8th — we don’t care. But before calculating any probabilities, we must first choose one view or the other because the probability of winning depends on how we define what a game is. It doesn’t matter which view you choose; both are reasonable definitions. But anything you might want to say about probabilities will not be meaningful to people unless you tell them which view you are using. Each view leads to a different set of values for the parameters of interest ( number of games played, probability to win, expectation, house edge, etc. ).
Here is a summary of what we will find when we analyze the don’t pass bet.
View 1: A Push Does Not End The Game
 P( win ) = 949 / 1925 = 0.492987….
 odds against winning ≈ 1.03 to 1.
 Expectation = bet * ( – 27 / 1925 )
 House Edge = 27 / 1925 ≈ 1.40%
View 2: A Push Ends The Game In A Tie
 P( win ) = 949 / 1980 = 0.47929….
 odds against winning ≈ 1.09 to 1.
 Expectation = bet * ( – 3 / 220 )
 House Edge = 3 / 220 ≈ 1.36%
DON’T PASS — VIEW 1 — CALCULATING THE PROBABILITY TO WIN
In this section we shall compute the probability of winning Don’t Pass bet assuming that a push does not end the current game. ( i.e. no game can end in a tie )
In a single game, the shooter could begin his play by rolling the number 12 many times in succession before finally having a come out roll with some number other than 12. For example, it is theoretically possible for the shooter to start out by rolling a 12 ten times in succession before finally having a come out of either 2, 3, 7, 11, or a point. Although the chances of this particular event are less than one in three quadrillion, such events could occur and must be taken into account.
Let’s begin our analysis by defining some helper events
 W1 = The event of winning immediately on the comeout roll by rolling either 2 or 3
 W2 = The event of establishing a point on the comeout roll and then winning by rolling a 7 before the point
 Y = W1 or W2
 T = The event of rolling a 12 on the comeout roll
Next, we find some preliminary probabilities:
P( W1 ) = P( 2 or 3 ) = 3 / 36 = 1 / 12
 P( W2 ) = 2 * [ P( 4 ) * P( 7 before 4 ) + P( 5 ) * P( 7 before 5 ) + P( 6 ) * P( 7 before 6 ) ]
 = 2 * [ ( 3 / 36 )*( 2 / 3 ) + ( 4 / 36 )*( 3 / 5 ) + ( 5 / 36 )*( 6 / 11 ) ]
 = 196 / 495
 P( Y ) = ( 1 / 12 ) + ( 196 / 495 ) = 949 / 1980 [ = (13)(73) / (20)(99) ]
Now let’s account for the possibility that play could begin with one or more rolls of 12.
 Let t = P( T ) = 1 / 36
 Let z = P( Y ) = 949 / 1980
 Let x = probability of winning a Don’t Pass bet [ using view #1 ]
DON’T PASS — VIEW 2 — CALCULATING THE PROBABILITY TO WIN
Now we will use View #2 to find the probability of winning a don’t pass bet.
P( win )  =  P( 2 or 3 on come out ) + P( point and make that point ) 
=  ( 3 / 36 ) + P( point and make it )  
=  ( 3 / 36 ) + 2 * [ P( 4 ) P( 7 before 4 ) + P( 5 ) P( 7 before 5 ) + P( 6 ) P( 7 before 6 ) ]  
=  ( 3 / 36 ) + ( 196 / 495 ) [ using the P( W2 ) result from the previous section ]  
=  949 / 1980 = 0.4792929….  
P( tie )  =  1 / 36 
P( lose )  =  1 – P( tie ) – P( win ) 
=  1 – ( 1 / 36 ) – ( 949 / 1980 ) = 244 / 495 [ = 0.492929…. ] 
Let’s find the expectation and house edge.
Let B  =  initial bet ( placed on the donâ€™t pass line ) 
E  =  Expectation = B * P( win ) + ( – B ) * P( lose ) + 0 * P( tie ) 
=  B * ( 949 / 1980 ) – B * ( 244 / 495 ) = B * ( – 3 / 220 )  
H  =  House Edge =  E  / B = 3 / 220 » 1.3636% 
MAKING FREE ODDS BETS DOES NOT CHANGE THE EXPECTATION
When a craps player bets the pass line and then establishes a point he is given the opportunity to make a free odds bet. If he decides to make this second bet then we can look at his total bet as being made up of two parts — the free odds part ( which is treated fairly by the house ) and the initial line bet, for which the game is biased in favor of the house.
Making free odds bets does not change the expectation because the free odds portion of the player’s total bet has an expected value of zero. Since the House Edge is given by H =  expectation / bet , the H will decrease if the expectation stays the same while the amount bet goes up. The following hypothetical example should help you to see this.
Example
Consider an imaginary game where the probability of winning is always 2 / 5, and  
the payoff odds are one to one.  
Letâ€™s find the expectation and the house edge.  
 
Now suppose that our imaginary game is changed so that at the start  
you can, if you wish, place two bets instead of just one. The first bet pays  
even money, but the second bet is a â€œfree oddsâ€ bet that pays off at  
whatever would be fair. Letâ€™s recompute E and H.  
Since p = 2 / 5, the odds against winning are 3 to 2 [ because 1 / p = 1 + g ].  
Let x = the size of our free odds bet.  
If we win, then the payoff for our free odds bet will be ( 3 / 2 ) * ( x ).  
The total amount bet is B + x, and we stand to win B + ( 3x / 2 ). 
So, E  =  ( B + 3x/2 ) ( 2 / 5 ) + ( 1 ) * ( B + x ) * ( 3 / 5 ) 
=  ( 2B / 5 ) + ( 3x / 5 ) + ( 3B / 5 ) + ( 3x / 5 )  
=  – B / 5 
The expectation is the same as it was before.  
The house edge is now given by  
 
Suppose x = 2 * B. Then H = ( 1 / 5 ) / 3 = 1 / 15 = 6.666….%  
In this imaginary game the ability to make so many free odds bets  
lowered the house edge from 20% to about 7%. 
Now back to reality.
COMBINING PASS LINE BETS WITH FREE ODDS BETS
For a player who makes only pass line bets, the house edge is about 1.41%. But if this player also makes free odds bets whenever possible, then the house edge will be reduced to a lower value, which we now want to calculate.
We want to find the house edge when free odds are offered at various rates, such as: 1X, 2X, 5X, 10X, 20X, and 100X.
For a pass line bet of size B, the casino limits on free odds bets are:
1X means the free odds bet cannot exceed 1 * B  
2X means the free odds bet cannot exceed 2 * B  
5X means the free odds bet cannot exceed 5 * B  
etc. 
Imagine that we are playing a series of craps games. We begin each game by putting B dollars on the pass line. Whenever a point is established, we make a free odds bet for the maximum amount permitted. Thus the amount we bet in each game is a variable. When the come out roll does not establish a point our bet stays at its initial value B, but when a point is established, our total bet is B plus the maximum amount allowed for a free odds bet.
Let N = the number of craps games we play [ some large fixed number ]
Let B = the dollar amount bet on the pass line at the start of each game
Let p = probability to win a pass line bet = 244 / 495 [ shown previously ]
Let L = probability to lose the pass line bet = 1 – p = 251 / 495
For later use, we note that p – L = – 7 / 495.
To make our work easier, we will use the following definitions:
α = size of the free odds bet when the point is 4 or 10  
β = size of the free odds bet when the point is 5 or 9  
γ = size of the free odds bet when the point is 6 or 8  
( In nearly all cases we will have α = β = γ = casinoâ€™s upper limit ) 
To find the house edge, we will first need to calculate the total amount of money we would expect to bet in N plays.
In an earlier section, we mentioned the fact that in a binomial experiment the expected number of successes is simply the product of the number of trials and the probability of getting a success on any one trial. So if we play craps N times, the expected number of times we would get a point of 4 on the come out roll is N * P( 4 ) = N * ( 3 / 36 ) = N / 12.
After similar calculations for the other points, we have:
average number of times that 4 is the point = N / 12  
average number of times that 5 is the point = N / 9  
average number of times that 6 is the point = 5N / 36  
average number of times that 8 is the point = 5N / 36  
average number of times that 9 is the point = N / 9  
average number of times that 10 is the point = N / 12 
In example 10, we found that in any one craps game, there is a twothirds probability for the shooter to establish a point. Therefore, the probability that no point is established is onethird. If we play N games, we would expect to have no point established in N / 3 games.
Let’s summarize our results in a table.
Point  Line Bet  Odds Bet  Sum of Bets  Nbr of Times We Bet That Sum  Product of Previous Two Columns 

