**Probability**

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**Theoretical Probability - **theoretical probability is a deductive inference about the degree of likelihood than an event will occur prior to the occurrence of that event. The contents of our reasoning are assumptions formed before our experience of the actual event. The more specific our assumptions are the more closely our conclusions will conform to the actual probability of the occurrence of an event. An important consideration in our calculations of theoretical probabilities is the element of randomness. If our judgment of the probability of an event occurring is truly theoretical then these predictions are based on the assumption that one outcome is as likely to occur as another.

**Empirical Probability – **Empirical probability is easily distinguished from theoretical probability. Empirical probability is determined by induction.** **That is, empirical probability is generated from experience. Empirical probability is a prediction that uses past behavior to predict the future. The past performance of an event is used to determine the likelihood that the event will occur in the same manner again. An easily understood example would be the horse races. Suppose there are seven horses in a race. Theoretically, given the assumption of equal likelihood, each horse would have a one in seven chance of winning. Empirically we know that this isn’t the case. In the past, certain horses performed better than others. Conditions on the track, genetics, the jockey riding the horse, the number of races previously run, etc. are all factors which could control the outcome of the race. These factors must be relevant factors. (Relevant to the situation being studied) The color of the jockeys uniform, the gender of the jockey, the horses name; do not generally have any affect on the probability of whether the horse will win or lose. Relevant factors are those factors that will affect the outcome of the race. Our assumption of equal likelihood does not apply in cases such as this.

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**Master Formula** - As in the example of the “horse race”, if there are seven horses and each horse has an equal opportunity to win then we can predict that any given horse has a one chance out of seven (1/7) chance of winning. In a similar example, if a coin is flipped and you were asked to predict the chances that a heads would land face up, it would be an obvious prediction to claim that there is a 50% chance of this occurring. (one chance out of two – ½) From these examples we can generate a master formula. The master formula is total number of favorable possibilities divided by the total number of possibilities.

__Total Favorable Possibilities__

**Total Possibilities**

**Single Events** - This formula has many applications. Concerning single events merely calculate the total possibilities and then determine the favorable (the event that is under investigation) possibilities. Plug these values into the master formula and you have the answer. For example, if you wish to find the probability of drawing a spade from a standard deck (52 cards) of cards you can construct the following formula. 13 cards are favorable (13 cards in a suit and 4 suits in a deck). There are 52 possible cards in a standard deck. Therefore the answer is 13/52 or ¼. (One chance out of four- 25%)

**Conjunctions of Multiple Events**

**Independent Events**- Often it is necessary to determine the probability of multiple events occurring. Events are said to be independent when the occurrence of the any of the events have no effect the probability of the occurrence of the other events. The easiest example is flipping a coin. What is the probability of flipping a coin twice and having it come up “heads” twice in a row? The chances it will come up “heads” the first time is ½ and the chance that it will come up “heads” the second time is also ½. To compute this conjunction of events we use the*rule of multiplication*–**“The probability that two or more events will occur together is the product of the fractions expressing the probability of each event”**. (Moore/McCann/McCann - p.100) Therefore ½ x ½ = ¼. The probability of flipping a coin twice and getting heads both times is one chance out of four (25%). Suppose that there are 5 multiple-choice questions on a test. Each question has 4 possible answers, one of which is correct. What is the probability that you can guess the correct answer to all of the problems? (¼ x ¼ x ¼ x ¼ x ¼ = 1/1024 = .098% - better study – less than 1% chance that you can guess your way through the test.) However there is one aspect of multiple-choice questions that can increase the probability of getting the correct answer. When reading a multiple-choice question certain answers are often eliminated immediately. When this occurs the probability that the answer can be “guessed” is increased. Instead of one chance out of four, the odds per event often increase to one chance out of three or maybe even two depending on how many possible answers you can rule out.) Try this problem out. The answer will be at the bottom of the page. The cafeteria is serving free food. The only catch is that you can’t choose what food is to be served to you. There is a choice of meat and a choice of dessert. For meat they will be serving hamburger, hotdogs and chili. For desert they will be serving ice-cream, Jell-O and chocolate cake. You hate chili and ice-cream. What is the probability that you will be served this meal?* (See answer at the bottom of the page)

