# How to Work Out Odds, Permutations and Combinations

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## What is Probability Theory

Probability theory is an interesting area of statistics concerned with the odds or chances of an event happening in a trial, e.g. getting a six when a dice is thrown. To work out odds, we also need to have an understanding of permutations and combinations. The math isn't terribly complicated, so read on and you might be enlightened!

## Definitions

Probability is a measure of the likelihood of an event occuring

Trial This is an experiment or test. e.g Throwing a dice or a coin

Outcome The result of a trial. e.g the number when a dice is thrown, or the card pulled from a shuffled pack.

Event An outcome of interest. E.g. getting a 6 in a dice throw or an ace when a card is drawn.

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## Waht is the Probability of an Event?

There are two types of probability, empirical and classical.

If A is the event of interest, then the probability of A occurring is denoted by P(A)

Empirical Probability This is determined by carrying out a series of trials. So for instance a batch of products is tested and the number of faulty items is noted plus the number of acceptable items.

If there are n trials

and A is the event of interest

Event A occurs x times

Then

P(A) = x / n

e.g. A sample of 200 products is tested and 4 faulty items are found. What is the probability of a product being faulty?

So x = 4 and n = 200

Therefore P(faulty item) = 4 / 200 = 0.02

Classical Probability This is a theoretical probability which can be worked out mathematically.

If A is the event, then

P(A) = Number of ways the event can occur / The total number of possible outcomes

Example 1: What are the chances of getting a 6 when a dice is thrown?

In this example, there is only 1 way a 6 can occur and there are 6 possible outcomes, i.e. 1, 2, 3, 4, 5 or 6

So P(6) = 1/6

Example 2: What is the probability of drawing a 4 from a pack of cards in one trial?

There are 4 ways a 4 can occur, i.e. 4 of hearts, 4 of spades, 4 of diamonds or 4 of clubs

Since there are 52 cards, there are 52 possible outcomes in 1 trial.

So P(4) = 4 / 52 = 1 / 13

## Expectation

Once a probability has been worked out, it's possible to get an estimate of how many events will likely happen in future trials. This is known as the expectation and is denoted by E.

If the event is A and the probability of A occurring is P(A), then for N trials, the expectation is:

E = P(A) N

For the simple example of a dice throw, the probability of getting a six is 1/6.
So in 60 trials, the expectation or number of expected 6s is:

E = 1/6 x 60 = 10

Remember, the expectation is not what will actually happen, but what is likely to happen. In 2 throws of a dice, the expectation of getting a 6 is:

E = 1/6 x 2 = 1/3

However as we all know, it's quite possible to get 2 sixes in a row, even though the probability is only 1 in 36 (see how this is worked out later). As N becomes larger, the actual number of events which happen will get closer to the expectation. So for example when flipping a coin, if the coin isn't biased, the number of heads will be closely equal to the number of tails.

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## Success or Failure

The probability of an event can range from 0 to 1.

Remember P(Event) = Number of ways the event can occur / The total number of possible outcomes

So for a dice throw

P(getting a number between 1 and 6) = 6 / 6 = 1 (since there are 6 ways you can get "a" number and 6 possible outcomes)

P(getting a 7) = 0 / 6 = 0 (there are no ways the event 7 can occur in any of the 6 possible outcomes)

P(getting a 5) = 1 / 6 (only 1 way of getting a 5)

If there are 999 failures in 100 samples

Empirical probability of failure = P(failure) = 999/1000 = 0.999

A probability of 0 means that an event will never happen.
A probability of 1 means that an event will definitely happen.

In a trial if event A is a success, then failure is not A (not a success)

And P(A) + P(not A) = 1

For mutually exclusive (they can't occur simultaneously) events A and B

P(A or B) = P(A) + P(B)

e.g. A dice is thrown and a card is drawn from a pack, what is the possibility of getting a 6 or an ace?

P(getting a six) is 1/6

There are 52 cards in a pack and 4 ways of getting an ace

P(getting an ace) is 4/52 = 1/13

So P(getting a six or getting an ace) = P(getting a six) + P(getting an ace) = 1/6 + 1/13

Remember in these type of problems, how the question is phrased is important. So the question was to determine the probability of one event occurring "or" the other event occurring and so the addition law of probability is used.

## Multiplication Law of Probability

For independent (the first trial doesn't affect the second trial) events A and B

P(A and B) = P(A) x P(B)

e.g. A dice is thrown and a card drawn from a pack, what is the probability of getting a 5 and a spade card?

P(getting a 5) = 1/6

There are 52 cards in the pack and 4 suits or groups of cards , aces, spades, clubs and diamonds. Each suit has 13 cards, so there are 13 ways of getting a spade.

So P(drawing a spade) = 13/52 = 1/4

P(getting a 5 and drawing a spade) = P(getting a 5) x P(drawing a spade) = 1/6 x 1/4 = 1/24

Again it's important to note that the word "and" was used in the question, so the multiplication law was used.