none  B  0  B  N / 3  BN / 3 
4  B  α  B + α  N / 12  ( N / 12 )( B + α ) 
5  B  β  B + β  N / 9  ( N / 9 )( B + β ) 
6  B  γ  B + γ  5N / 36  ( 5N / 36 )( B + γ ) 
8  B  γ  B + γ  5N / 36  ( 5N / 36 )( B + γ ) 
9  B  β  B + β  N / 9  ( N / 9 )( B + β ) 
10  B  α  B + α  N / 12  ( N / 12 )( B + α ) 
Adding the values in the last column of the table will give us the total amount bet.
Let T  =  Total amount of money bet in N plays 
=  BN / 3 + 2 * [ ( N / 12 ) ( B + α ) + ( N / 9 ) ( B + β ) + ( 5N / 36 ) ( B + γ ) ]  
=  BN / 3 + 2N * [ ( 1 / 12 ) ( B + α ) + ( 1 / 9 ) ( B + β ) + ( 5 / 36 ) ( B + γ ) ]  
=  BN / 3 + ( 2N / 36 ) * [ 3 ( B + α ) + 4 ( B + β ) + 5 ( B + γ ) ]  
=  BN / 3 + ( N / 18 ) * ( 12B + 3α + 4β + 5γ )  
=  ( BN / 3 ) + ( 12BN / 18 ) + ( N / 18 ) ( 3α + 4β + 5γ )  
=  BN + ( N / 18 ) ( 3α + 4β + 5γ ) 
Thus, the average bet in N plays is T / N = B + ( 1 / 18 ) * ( 3α + 4β + 5γ )
Since free odds betting does not change the expectation, we know that  
E  =  Player’s Expectation 
=  same as it would be without any free odds bet = ( B * p ) + ( – B * L )  
=  B * ( p – L )  
=  B * ( – 7 / 495 ) [ recall that we computed p – L earlier ]  
Since E < 0, we are expecting a loss, whose magnitude is ( B ) * ( 7 / 495 ). 
Let H  =  House Edge 
=  Average Loss per Average Bet  
=  ( B ) * ( 7 / 495 ) divided by B + ( 1 / 18 )( 3α + 4β + 5γ )  
=  7 / 495 / [ 1 + ( 1 / 18B ) ( 3α + 4β + 5γ ) ] 
Some Example Cases of Pass Line Bets Combined With Free Odds Bets
Case 1: Free odds = 1X
Free odds = 1X → α = β = γ = B  
H = House Edge =  7 / 495 1 + ( 12 / 18 )  = 21 / ( 5 * 495 ) → H ≈ 0.8485 % 
Case 2A: Free odds = 2X
Free odds = 2X → α = β = γ = 2B  
H = House Edge =  7 / 495 1 + ( 24 / 18 )  = 3 / 495 → H ≈ 0.6061 % 
Case 2B: Free odds = “Full Double Odds”
Free odds = “Full Double Odds” → B = 2 γ = 5 α = β = 4  
H = House Edge =  7 / 495 1 + ( 53 / 36 )  = 252 / ( 89 * 495 ) → H ≈ 0.5720 % 
Case 3: Free odds = 3X
Free odds = 3X → α = β = γ = 3B  
H = House Edge =  7 / 495 1 + 3 ( 12 / 18 )  = 7 / ( 3 * 495 ) → H ≈ 0.4714 % 
Case 5: Free odds = 5X
Free odds = 5X → α = β = γ = 5B  
H = House Edge =  7 / 495 1 + 5 ( 12 / 18 )  = 21 / ( 13 * 495 ) → H ≈ 0.3263 % 
Case 10: Free odds = 10X
Free odds = 10X → α = β = γ = 10B  
H = House Edge =  7 / 495 1 + ( 12(10) / 18 )  = 21 / ( 23 * 495 ) → H ≈ 0.1845 % 
Case 20: Free odds = 20X
Free odds = 20X → α = β = γ = 20B  
H = House Edge =  7 / 495 1 + ( 12(20) / 18 )  = 21 / ( 43 * 495 ) → H ≈ 0.09866 % 
Case 100: Free odds = 100X
Free odds = 100X → α = β = γ = 100B  
H = House Edge =  7 / 495 1 + ( 12(100) / 18 )  = 21 / ( 203 * 495 ) → H ≈ 0.02090 % 
Now let’s see what happens to the house edge when a player combines free odds bets with Don’t pass bets. We will first use View #1 and then View #2.
VIEW #1 OF COMBINING DON’T PASS BETS WITH FREE ODDS BETS
We want to compute the house edge for players who make only don’t pass bets and free odds bets.
We imagine that we are playing a series of N craps games; and, since we are using the View #1 definition of a game, the occurrence of a push does not end the current game.
We begin each game by putting the amount B on the don’t pass line. If a point is established, we bet as much as we can in free odds. Whenever a 12 is rolled on the come out, we simply roll the dice again, as often as necessary. A new game doesn’t start until our line bet has been won or lost.
We use the following definitions:
N = the number of craps games we play [ some large fixed number ]  
B = the dollar amount bet on the donâ€™t pass line at the start of each game  
p = probability to win a donâ€™t pass bet ( according to View #1 ) = 949 / 1925  
L = probability to lose = 1 – p = 976 / 1925 
For later use, we note that p – L = – 27 / 1925
α = size of the free odds bet when the point is 4 or 10  
β = size of the free odds bet when the point is 5 or 9  
γ = size of the free odds bet when the point is 6 or 8 
The first thing we shall do is determine the probability of establishing a point in any one game.
Consider the possibility of getting 4 as our point.
Let F be the event of rolling a 4, and let V be the event of rolling a 12.
The events that could establish 4 as a point are those in the sequence:
F, VF, VVF, VVVF, VVVVF, VVVVVF, ….
Let a = P( 4 ) = 3 / 36, and let r = P( V ) = 1 / 36.
P( point is 4 )  =  P( F ) + P( VF ) + P( VVF ) + P( VVVF ) + …. 
=  a + ( a * r ) + a * ( r^{2} ) + a * ( r^{3} ) + …. 
Using the formula for the sum of an infinite geometric progression, we have
P( point is 4 ) = a / ( 1 – r ) = 3 / 35.
After similar calculations for the other points, we have:
P( point is 4 )  =  3 / 35 
P( point is 5 )  =  4 / 35 
P( point is 6 )  =  5 / 35 
P( point is 8 )  =  5 / 35 
P( point is 9 )  =  4 / 35 
P( point is 10 )  =  3 / 35 
Summing the above values, we see that the probability of establishing a point is 24 / 35. Therefore, in any one game, the probability that no point is established is 1 – ( 24 / 35 ) = 11 / 35.
So, in N games, we expect no point to be established approximately (11*N) / 35 times.
Letâ€™s use a table to summarize what weâ€™ve found so far.
Point  Line Bet  Odds Bet  Sum of Bets  Nbr of Times We Bet That Sum  Product of Previous Two Columns 