**Conjunctions of Dependent Events**

- A conjunction of events is said to be dependent when the outcome of one or more of the calculations is affected by an event that occurred previously. An event is said to be dependent on one or more prior events when its probability is affected by the outcome of these events. Consider the following. Suppose you are dealt three cards from a standard deck of playing cards. What is the chance that you are dealt all red cards? Use the master formula. First card dealt 26/52 (1/2), second card dealt 25/51( there are only 25 red cards left and only 51 total cards left – we must assume that the first card dealt was red). The probability that a red card would be dealt the third time is 24/50. We have just applied the
*rule of dependency***“When events are dependent, the fractions expressing their probability must be adjusted for the effects of previous events.”**(Moore/McCann/McCann - p.101) The calculations are 1/2 x 25/51x 24/50 = 2/17. There are two chances out of seventeen that you will be dealt three black cards in a row. If you’ll notice this is slightly worse odds than you would have if you had returned the cards to the deck after each draw (1/16 :: 1/17

It is often possible to enumerate all the possible combinations of certain groups of objects or events. There are other times that a mathematical formula will be necessary for these computations. Suppose we have three coins. (Penny, Nickel, and a Dime) What is the probability that if all three coins were tossed in the air that when they landed there would be two heads and one tails showing? We first enumerate all the possibilities.

Penny |
Nickel |
Dime |

H |
H |
H |

H |
H |
T |

H |
T |
T |

H |
T |
H |

T |
T |
T |

T |
T |
H |

T |
H |
H |

T |
H |
T |

You can use the diagonal chart to the right of the table of total possibilities to generate the table. This may be useful if the question involved using 4 or more coins. Just start at the first “head”, then follow the line to the end being sure to follow all the branches. Repeat this procedure with the “tail” line. There is a mathematical formula by which to extract the number of possibilities but it doesn’t tell you what the possibilities are. (formula for this type of combination calculation - n= number of objects being used – 2^n ) You need to know what the possibilities are in order to use the master formula. In order to answer our question use the master formula. Total possibilities (2^3) is eight. Total favorable possibilities (I have put these in italics and red). Answer - **3/8**

Various mathematical formulas can be generated to calculate the total number of possible arrangements of objects. Be sure to visit some of the websites listed under permutations, combinations, probability, etc. for further study of areas that may interest you. One of the more interesting calculations of probability involves calculations for your lottery choices. State run lotteries are ever present. They spend quite a few dollars in an attempt to convince players that an individual’s odds are good enough to win and that playing the lottery is a wise investment both of time and money. Let’s examine this proposition. Calculate the odds of winning the “pick three”. In Illinois Pick three you choose three digits from 0-9. They can be repeated. Ultimately you are choosing a number from 000-999. This would mean that your odds of winning would be 1/1000. This is at least a number that one can imagine. Let’s compare this to another set of probabilities. What is the chance that if I threw 9 coins in the air that all 9 coins would land heads up? That seems pretty far fetched and no one would ever bet on such a gamble – or would they. The probability of the 9 coins landing heads up is - ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ = 1/512. Hmmmmmm nearly twice as good as “pick three”. Our intuitive estimate of probability often betrays us. Later I will discuss the **gamblers fallacy **and how it plays into our mistaken intuitions and calculations of theoretical probability.

There are other forms of probability calculations regarding the lottery that also interest us. These calculations involve sequences. Suppose you have 5 objects. How many different ways can you pair these items together? Look at the table below.

A |
B |
C |
D |
E |

Baseball Glove |
Spikes |
Bat |
Bases |
Catchers Mask |

Baseball Glove/Spikes

Baseball Glove/Bat

Baseball Glove/Bases

Baseball Glove/Catchers Mask

Spikes/Bat

Spikes/Bases

Spikes/Catchers Mask

Bat/ Bases

Bat/Catchers Mask

Bases/Catchers Mask

There are 10 possible pairings of these items and these pairing (sequence of two) are fairly easy to enumerate. However in the lottery you pick a sequence of six numbers from a possible 50 (1-50). These numbers are not repeated. The question is - what is the probability of picking a sequence of six numbers from the fifty available. For this we need a mathematical formula to calculate all the permutations. Lets start with the example above since we know it is correct. I will express the relationship being described as follows.

