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| This Site : | Bottom-of-Page | Home | Part 1 | Part 2 | Other Topics |
| 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.
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 )
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 / 13 |
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.
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. | ||||||||
|
||||||||
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 ]
There are 38 numbers on an American roulette wheel. The numbers zero and double-zero 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.
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 vice-versa :
Simply notice that 1/p = (a + b) / a = a/a + b/a = 1 + b/a = 1 + g.
| Tip: | You should memorize the formula |
|
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. |
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.
The term “odds” has at least three uses:
Odds and probability are just two different ways of measuring the same thing.
Can you think of other situations where we sometimes measure or view the same concept in different ways ?
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.
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 favor. 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 |
If 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.
Show that if a game is fair then the payoff odds must exactly match the odds against winning.
Suppose that the probability to win a certain game is exactly 1 / 10, and the payoff odds are ten to one. Is this a fair game ?
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.
To illustrate house edge, we will find the expectation and the house edge when you bet on
any particular number in roulette. But first, take a look at the following image showing
a roulette table layout with a few example bets. The light blue circles represent
chips, and the letter inside each circle indicates what type of bet that chip placement
signifies. For example, the blue chip labeled A shows how you would make the five number
bet ( which is a bad bet ).
Note: The author doesn't really know how to make gif files; and so, some of the horizontal lines got lost for some reason when the following image was created.
Now let's find the expectation and house edge for the case where you have bet on any one particular number.
Whenever you bet on a particular number, the payoff odds are 35 to 1.
Let B = the size of your bet [ for example, B might be $2 ]
Let W = amount of money you receive if you win your bet.
Note that there are only two outcomes for your bet:
either you will win 35 times your bet or
you will lose your bet.
Thus the random variable W can take on only two possible values:
either W will be 35*B or else W will be ( - B ).
To get the expectation, we must multiply each possibility by its associated probability and add the products.
Let p = probability to win = 1 / 38
Then 1 - p = probability to lose = 37 / 38.
So the expectation is simply ( 35B ) * p + ( - B ) * ( 1 - p )
E = expectation = ( 35B ) * ( 1 / 38 ) + ( - B ) * ( 37 / 38 ) = - 2B / 38 = - B / 19
E / B = - 1 / 19
H = House Edge = | E / B |
H = 1 / 19 = ( 1 / 19 ) * 100 % ≈ 5.263%
So, when you bet on any particular number in roulette, the house edge is roughly 5.25%.
In fact, it can easily be shown that, in roulette, the house edge is about 7.89% for the five number bet and is roughly 5.25% for all other bets.
For comparison, we will see that in craps the house edge on line bets is less than 1.5%.
Suppose that a casino lets you bet on the outcome of flipping a balanced coin -- heads you win, tails you lose. If it costs one dollar to play, but the payoff odds are only one to two, then what would be the house edge for this ( unfair ) game ?
Of course house edge is not the only factor to consider when you are deciding whether to play a particular game. If the probability of winning is very small then you might not want to play even if your expectation is a positive number.
Before we study craps, let’s take a quick look at lotteries. If you don't care about lotteries then skip example 6, exercise 4, and exercise 5.
The lottery numbers are 1, 2, 3, ..., 49; and six of these are drawn at random. For simplicity, we assume that there is only one prize, and you win it only if your choice of six numbers matches the six numbers drawn. The house edge depends on how much money you get for winning.
What is the house edge if you buy one ticket, which costs one dollar, and the prize is ten million dollars ? ( Ignore the possibility of winning smaller prizes; assume you either win $10,000,000 or lose $1. And assume you owe no taxes on the winnings. )
We first need to know how many ways the six numbers can be drawn. In general, the number of ways to choose r things from a set of n things is given by the following formula, whose left hand side is pronounced "n binomial r" :
| ( | n | ) |
|
|||||
| r |
|
( Instead of using the above formula,
you can sometimes use Pascal's Triangle
to evaluate binomial coefficients. )
Let c = nbr of ways to choose 6 numbers from 49.
For n = 49 and r = 6, you can use a calculator to find that
| c = | ( | 49 | ) |
|
= 13,983,816 | |||||
| 6 |
Let p = the probability of
winning the lottery = 1 / c
Let Z = the prize = 107 [ in dollars ]
Since a ticket costs one dollar, | E | / bet = | E |.
| E | = | expectation | = | Zp + ( - 1 ) * ( 1 - p ) | |
| = | Zp - 1 + p | ||||
| = | p * ( Z + 1 ) - 1 | ||||
| = | ( -1 ) + ( Z + 1 ) / c | ||||
| = | ( -1 ) + ( 1 + 107 ) / 13,983,816 | ||||
| = | - 0.2848875.... | ||||
H = house edge = | E | / bet ≈ 28.49 %
Thus, for a ten million dollar tax-free prize the house edge is more than 25%.
What would the lottery prize need to be for our expectation to be positive ?
An activity is called a binomial experiment if it has these four features:
If we play the lottery a fixed number of times, we have a binomial experiment in which “success” means winning the prize.
In n trials, the total number of successes could be 0 or 1 or 2 or 3 or ... or n.
The probability to get any number of successes can be found by this formula :
| probability of getting exactly x successes in n trials = | ( | n | ) | p x ( 1 - p ) ( n - x ) | ||
| x |
Substituting 0 for x, we see that the probability of getting
no successes is ( 1 - p ) n .
And so, the probability of getting at least
one success is 1 - ( 1 - p ) n .
How many trials are needed to ensure that the probability of winning the lottery at least once is greater than one-half ?
Here is an important fact that we will use later about binomial experiments.