none  B  0  B  11 N / 35  ( 11 BN / 35 ) 
4  B  α  B + α  3N / 35  ( 3N / 35 )( B + α ) 
5  B  β  B + β  4N / 35  ( 4N / 35 )( B + β ) 
6  B  γ  B + γ  5N / 35  ( 5N / 35 )( B + γ ) 
8  B  γ  B + γ  5N / 35  ( 5N / 35 )( B + γ ) 
9  B  β  B + β  4N / 35  ( 4N / 35 )( B + β ) 
10  B  α  B + α  3N / 35  ( 3N / 35 )( B + α ) 
To find the total amount weâ€™d expect to bet in N plays, we add the values in the last column of the above table.
Let T = Total amount of money bet in N plays
T  =  ( 11 BN / 35 ) + 2 * [ ( 3N / 35 ) ( B + α ) + ( 4N / 35 ) ( B + β ) + ( 5N / 35 ) ( B + γ ) ] 
=  ( 11 BN / 35 ) + ( 2N / 35 ) * [ 3 ( B + α ) + 4 ( B + β ) + 5 ( B + γ ) ]  
=  ( 11 BN / 35 ) + ( 2N / 35 ) * [ 12B + 3α + 4β + 5γ ]  
=  BN + ( 2N / 35 ) ( 3α + 4β + 5γ ) 
The average amount bet in N plays is T / N = B + ( 2 / 35 ) ( 3α + 4β + 5γ )
Since free odds bets do not change the expectation, we can compute the average loss from the expectation:
E  =  expectation = expected winnings = Bp + ( – B ) * L = B * ( p – L ) 
=  B * ( – 27 / 1925 ) → expected loss = 27B / 1925 
Putting the above two pieces together, we find:
H = house edge = average loss / average bet


We want to find the house edge when free odds are offered at various rates, such as 1X, 2X, 3X, 5X, 10X, 20X, and 100X.
For a PASS line bet of size B, the casino limits on free odds bets are:
1X means the free odds bet cannot exceed 1 * B  
2X means the free odds bet cannot exceed 2 * B  
etc. 
But for a DONâ€™T pass line bet, we need to know the shooterâ€™s point before we can compute the maximum size of our free odds bet.
Recall these results from part 1 :
m  =  maximum possible don’t pass free odds bet  
B  =  line bet = amount we already put on the don’t pass line  
f  =  odds factor ( e.g. 2X odds → f = 2 )  
y  =  payoff odds  
Full Double Odds in effect  →  m = 3 * B  
Full Double Odds not in effect  →  m = ( f * B ) / y 
Here are the payoff odds :
If the point is 4 or 10 then  
 
If the point is 5 or 9 then  
 
If the point is 6 or 8 then  

Define α, β, and γ as before.
α = size of the free odds bet when the point is 4 or 10  
β = size of the free odds bet when the point is 5 or 9  
γ = size of the free odds bet when the point is 6 or 8 
Let’s use “FDO yes” to mean that Full Double Odds are in effect and
“FDO no” to mean that Full Double Odds are not in effect.
point is 4 or 10  →  “FDO no” and y = 1 / 2  
→  α = fB / y = 2fB  
point is 5 or 9  →  “FDO no” and y = 2 / 3  
→  β = fB / y = ( 3fB ) / 2  
point is 6 or 8 and “FDO yes”  →  γ = 3B  
point is 6 or 8 and “FDO no”  →  γ = fB / y = 6fB / 5 
If full double odds do not apply then 3α + 4β + 5γ = 6fB + 6fB + 6fB = 18fB
and the house edge is given by
H = House Edge  =  27 / 1925 1 + ( 2 / 35B ) ( 3x + 4y + 5z )  =  27 / 1925 1 + ( 2 / 35B ) ( 18fB )  
=  27 * 35 1925 * ( 35 + 36f )  =  27 55 * ( 35 + 36f )  
=  27 / 55 35 + 36f  assuming no full double odds 
If full double odds do apply then
B  =  line bet = 2k for some integer k  
α  =  size of the free odds bet when the point is 4 or 10  
=  fB / ( 1/2 ) = 2fB = 2f * 2k = 4fk  
β  =  size of the free odds bet when the point is 5 or 9  
=  fB / ( 2/3 ) = 3fB / 2 = (3f/2) * 2k = 3fk  
γ  =  size of the free odds bet when the point is 6 or 8  
=  3B = 3 * 2k = 6k  
3α + 4β + 5γ  =  24fk + 30k = k * ( 30 + 24f )  
H = House Edge  =  27 / 1925 1 + ( 2 / 35B ) ( 3α + 4β + 5γ )  =  27 / 1925 1 + ( 2 / 35B ) * k * ( 30 + 24f )  
=  27 / 1925 1 + ( 30 + 24f ) / 35  =  ( 27 / 1925 ) * 35 35 + ( 30 + 24f )  
=  27 * 35 1925 * ( 65 + 24f )  =  27 55 * ( 65 + 24f )  
=  27 / 55 65 + 24f  assuming full double odds 
Don’t Pass + Free Odds Using View #1
Case 1  Odds = 1X  →  f = 1  →  H =  ( 27 / 55 ) / 71  ≈  0.69142% 
Case 2a  Odds = 2X “FDO no”  →  H =  ( 27 / 55 ) / ( 35 + 72 )  ≈  0.45879%  
Case 2b  Odds = 2X “FDO yes”  →  H =  (27/55) / ( 65 + 48 )  ≈  0.434433%  
Case 3  Odds = 3X  →  f = 3  →  H =  ( 27 / 55 ) / ( 35 + 108 )  ≈  0.34329% 
Case 4  Odds = 4X  →  f = 4  →  H =  ( 27 / 55 ) / ( 35 + 144 )  ≈  0.27425% 
Case 5  Odds = 5X  →  f = 5  →  H =  ( 27 / 55 ) / ( 35 + 180 )  ≈  0.2283298% 
Case 10  Odds = 10X  →  f = 10  →  H =  ( 27 / 55 ) / ( 35 + 360 )  ≈  0.12428% 
Case 20  Odds = 20X  →  f = 20  →  H =  ( 27 / 55 ) / ( 35 + 720 )  ≈  0.065021% 
Case 100  Odds = 100X  →  f = 100  →  H =  ( 27 / 55 ) / ( 35 + 3600 )  ≈  0.013505% 
THE SIDE BETS IN CRAPS
Now let’s look at some of the side bets a craps player can make.
PLACE BETS AND BUY BETS
With place bets and buy bets, you are betting that the shooter will roll your chosen number before he rolls a 7. If you buy a number, you must pay the house a commission along with your bet. Then if you win, you will be paid at fair odds. When you place a number, there is no commission, but the payoff odds are biased against you. We will find that when you buy a number the house edge is the same no matter which number you buy; however, when you place a number the house edge depends on which number is being placed.
PLACE BETS
You can make a place bet on any of the â€œpointâ€ numbers ( 4 5 6 8 9 10 ). When you place a number, you are betting that the shooter will roll that number before he rolls a 7. ( It doesn’t matter what the shooter’s point actually is. )
Point  Payoff  House Edge 