** 5,2** (This represents the problem being solved)

In order to solve the problem I must introduce the mathematical term – factorial.

Factorial (def) **n****!=( n) (n- 1)(n-2) · · ·1 where n is an integer greater than 0**

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** The “!” after a number means factorial**

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**Example**: 6! = 6x5x4x3x2x1 = 720

**General formula** (for the calculation of the number of sequences of a given size in a particular domain)

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__ n! __** n = total number of elements **

**k! (n-k)! k = number of elements in the permutation**

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__ 5! __** = 5! = 5x4x3x2x1 = 20 = 10**

**2!(5-2)! 2!(3!) (2x1)(3x2x1) 2**

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A more extreme calculation would be the calculations for determining the probability of winning the lottery

*50, 6 n=50, k=6*

__ 50! __** = 50! = (50)(49)(48)(47)(46)(45)(44)(43)……..(1) =**

**6! (50-6)! 6!(44!) (6)(5)(4)(3)(2)(1) X (44)(43)(42)………(1)**

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__(50)(49)(48)(47)(46)(45)__** = 11,441,304,000 = 15,890,700**

** (6)(5)(4)(3)(2)(1) 720 1**

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There is some salvation. (hehe) In the Illinois lottery when you buy a $1.00 ticket you get two choices. That would increase your odds of winning to 1 chance out of 7,945,350. Good odds? Revisit the discussion about throwing coins in the air and having them all land heads up. How many coins would you have to throw into the air to get a probability that resembled the odds of the lottery? Check out this web page – some interesting calculations - http://skepdic.com/gamblers.html

One final calculation – What are the odds you will role a 7 at the craps tables. What are the odds you will roll an 11. If you add these odds together you will have the odds of winning on your first roll at the tables. First find all the possibilities of a roll of two dice. (Special note: an easy way to spot loaded – unfair dice is to examine the top and bottom of the dice. Add the face showing on top to the face of the dice sitting on the table – these two numbers must always add up to seven)

Possible combinations: 1-1, 1-2, 1-3, 1-4, 1-5,1-6, 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 3-1, 3-2,

3-3, 3-4, 3-5, 3-6, 4-1, 4-2, 4-3, 4-4, 4-5, 4-6, 5-1, 5-2, 5-3, 5-4, 5-5, 5-6, 6-1,6-2,6-3, 6-4, 6-5, 6-6.

The probability of rolling a “seven” is 6/36

The probability of rolling an “eleven” is 2/36

The probability of winning on the first roll is 8/36 or 2/9

**Gamblers fallacy**

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There are two ways to figure the probability of an event occurring – theoretical probability and empirical probability. One takes into account the past performance of the event/objects in question the other bases it computations on the number of possible favorable outcomes divided by the total possibilities. There is no third manner in which to evaluate the probability of an event occurring. Whenever concepts such as “evening out” , “my luck has to change” or “I’m due” are invoked an error in reasoning (fallacy) has occurred. People who argue that they play the same number in the lottery, year after year, because that series of numbers “must come in some time” have committed the gamblers fallacy. Theoretically that number has exactly the same chance to be registered today that it had a year ago. That is assuming that the game is fair. If the game is not fair and a series of numbers has a better chance of coming up the we must employ our tools regarding empirical probability. The gambler's fallacy carries with it an embedded presupposition that the chances for something with a fixed probability can increase or decrease over a period of time. A belief that success or failure will change the calculations of the theoretical probability

__gamblers' fallacy__

*noun*

(*Psychol*) the fallacy that in a series of chance events the probability of one event occurring increases with the number of times another event has occurred in succession **(Source: The Collins English Dictionary © 2000 HarperCollins Publishers:)**

good discussion of the gamblers fallacy - http://www.wikipedia.org/wiki/Gamblers_fallacy

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***** chili and ice-cream probability = 1/3 x 1/3 = 1/9