( This fact is proved in many elementary statistics books. )
| Fact : | |
| Let X be the number of successes obtained in the n trials of a binomial experiment. | |
| Let p be the probability of getting a success on any one trial. | |
|
Then the expected value
[ i.e. “average value" ] of the random variable X is simply the product of n and p. |
|
For example, suppose a binomial experiment consists of flipping a coin 100 times; and, suppose that the coin is biased so that on any one flip there is a 70% chance of getting heads. If we repeated this experiment of making 100 flips and writing down the number of heads obtained in the 100 flips, then, on the average, the number of heads in each set of flips would tend to be close to 100 * 0.7 = 70.
Another example: Roll a pair of dice 600 times, and call it a success when you roll a 4.
Then p = 1 / 12 ( shown in example 7), and the expected nbr of successes is
600 * ( 1 / 12 ) = 50. In 600 rolls of the dice, 4 should occur about 50 times.
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.
Find the probability of rolling any particular number with a pair of balanced dice
Using the method employed in example 7, 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 |
When you go to a casino you probably don’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 let’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:
| Correction: | ||
|
Ignore these steps and skip down a few
lines to the Better Method paragraph. |
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 don’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.
Unless we say otherwise, we will always use the word “or” to mean "and/or" [ i.e “one or the other or both”]. When events consist of elementary outcomes and we want to combine the events using ANDs and ORs, the following “sum” 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 “A or B”. Then dividing each item by the total number of outcomes possible gives us the probabilities shown in the sum rule.
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.
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 A1, A2, ..., An
that are pairwise mutually exclusive, we have this rule:
P( A1 or A2 or ... or An )
= P( A1 ) +
P( A2 ) + ... + P( An ).
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. )
Two events can be either independent or dependent. Being independent means that the occurrence or non-occurrence 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 “rolling a 7” and “rolling a 9” 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.
When rolling dice repeatedly, what is the probability of rolling a 7 twice in a row ?
Our intuition tells us that the dice don’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.
Suppose we roll a pair of dice one time.
Let S be the event of getting a 7.
Let T be the event of getting either 7 or 9.
Are the events S and T dependent or independent ?
What is the probability that a card drawn from a deck is either red or an ace ?
Is drawing an ace independent of drawing a red card ?
To help us compute the odds against making various points in a craps game we need to know a little bit about conditional probability.
The probability of an event can change if we are given new information that we didn’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 ]
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 shooter’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.
In a craps game what are the odds against rolling a 9 before a 7 ?
Earlier we said that two events are independent if the occurrence or non-occurrence 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 non-zero probabilities then the following 5 statements are equivalent:
| 1. | A is independent of B | |
| 2. | B is independent of A | |
| 3. | P( A ) = P( A | B ) | |
| 4. | P( B ) = P( B | A ) | |
| 5. | P( A and B ) = P( A ) * P( B ) |
If A and B are not independent then they are dependent.
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. )
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.
P( win ) = W( 5 ) / ( W( 5 ) + W( 7 ) ) = 4 / ( 4 + 6 ) = 2 / 5
1 / p = 1 + g → 5 / 2 = 1 + g → g = 3 / 2
P( win ) = W( 6 ) / ( W( 6 ) + W( 7 ) ) = 5 / ( 5 + 6 ) = 5 / 11
1 / p = 1 + g → 11 / 5 = 1 + g → g = 6 / 5
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
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 ).
Show that in craps the house edge for a pass line bet is about 1.41%
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 ?
Show that about 45% of the winning pass line bets are made
on the come out roll.
Hint: P( natural | win ) = ?
Consider the following simple game.
Use two decks of cards to create a new deck consisting of
any 36 red cards plus any 35 black cards. Shuffle this new
71-card deck and draw a card at random. If the chosen card
is black, you win; if it's red you lose.
Show that the probability of winning this simple game is about the same as that for winning a pass line craps bet.
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 ?
Before we compute the probability of winning a don’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.
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 non-zero 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*(r2) + a*(r3) + a*(r4) + ... + a*( rn - 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*r2 + a*r3 + a*r4 + ... + a*r(n-1) + a*rn |
subtracting gives S(n) - r * S(n) = a - a*(rn)
Then factoring both sides gives ( 1 - r ) * S(n) = a * ( 1 - rn )
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 - rn ) / ( 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, rn 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 "S-sub-infinity" ) 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 + ar2
+ ar3 + ....
= a / ( 1 - r )
when 0 < r < 1
|
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 don’t pass bet ? The answer is not as straight forward as we’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
don’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% |
In this section we shall compute the probability of winning a “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 come-out roll by rolling either 2 or 3 |
| W2 | = | The event of establishing a point on the come-out roll and then winning by rolling a 7 before the point |
| Y | = | W1 or W2 |
| T | = | The event of rolling a 12 on the come-out 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 ] |
| Then x | = | P( Y or TY or TTY or TTTY or TTTTY or .... ) |
| = | z + t*z + (t2)*z + (t3)*z + (t4)*z + .... | |
| = | z * ( 1 + t + t2 + t3 + t4 + .... ) | |
| = | z * [ 1 / ( 1 - t ) ] | |
| = | ( 949 / 1980 ) * ( 36 / 35 ) = ( (13)(73) / (20)(99) ) * ( 36 / 35 ) | |
| = | 949 / 1925 = 0.492987012987.... |
Using View #1 [ game cannot end in a tie ], find the house edge when a craps player bets the don't pass line.
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 don’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% |
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.
| Consider an imaginary game where the probability of winning is always 2 / 5, and | ||||||||||
| the payoff odds are one to one. | ||||||||||
| Let’s find the expectation and the house edge. | ||||||||||
|
||||||||||
| Now suppose that our imaginary game is changed so that | ||||||||||