4 or 10  9 to 5  ≈ 6.67% 
5 or 9  7 to 5  4.00% 
6 or 8  7 to 6  ≈ 1.52% 
Using the payoffs from the above table, let’s find the expectation ( E ) and
verify the stated house edge ( HE ) values.
Case 1 Point = 4 or 10
p  =  P( win ) = P( 4 before 7 ) = P( 4  4 or 7 ) = 3 / ( 3 + 6 ) = 1 / 3  
B  =  amount bet  
E  =  ( 9 / 5 ) B ( 1 / 3 ) + ( – B ) ( 2 / 3 )  
=  B * [ ( 9 / 5 )( 1 / 3 ) – ( 2 / 3 ) ] = – B / 15  
H  =   E  / B = 1 / 15 ≈ 6.6666….% 
Case 2 Point = 5 or 9
p  =  P( win ) = P( 5 before 7 ) = P( 5  5 or 7 ) = 4 / ( 4 + 6 ) = 2 / 5  
B  =  amount bet  
E  =  ( 7 / 5 ) B ( 2 / 5 ) + ( – B ) ( 3 / 5 )  
=  B * [ ( 14 / 25 ) – ( 15 / 25 ) ] = – B / 25  
H  =   E  / B = 1 / 25 = 4.00% 
Case 3 Point = 6 or 8
p  =  P( win ) = P( 6 before 7 ) = P( 6  6 or 7 ) = 5 / ( 5 + 6 ) = 5 / 11  
B  =  amount bet  
E  =  ( 7 / 6 ) B ( 5 / 11 ) + ( – B ) ( 6 / 11 )  
=  B * [ ( 35 / 66 ) – ( 36 / 66 ) ] = – B / 66  
H  =   E  / B = 1 / 66 ≈ 1.51515….% 
Conclusion: It’s ok to place the 6 or the 8, but you will probably want to avoid all other place bets. If you place the 6 or 8, your bet should be an integer multiple of six dollars so that your potential winnings will be a whole number of dollars.
BUY BETS
Buying a number is similar to placing a number. When you buy a particular number ( 4 5 6 8 9 10 ), you are betting that the shooter will roll that number before he rolls a 7. If you win you will be paid off at fair odds.
To buy a number you must give the dealer your bet plus another payment called a vigorish [ vig, for short ]. The vig is the upfront payment you must make in exchange for getting mathematically fair payoff odds.
Usually, the vig is 5% of your bet, but not all casinos treat the vig in the same way. Check the house rules where you play to see if the vig is rounded up or down for bets not evenly divisible by 20.
Let’s find the expectation ( E ) and the house edge ( H ) for any BUY bet.
We need a few helper definitions:
p  =  P( win )  
g  =  odds against winning [ 1 / p = 1 + g ]  
B  =  amount bet [ not counting the vig ]  
V  =  vig  
G  =  amount [ in dollars ] you give to the dealer = B + V  
R  =  amount the dealer returns to you if you win = B + gB  
w  =  amount you stand to win = R – G = gB – V 
Now let’s apply the definition of Expectation:
E  =  wp + (G) ( 1 – p )  
=  wp + ( 1 ) ( B + V ) ( 1 – p ) = wp + ( B + V ) ( p – 1 )  
=  wp + Bp + Vp – B – V 
But w = gB – V  →  E  =  ( gB – V ) p + Bp + Vp – B – V  
=  gBp – Vp + Bp + Vp – B – V  
=  ( gBp + Bp ) – B – V 
Now use 1 / p = 1 + g → 1 = p + gp → B = Bp + gBp
Substituting B for ( gBp + Bp ), we find
E  =  B – B – V = – V 
H  =   E  / bet = V / ( B + V ) = 1 / ( 1 + B / V ) 
V  =  B / 20 → H = 1 / ( 1 + 20 ) = 1 / 21 = 4.76190….% 
So the house edge for a buy bet is about 4.76%, regardless of which number you buy.
LAY BETS
A lay bet is the revers of a buy bet.
When you buy a number, you are betting that the shooter will roll that number before rolling a 7. When you lay a number, you are betting that the shooter will roll a 7 before rolling that number. In both cases you must pay a vig when placing your bet, and the payoff will then be at fair odds.
With buy bets, the vig is usually 5% of the size of your bet.
With lay bets, the vig is usually 5% of the amount you stand to win.
No matter which number you lay, your probability of winning will be greater than onehalf; therefore, you will have to bet more money than you stand to win.
To avoid potential round off losses, you should ensure that the amount you stand to win is some integer multiple of twenty dollars ( because you find 5% of a number by dividing it by 20 ).
p  =  P( win )  
g  =  odds against winning [ Recall that 1 / p = 1 + g ]  
B  =  size of bet [ not counting the vig ]  
V  =  vig = ( 1 / 20 ) * ( amount you stand to win ) = ( 1 / 20 ) ( gB ) = gB / 20 
Laying the 4 or 10  →  g = 1 / 2  →  V = B / 40 
Laying the 5 or 9  →  g = 2 / 3  →  V = B / 30 
Laying the 6 or 8  →  g = 5 / 6  →  V = B / 24 
Thus, if you want to ensure that the vig is a whole number of dollars you should bet as follows:
Laying the 4 or 10  →  B = multiple of $40  →  Minimum bet + vig = $41 
Laying the 5 or 9  →  B = multiple of $30  →  Minimum bet + vig = $31 
Laying the 6 or 8  →  B = multiple of $24  →  Minimum bet + vig = $25 
Letâ€™s find the expectation ( E ) and the house edge ( H ).
From our analysis of the BUY bet, we have E = – V.
Thus H =  E  / total bet =  E  / ( B + V ) = V / ( B + V ) = 1 / ( 1 + ( B / V ) )
But V = gB / 20 → H = 1 / ( 1 + ( 20 / g ) )
Laying the 4 or 10 → H = 1 / 41 » 2.439%
House Edge =  E  / total bet =   E  V + B  =  V V + B  
=  1 1 + ( B / V ) 
But V = gB / 20 → H = 1 / ( 1 + ( 20 / g ) )
Laying the 4 or 10  →  House Edge  =  1 / 41  ≈  2.439% 
Laying the 5 or 9  →  House Edge  =  1 / 31  ≈  3.2258% 
Laying the 6 or 8  →  House Edge  =  1 / 25  ≈  4.0% 
PLACETOLOSE BETS
The bet wins if a 7 is rolled before the chosen number. The payoff odds are as follows:
Number  Payoff Odds 

4 or 10  5 to 11 
5 or 9  5 to 8 
6 or 8  4 to 5 
Case 1 The number is 4 or 10  
P( win )  =  P( 7 before 4 ) = 6 / ( 6 + 3 ) = 2 / 3  
E  =  expectation  
=  ( 5B / 11 ) ( 2 / 3 ) + ( – B ) ( 1 / 3 ) = – B / 33  
H  =  house edge  
=   E  / B = 1 / 33 ≈ 3.03% 
Case 2 The number is 5 or 9  
P( win )  =  P( 7 before 5 ) = 6 / ( 6 + 4 ) = 3 / 5  
E  =  expectation  
=  ( 5B / 8 ) ( 3 / 5 ) + ( – B ) ( 2 / 5 ) = – B / 40  
H  =  house edge  
=   E  / B = 1 / 40 = 2.50% 
Case 3 The number is 6 or 8  
P( win )  =  P( 7 before 6 ) = 6 / ( 6 + 5 ) = 6 / 11  
E  =  expectation  
=  ( 4B / 5 ) ( 6 / 11 ) + ( – B ) ( 5 / 11 ) = – B / 55  
H  =  house edge  
=   E  / B = 1 / 55 ≈ 1.82% 
BIG SIX
This bet wins if the shooter rolls a 6 before rolling a 7.
It pays even money.
P( win )  =  P( 6 before 7 )  =  5 / 11 
E  =  B ( 5 / 11 ) + ( – B ) ( 6 / 11 )  =  – B / 11 
House Edge  =   E  / B = 1 / 11  ≈  9.09% 
You should never make this bet. If you want to bet on the number 6, you should avoid this bet and place the 6 instead. That bet pays 7 to 6 and has a house edge of only 1.52%.
BIG EIGHT
In online craps, this bet wins if the shooter rolls an 8 before rolling a 7.
It pays even money.
P( win )  =  P( 8 before 7 )  =  5 / 11 
E  =  B ( 5 / 11 ) + ( – B ) ( 6 / 11 )  =  – B / 11 
House Edge  =   E  / B = 1 / 11  ≈  9.09% 
You should never make this bet. If you want to bet on the number 8, you should avoid this bet and place the 8 instead. That bet pays 7 to 6 and has a house edge of only 1.52%.
THE HORN BET
This is a one roll bet where you win if the outcome is any one of these four numbers : 2 3 11 12.
You make four bets at once by placing equal amounts of money on each of the four numbers. The next time the shooter rolls the dice, one of two things will happen:
either you will lose all four of your bets or else one of your bets will win, and the other three will lose.
Rolling 2 or 12 pays 30 to 1.
Rolling 3 or 11 pays 15 to 1.
Letâ€™s find the expectation ( E ) and the house edge ( H ).
Let B = size of each bet. [ The total amount bet is 4 * B. ]
If one of our 4 bets wins, then the other 3 must lose.
Roll 2 or 12 → win 30B but lose 3B → net win is 27B.
Roll 3 or 11 → win 15B but lose 3B → net win is 12B.
Since P(2) = P(12) and P(3) = P(11), we have
E  =  expectation = 2 * [ ( 27B ) P( 2 ) + ( 12B ) P( 3 ) ] + ( – 4B ) P( lose )  
=  2 * [ ( 27B ) * ( 1 / 36 ) + ( 12B ) * ( 2 / 36 ) ] + ( – 4B ) ( 5 / 6 )  
=  ( 2B / 36 ) * [ 27 + 24 ] + ( – 10B / 3 )  
=  ( B / 18 ) * ( 51 ) + ( – 10B / 3 )  
=  ( 17B / 6 ) + ( – 20B / 6 ) = – B / 2  
H  =  house edge  
=   E / total_bet  * 100% = ( B / 2 ) / 4B  
=  ( 1 / 8 ) = 12.50% 
THE FIELD BET
This is a one roll bet where you win if the shooterâ€™s next roll is
any one of these seven numbers : 2 3 4 9 10 11 12.
You lose on 5 6 7 8.
( Some casinos use 5 instead of 9 as one of the winning field numbers. )
Rolling 2 pays 2 to 1.
Rolling 12 pays either 2 to 1 or 3 to 1, depending on the casino.
The Wizard of Odds uses these terms to characterize casinos:  
Loose casino  ↔  pays 3 to 1 for rolling 12  
Tight casino  ↔  pays 2 to 1 for rolling 12 
Let’s find the expectation ( E ) and the house edge ( H ).
Let  B  =  size of bet 
Let  kB  =  amount won for rolling 12 [ k can be either 2 or 3 ] 
E = Expectation  =  ( 2B ) * P( 2 ) + 
( B ) * P( 3 or 4 or 9 or 10 or 11 ) +  
( k*B ) * P( 12 ) +  
( – B ) * P( 5 or 6 or 7 or 8 )  
E / B  =  ( 2 ) * ( 1 / 36 ) + 
( 14 / 36 ) +  
( k / 36 ) +  
( – 20 / 36 )  
=  ( k – 4 ) / 36 
case 1 Loose casino  
k = 3 →  E / B  = 1 / 36 → H = 2.777….%  
case 2 Tight casino  
k = 2 →  E / B  = 2 / 36 → H = 5.555….% 
Thus the house edge is either 2.8% or 5.6%, depending on the casino.
THE PROPOSITION BETS
A list of the most stupid bets in craps would probably include the big 6, the big 8, the horn, the field, and all the proposition bets.
The proposition bets are the ones advertised in the center of the craps table layout. In this section we describe some of those bets and calculate the house edge for each one.
All proposition bets are oneroll bets, except for the four hardway bets.
Any Craps
This is a one roll bet, which you win if the shooter’s next roll is 2, 3, or 12.
The payoff odds are 7 to 1.
Let B = size of bet.
E = ( 7B ) ( 4 / 36 ) + ( – B ) ( 32 / 36 ) = – 4B / 36 = – B / 9
House Edge =  E  / B = 1 / 9 ≈ 11.1%
CrapsEleven ( C & E )
This is a one roll bet.
If the shooter rolls 2, 3, or 12 you win, and the payoff is 3 to 1.
If the shooter rolls 11 you win, and the payoff is 7 to 1.
If the shooter rolls anything else you lose your bet.
Let B = size of bet.
E = ( 7B ) * ( 1 / 18 ) + ( 3B ) * ( 1 / 9 ) + ( – B ) * ( 5 / 6 ) = B ( – 1 / 9 )
H =  E  / B = 1 / 9 ≈ 11.1%
Craps 2
This is a one roll bet, which you win if the shooter’s next roll is 2.
The payoff odds are 30 to 1.
Let B = size of bet.
E  =  expectation  
=  ( 30B ) * ( 1 / 36 ) + ( – B ) * ( 35 / 36 ) = – 5B / 36 )  
H  =  house edge  
=  5 / 36 ≈ 13.89% 
Craps 12
This is a one roll bet where you win if the shooter’s next roll is 12.
The payoff odds are 30 to 1.
Let B = size of bet.
E  =  expectation  
=  ( 30B ) * ( 1 / 36 ) + ( – B ) * ( 35 / 36 ) = – 5B / 36 )  
H  =  house edge  
=  5 / 36 ≈ 13.89%. 
Big Red [ Any 7 ]
This is a one roll bet, which you win if the shooter’s next roll is 7.
The payoff odds are 4 to 1.
Let B = size of bet.
E  =  expectation  
=  ( 4B ) * ( 1 / 6 ) + ( – B ) * ( 5 / 6 ) = – B / 6 )  
H  =  house edge = 1 / 6 » 16.7%. 
( This is one of the worst craps bets you can make. )
Hard 4
This is a multiroll bet, which you win if the shooter rolls a hard four before rolling either a seven or an easy four.
The payoff odds are 7 to 1.
F  =  the event of rolling a hard four = rolling 2 and 2  
Y  =  the event of rolling an easy four = rolling 1 and 3  
S  =  the event of rolling a seven  
Z  =  S or Y  
P( F )  =  1 / 36  
P( Y )  =  2 / 36  
P( S )  =  6 / 36  
P( Z )  =  P( S ) + P( Y ) = 8 / 36  
P( win )  =  P( F before Z ) = P( F  F or Z )  
=  P( F and ( F or Z ) ) / P( F or Z )  
=  P( F ) / ( P( F ) + P( Z ) )  
=  ( 1 / 36 ) / ( ( 1 / 36 ) + ( 8 / 36 ) )  
=  1 / 9  
B  =  size of bet  
E  =  expectation  
=  ( 7B ) ( 1 / 9 ) + ( – B ) ( 8 / 9 ) = – B / 9  
H  =  house edge  
=   E  / B = 1 / 9 ≈ 11.1% 
Hard 10
This is a multiroll bet, which you win if the shooter rolls a hard ten before rolling either a seven or an easy ten.
Hard ten = 5 and 5; Easy ten = 4 and 6
The payoff odds are 7 to 1.
The house edge is the same as for the hard four ( ≈ 11.1% ).
Hard 6
This is a multiroll bet, which you win if the shooter rolls a hard six before rolling either a seven or an easy six.
Hard six = 3 and 3; Easy six = 1 and 5 or 2 and 4
The payoff odds are 9 to 1.
X  =  the event of rolling a hard six = rolling 3 and 3  
Y  =  the event of rolling an easy six = rolling 1 and 5 or 2 and 4  
S  =  the event of rolling a seven  
Z  =  S or Y  
P( X )  =  1 / 36  
P( Y )  =  4 / 36  
P( S )  =  6 / 36  
P( Z )  =  P( S ) + P( Y ) = 10 / 36  
P( win )  =  P( X before Z ) = P( X  X or Z )  
=  P( X and (X or Z) ) / P( X or Z )  
=  P( X ) / ( P(X) + P(Z) )  
=  ( 1 / 36 ) / ( ( 1 / 36 ) + ( 10 / 36 ) )  
=  1 / 11  
B  =  size of bet  
E  =  expectation  
=  ( 9B ) ( 1 / 11 ) + ( – B ) ( 10 / 11 ) = – B / 11  
H  =  house edge =  E  / B = 1 / 11 » 9.09% 
Hard 8
This is a multiroll bet, which you win if the shooter rolls a hard eight before rolling either a seven or an easy eight.
Hard eight = 4 and 4; Easy eight = 2 and 6 or 3 and 5
The payoff odds are 9 to 1.
The house edge is the same as for the hard six ( ≈ 9.09% ).
Items The Shooter’s Baby Might Need
If you listen very carefully at the craps table you might hear the shooter shout out  
“Baby need a new shoes !”  
as the dice are tossed. 
Perhaps you can improve your luck by informing the dice of your baby’s requirements.
Items many of us could use include:
 
 
 
 
 
 
 
 
 
 

Additional Craps Related Math Exercises
Exercise
If a point is established on the come out roll, then on average, how many additional rolls of the dice will be needed for the shooter’s line bet to be resolved ?
suggestion: Split into 3 cases; and, for each case, find the probability that consecutive rolls of the dice will never yield the point or a 7
Case 1 The shooter’s point is 4 or 10
To be specific, let’s assume that the shooter’s point is 4.
Suppose the shooter needs to roll the dice N additional times after the come out roll
before his line bet is decided.
If a roll of the dice resolves the line bet, we will say that the roll was a “success”;
otherwise, we will call the roll a “failure”.
In this particular case, success means rolling 4 or 7, and failure means rolling anything else.
We can easily compute the probability of getting N consecutive failures.
On each roll we have:
P( failure ) = 1 – [ P(4) + P(7) ] = 1 – [ (3/36) + (6/36) ] = 3 / 4
Since the dice rolls are independent events, the probability of not getting a 4 or a 7 in two consecutive rolls is ( 3 / 4 )^{2 }. The probability for three consecutive failures is ( 3 / 4 )^{3 }, etc.
Let’s make a table showing the probability for never getting a 4 or a 7 in N consecutive rolls of the dice.
Rolls 1 thru N  Probability of not getting 4 or 7 in N consecutive rolls 

N = 1  0.75^{1} = 0.7500 
N = 2  0.75^{2} = 0.5625 
N = 3  0.75^{3} ≈ 0.4219 
N = 4  0.75^{4} ≈ 0.3164 
N = 5  0.75^{5} ≈ 0.2373 
Scan down the 2nd column of the above table, and notice that there is a better than 5050 chance that we will not get a 4 or 7 during the first two rolls. But by the 3rd roll this turns around. The probability of 3 consecutive failures is less than one half ( only 42.19% ).
Since 1 – 0.4219 = 0.5781, there is a 57.81% chance that we will not get 3 consecutive failures.
So, when the shooter’s point is 4 or 10, there is a 57.8% probability that at most 3 rolls will be needed to decide the outcome of his line bet.
Case 2 The shooter’s point is 5 or 9
To be specific, let’s assume that the shooter’s point is 5.
The probability of not getting a 5 or a 7 in any roll is
1 – [ P(5) + P(7) ] = 1 – [ (4/36) + (6/36) ] = 26 / 36 ≈ 0.7222.
Rolls 1 thru N  Probability of not getting 5 or 7 in N consecutive rolls 

N = 1  0.7222^{1} = 0.7222 
N = 2  0.7222^{2} ≈ 0.5216 
N = 3  0.7222^{3} ≈ 0.3767 
N = 4  0.7222^{4} ≈ 2721 
In the 2nd column of the table the first entry less than one half occurs at N = 3. So it is unlikely to get 3 consecutive failures.
Since 1 – 0.3767 = 0.6233, there is a 62.3% chance that a 5 or 7 will occur within the first 3 rolls after the shooter has established a point of 5.
So, when the point is 5 or 9, there is a 62.3% probability that 3 or fewer rolls will be needed to decide the outcome of the shooter’s line bet.
Case 3 The shooter’s point is 6 or 8
To be specific, let’s assume that the shooter’s point is 6.
The probability of not getting a 6 or a 7 in any roll is
1 – [ P(6) + P(7) ] = 1 – [ (5/36) + (6/36) ] = 25 / 36 ≈ 0.6944.
Rolls 1 thru N  Probability of not getting 6 or 7 in N consecutive rolls 

N = 1  0.6944^{1} = 0.6944 
N = 2  0.6944^{2} ≈ 0.4823 
N = 3  0.6944^{3} ≈ 0.3349 
N = 4  0.6944^{4} ≈ 0.2326 
Since 1 – 0.4823 = 0.5177, there is a 51.8% chance that we will not have 2 consecutive failures.
There is a 51.8% chance that a 6 or 7 will occur within the first 2 rolls after the shooter has established a point of 6 ( or 8 ).
In other words, when the point is 6 or 8, there is a 51.8% probability that 2 or fewer additional rolls after the comeout will be needed to decide the outcome of the shooter’s line bet.
Summary:
point is 4 or 10  →  usually [ about 58% of the time ] need no more than 3 additional rolls  
point is 5 or 9  →  usually [ about 62% of the time ] need no more than 3 additional rolls  
point is 6 or 8  →  usually [ about 52% of the time ] need no more than 2 additional rolls 
Exercise
How long can a craps game last ? Fill in the blanks with valid integer values:
 When the come out roll establishes a point of ____ or ____
 there is less than a one percent chance that an additional ____ or more rolls
 of the dice will be needed to decide the outcome of the shooter’s line betCase 1 The shooter’s point is 4 or 10
To be specific, assume the shooter’s point is 4.
Suppose the shooter rolls the dice N additional times after the come out roll.
Each roll of the dice can be labelled as either a “success” or a “failure”,
depending on whether or not the roll determines the outcome of the shooter’s line bet.
For this case:
success  means rolling either 4 or 7  
failure  means rolling any number except 4 or 7 
We can easily compute the probability of getting N consecutive failures.
On each roll we have:
P( failure ) = 1 – [ P(4) + P(7) ] = 1 – [ (3/36) + (6/36) ] = 3 / 4
Since the dice rolls are independent events, the probability of not getting a 4 or a 7 in N consecutive rolls is ( 3 / 4 )^{N } for N = 1, 2, 3, ….
Here is a table showing the probability for never getting a 4 or a 7 in N consecutive rolls of the dice.
Rolls 1 thru N  Probability of not getting 4 or 7 in N consecutive rolls 

N = 1  0.75^{1} = 0.75 
N = 2  0.75^{2} = 0.5625 
N = 3  0.75^{3} ≈ 0.4219 
N = 4  0.75^{4} ≈ 0.3164 
N = 5  0.75^{5} ≈ 0.2373 
N = 6  0.75^{6} ≈ 0.1780 
N = 7  0.75^{7} ≈ 0.1335 
N = 8  0.75^{8} ≈ 0.1001 
N = 9  0.75^{9} ≈ 0.0751 
N = 10  0.75^{10} ≈ 0.0563 
N = 11  0.75^{11} ≈ 0.0422 
N = 12  0.75^{12} ≈ 0.0317 
N = 13  0.75^{13} ≈ 0.0238 
N = 14  0.75^{14} ≈ 0.0178 
N = 15  0.75^{15} ≈ 0.0134 
N = 16  0.75^{16} ≈ 0.01002 
N = 17  0.75^{17} ≈ 0.0075 
N = 18  0.75^{18} ≈ 0.0056 
Scanning down the 2nd column of this table, we see that the probability for getting consecutive failures doesn’t fall below 1% until we reach N = 17.
So, when the shooter’s point is 4 or 10, there is less than a 1% chance that 17 or more additional rolls of the dice will be needed to resolve the shooter’s line bet.
Case 2 The shooter’s point is 5 or 9
To be specific, let’s assume that the shooter’s point is 5.
Define “success” and “failure” by:
success  means rolling either 5 or 7  
failure  means rolling any number except 5 or 7 
On each roll we have:
P( failure ) = 1 – [ P(5) + P(7) ] = 1 – [ (4/36) + (6/36) ] = 13 / 18 ≈ 0.7222
Rolls 1 thru N  Probability of not getting 5 or 7 in N consecutive rolls 

N = 1  0.7222^{1} = 0.7222 
N = 2  0.7222^{2} = 0.5216 
N = 3  0.7222^{3} ≈ 0.3767 
N = 4  0.7222^{4} ≈ 0.2721 
N = 5  0.7222^{5} ≈ 0.1965 
N = 6  0.7222^{6} ≈ 0.1419 
N = 7  0.7222^{7} ≈ 0.1025 
N = 8  0.7222^{8} ≈ 0.0740 
N = 9  0.7222^{9} ≈ 0.0535 
N = 10  0.7222^{10} ≈ 0.0386 
N = 11  0.7222^{11} ≈ 0.0279 
N = 12  0.7222^{12} ≈ 0.0201 
N = 13  0.7222^{13} ≈ 0.0145 
N = 14  0.7222^{14} ≈ 0.0105 
N = 15  0.7222^{15} ≈ 0.0076 
N = 16  0.7222^{16} ≈ 0.0055 
When the shooter’s point is 5 or 9, there is less than a 1% chance that 15 or more additional rolls of the dice will be needed to resolve the shooter’s line bet.
Case 3 The shooter’s point is 6 or 8
To be specific, let’s assume that the shooter’s point is 6.
Define “success” and “failure” by:
success  means rolling either 6 or 7  
failure  means rolling any number except 6 or 7 
On each roll we have:
P( failure ) = 1 – [ P(6) + P(7) ] = 1 – [ (5/36) + (6/36) ] = 25 / 36 ≈ 0.6944
Rolls 1 thru N  Probability of not getting 6 or 7 in N consecutive rolls 

N = 1  0.6944^{1} = 0.6944 
N = 2  0.6944^{2} = 0.4823 
N = 3  0.6944^{3} = 0.3349 
N = 4  0.6944^{4} = 0.2326 
N = 5  0.6944^{5} = 0.1615 
N = 6  0.6944^{6} = 0.1122 
N = 7  0.6944^{7} = 0.0779 
N = 8  0.6944^{8} = 0.0541 
N = 9  0.6944^{9} = 0.0376 
N = 10  0.6944^{10} = 0.0261 
N = 11  0.6944^{11} = 0.0181 
N = 12  0.6944^{12} = 0.0126 
N = 13  0.6944^{13} = 0.0087 
N = 14  0.6944^{14} = 0.0061 
Scanning down the 2nd column of the above table, we see that the probability for getting consecutive failures doesn’t fall below 1% until we reach N = 13.
Summary:
When the come out roll establishes a point of 4 or 10  
there is less than a one percent chance that an additional 17 or more rolls  
of the dice will be needed to decide the outcome of the shooter’s line bet.  
When the come out roll establishes a point of 5 or 9  
there is less than a one percent chance that an additional 15 or more rolls  
of the dice will be needed to decide the outcome of the shooter’s line bet.  
When the come out roll establishes a point of 6 or 8  
there is less than a one percent chance that an additional 13 or more rolls  
of the dice will be needed to decide the outcome of the shooter’s line bet. 
point is 4 or 10  →  rarely [ less than 0.75% of the time ] need more than 17 additional rolls  
point is 5 or 9  →  rarely [ less than 0.76% of the time ] need more than 15 additional rolls  
point is 6 or 8  →  rarely [ less than 0.87% of the time ] need more than 13 additional rolls 
Exercise
When the shooter makes a pass line bet, what is the average number of times that he will roll the dice before his pass line bet is resolved ? ( This exercise is not easy and its solution might require a bit of calculus. )
To solve this problem, I created a formula that uses a power series, which is
too tedious to evaluate without the help of a program. Perhaps you can find a better solution.
Using a java program to evaluate my formula I found that the average number of rolls is about 3.38.
Here is my solution:
Let the random variable X denote the number of times the dice must be rolled
to resolve the shooter’s pass line bet.
Note that X can take on any positive integer value, and
we want to find E = the expected value of X = 1*P( X=1) + 2*P(X=2) + 3*P(X=3) + ….
In order to evaluate this infinite sum, let’s begin by looking at a few special cases.
Case 1 The game ends in just 1 roll.
This can only happen when the come out roll is 2, 3, 12, 7, or 11.
The probability for this event is 12 / 36 = 1 / 3.
So, E = 1*(1/3) + more_terms
Let’s temporarily put aside our interest in E and, instead, focus our attention
on how we will compute the probability that exactly k rolls of the dice will
be needed, where k = 1 or 2 or 3 or ….
Note:  At this point most readers should skip over the following details and jump ahead to Case k. 
Case 2 The game ends after exactly 2 rolls.
This can only happen when the come out roll establishes a point and
the next roll is either that particular point or a 7.
In other words, we’d have to :
Come out on 4 and then  
roll 4 or 7  
Or come out on 5 and then  
roll 5 or 7  
Or come out on 6 and then  
roll 6 or 7  
Or come out on 8 and then  
roll 8 or 7  
Or come out on 9 and then  
roll 9 or 7  
Or come out on 10 and then  
roll 10 or 7 
Case 3 The game ends after exactly 3 rolls.
This can only happen when the come out roll establishes a point and
the next roll is neither that point nor a 7 and
the third roll is either the established point or a 7.
That requires that we :
Come out on 4 then  
roll anything except 4 or 7 and then  
roll 4 or 7  
Or come out on 5 then  
roll anything except 5 or 7 and then  
roll 5 or 7  
Or come out on 6 then  
roll anything except 6 or 7 and then  
roll 6 or 7  
Or come out on 8 then  
roll anything except 8 or 7 and then  
roll 8 or 7  
Or come out on 9 then  
roll anything except 9 or 7 and then  
roll 9 or 7  
Or come out on 10 then  
roll anything except 10 or 7 and then  
roll 10 or 7 
Jumping ahead a little bit, we consider the requirements for case 6.
Case 6 The game ends after exactly 6 rolls.
This can only happen when the come out roll establishes a point and
the next 6 – 2 rolls yield neither that point nor a 7 and
the sixth roll is either the established point or a 7.
Now look at the most general case for which the game doesn’t end on the come out roll.
Case k The game ends after exactly k rolls, where k > 1.
This can only happen when the come out roll establishes a point and
the next k2 rolls are neither that point nor a 7 and
the kth roll is either the established point or a 7.
( Note that since k > 1, k2 ≥ 0. )
In other words, for exactly k > 1 rolls, we need one of these 6 scenarios :
Come out on 4 then  
roll anything except 4 or 7 for k – 2 consecutive rolls and then  
roll 4 or 7  
Or come out on 5 then  
roll anything except 5 or 7 for k – 2 consecutive rolls and then  
roll 5 or 7  
Or come out on 6 then  
roll anything except 6 or 7 for k – 2 consecutive rolls and then  
roll 6 or 7  
Or come out on 8 then  
roll anything except 8 or 7 for k – 2 consecutive rolls and then  
roll 8 or 7  
Or come out on 9 then  
roll anything except 9 or 7 for k – 2 consecutive rolls and then  
roll 9 or 7  
Or come out on 10 then  
roll anything except 10 or 7 for k – 2 consecutive rolls and then  
roll 10 or 7 
We only need to analyze the first 3 scenarios and then use matching probabilities to handle the last 3.
Suppose the come out roll is a 4.  
The probability of rolling a 4  and then  
k2 instances of neither 4 nor 7  and then  
a 4 or 7 is given by  
P(4) * ( 1 – P(4 or 7) )^{k2} * P( 4 or 7 )  
which equals  
(3/36) * (27/36)^{k2} * (9/36)  
Suppose the come out roll is a 5.  
The probability of rolling a 5  and then  
k2 instances of neither 5 nor 7  and then  
a 5 or 7 is given by  
P(5) * ( 1 – P(5 or 7) )^{k2} * P( 5 or 7 )  
which equals  
(4/36) * (26/36)^{k2} * (10/36)  
Suppose the come out roll is a 6.  
The probability of rolling a 6  and then  
k2 instances of neither 6 nor 7  and then  
a 6 or 7 is given by  
P(6) * ( 1 – P(6 or 7) )^{k2} * P( 6 or 7 )  
which equals  
(5/36) * (25/36)^{k2} * (11/36)  
Since P(8) = P(6) and P(9) = P(5) and P(10) = P(4), we can simply multiply
the sum of the results of the above 3 scenarios by 2 to get our final result.
The probability for using exactly k rolls is
2 [ (3/36)*(27/36)^{( k – 2 )} *(9/36) + (4/36)*(26/36) ^{( k – 2 ) }*(10/36) + (5/36)*(25/36) ^{( k – 2 ) }*(11/36) ]
Which equals
(2/36) * [ (27/36) ^{( k – 1 ) } + (40/36)*(26/36)^{( k – 2 ) } + (55/36)*(25/36)^{( k – 2 ) } ]
and is valid for k ≥ 2.
By reducing fractions we can simplify this expression slightly to get
(1/18) * [ (3/4) ^{( k – 1 ) } + (10/9)*(13/18)^{( k – 2 ) } + (55/36)*(25/36)^{( k – 2 ) }
Define the function h by h(x) = the probability that the random variable X takes on the value x.
Then h(x) = (1/18) * [ (3/4) ^{( x – 1 ) } + (10/9)*(13/18)^{( x – 2 ) } + (55/36)*(25/36)^{( x – 2 ) }
Refer back to case 1 and see that
E = 1*(1/3) + the sum of all x*h(x) values as x runs from 2 thru infinity.
If you’ve taken some calculus courses then you might have seen the “ratio test”,
which can sometimes tell us whether an infinite series converges or diverges.
The ratio test shows that our expression for E must converge.
Since I don’t know what the series converges to, I wrote a java program which
uses several large values for N to compute the sum of the first N terms in the
series portion of the value for E and then adds 1*(1/3) to that, giving us an estimate for E.
In order to send the results to a text file instead of to my computer’s screen,
I used a modified version of the the java program shown at a link near the top of this page.
It took only a second or two to produce these results:
nbr of iterations is 10  approx mean is 2.9005061385337125  
nbr of iterations is 80  approx mean is 3.375757573735532  
nbr of iterations is 1000  approx mean is 3.375757575757575  
nbr of iterations is 10000  approx mean is 3.375757575757575 
Exercise
Suppose you bet the pass line, and a point is established on the come out roll. Can you now remove your pass line bet ?
No, you cannot change your mind and take back the pass line bet after a point is established.
If the point is 4 or 10, then your probability of winning is only 3 / ( 3+6) = 1 / 3. So, the casino will not want to give up its 2 / 3 probability of winning your line bet.
If the point is 5 or 9, then the casino’s probability of winning is 1 – 4 / ( 4+6) = 3 / 5. So, the casino should refuse to let you take back your bet.
If the point is 6 or 8, then the casino’s probability of winning is 1 – 5 / ( 5+6) = 6 / 11. So, again it is to the casino’s advantage to not let you take back your bet.
Exercise
Suppose you bet the don’t pass line, and a point is established on the come out roll. Can you now remove your don’t pass bet ?
Yes, you can remove it, but you should not do so.
If the point is 4 or 10, then your probability of winning is 6 / ( 6+3) = 2 / 3.
So, the casino would be happy to see you give up your advantage by taking back the bet you placed on the don’t pass line.
If the point is 5 or 9, then the casino’s probability of winning is only 1 – 6 / ( 6+4 ) = 2 / 5. So, the casino would gladly let you take back your bet.
If the point is 6 or 8, then the casino’s probability of winning is 1 – 6 / ( 6+5 ) = 5 / 11. So, again it is to the casino’s advantage to allow you to take back your don’t pass bet.
Exercise
Can you make a pass line bet after the shooter has established a point ?
Yes, you can but you shouldn’t. You should make a come bet instead.
If the point is 4 or 10, then your probability of winning would only be 3 / ( 3+6 ) = 1 / 3. But a come bet would have the same probability of winning as a “proper” pass line bet ( viz. 244 / 495 ≈ 0.493 ).
If the point is 5 or 9, then your probability of winning would only be 0.40.
And if the point is 6 or 8, then your probability of winning would only be 5 / 11 ≈ 0.45
Exercise
Can you make a don’t pass bet after the shooter has established a point ?
No, because that would shift the bias to your favor.
If the point is 4 or 10, then your probability of winning would be 1 – 3 / ( 3+6 ) = 2 / 3.
If the point is 5 or 9, then your probability of winning would be 1 – 4 / 10 = 0.60.
And if the point is 6 or 8, then your probability of winning would be 1 – 5 / 11 = 6 / 11 ≈ 0.545
If you could make don’t pass bets after a point is established then everybody would do this and the casino would quickly go broke.
Mathematics Reference Links
The Wizard of Odds : One of the best web sites for expert information on all the major casino games
Math Topics ( e.g. Craps ) by Dr. Math: For CRAPS articles, go to the site and type CRAPS in the SEARCH THE ARCHIVE box
Continued Fractions by Wikipedia : Nice article from a free encyclopedia
Perpetual Calendars by Dr. Math : Another version of the Reverend Zeller formula
Virtual Laboratories in Probability And Statistics : This page from the Math Department at the University of Alabama in Huntsville contains many lessons in Probability & Statistics and includes several java applets.
Mathematics Reference Books
( These are old but useful books. )
An Introduction to the Theory of Numbers
 by Ivan Niven and Herbert Zuckerman©
 1960 John Wiley & Sons, Inc.
 See chapter 7 for continued fractions and chapter 4 for bracket function
Mathematical Statistics 2nd edition
 by John E. Freund©
 1971 PrenticeHall, Inc.
 Chapter 1 has good information about Permutations and Combinations
Continued Fractions ( paperback )
 by C. D. Olds©
 1963 Yale University ( A volume in the New Mathematical Library )
 Targeted to high school and college students but some tough reading
An Introduction To Continued Fractions ( paperback )
 by Charles G. Moore©
 1964 The National Council of Teachers of Mathematics, Inc.
 Chapter 6 shows some practical applications, e.g. gear ratio problems
Elementary Number Theory
 by J. V. Uspensky and M. A. Heaslet©
 1930 McGrawHill Book Company, Inc.
 The appendix ( pages 206211 ) has a proof of the Rev Zeller formula for perpetual calendar calculations