Independent and Dependent Events

Events are independent when the occurrence of one event doesn't affect the probability of the other event.

So if a card is drawn from a pack, the probability of an ace is 4/52 = 1/13

If the card is replaced, the probability of drawing an ace is still 1/13

Events are dependent if the occurrence of the first event affects the probability of occurrence of the second event

If an ace is drawn from a pack and not replaced, there are only 3 aces left and 51 cards remaining, so the probability of drawing a second ace is 3/51

For two events A and B where B depends on A, the probability of Event B occurring after A is denoted by P(B|A)

## Permutations and Combinations

To solve more difficult problems and derive an expression for the probability of a general binomial distribution, we need to understand the concept of permutations and combinations. I won't go into the mathematics of the derivation, but basically the expression is derived from the equation for working out combinations.

## A Permutation is an Arrangement

A permutation is a way of arranging a number of objects. So for instance if you have the letters A, B, and C then all the possible permutations are:

ABC, ACB, BAC, BCA, CAB, CBA

Note that BA is a different permutation to AB

If you have n objects, there are n factorial number of ways of arranging them, written as n!

n! = n x (n-1) x (n-2) .... x 3 x 2 x 1

The reason for this is because for the first position, there are n choices, and for each of these choices, there are (n-1) choices for the second place (because 1 choice was used up for the first place), and for each of the choices in the first two places, (n-3) choices for the third place and so on.

In the example above, the 3 letters A, B, C could be arranged in 3! = 3 x 2 x 1 = 6 ways

In general if n objects are selected r at a time then, the number of permutations is:

n! / (n-r)!

This is written as nPr

e.g. 2 letters are chosen from the set of letters A, B, C, D

How many ways can the 2 letters be arranged?

There are 4 letters so n =4 and r = 2

nPr = 4P2 = 4! / (4 - 2)! = 4! / 2! = 4 x 3 x 2 x 1 / 2 x 1 = 12

## A Combination is a Selection

A combination is a way of selecting objects from a set without regard to the order of the objects. So again if we have the letters A, B, and C and select 3 letters from this set, there is only 1 way of doing this, i.e. select ABC.
If we select 2 letters at a time from ABC, all the possible selections are:

AB, AC and BC

Remember, BA is the same selection as AB etc.

In general if you have n objects in a set and make selections r at a time, the total possible number of selections is:

nCr = n! / ((n - r)! r!)

e.g. 2 letters are chosen from the set ABCD. How many combinations are possible?

There are 4 letters ao n = 4 and r = 2

nCr = 4C2 = 4! / ( (4 - 2)! x 2!) = 4! / (2! x 2!)

= 4 x 3 x 2 x 1 / ( (2 x 1) x (2 x 1) ) = 6

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## General Binomial Distribution

In a trial, an event could be getting heads in a coin throw or a six in a throw of a dice.

If the occurrence of an event is defined as a success, then

Let the probability of success be denoted by p

Let the probability of non occurrence of the event or failure be denoted by q

p + q = 1

Let the number of successes be r

And n is the number of trials

Then

e.g. What are the chances of getting 3 sixes in 10 throws of a dice?

There are 10 trials and 3 events of interest, i.e. successes so:

n = 10

r = 3

Probability of getting a 6 in a dice throw is 1/6, so:

p = 1/6

Probability of not getting a dice throw is:

q = 1 - p = 5/6

P(3 successes) = 10! / ((10 - 3)! 3!) x (5/6)(10 - 3) x (1/6)3

= 10! / (7! x 3!) x (5/6)7 x (1/6)3

= 3628800 / (5040 x 6) x (78125 / 279936) x (1/216)

= 0.155

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## Winning the Lottery! - How to Work out the Odds

We would all like to win the lottery, but the chances of winning are only slightly greater than 0. However "If you're not in, you can't win", and a slim chance is better than none at all!

Take for example the California State Lottery. A player must choose 5 numbers between 1 and 69, and 1 Powerball number between 1 and 26. So that is effectively a 5 number selection from 69 numbers and a 1 number selection from 1 to 26. To calculate the odds, we need to work out the number of combinations, not permutations, since it doesn't matter what way the numbers are arranged to win.

The number of combinations of r objects is nCr = n! / ((n - r)! r!)

n = 69

and

r = 5

and nCr = 69C5 = 69! / ( (69 - 5)! 5!) = 69! / (64! 5!) = 11,238,513

So there are 11,238,513 possible ways of picking 5 numbers from a choice of 69 numbers.

Only 1 Powerball number is picked from 26 choices, so there are only 26 ways of doing this.

For every possible combination of 5 numbers from the 69, there are 26 possible Powerball numbers, so to get the total number of combinations, we multiply the two combinations.

So the total possible number of combinations = 11,238,513 x 26 = 292,201,338 or roughly 293 million, and the probability of winning is 1 in 293 million.

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Was the info in this article useful and instructive? How can I improve it? Would you like to ask me any questions?

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Thanks!

## More by this Author

Jodah 10 months ago from Queensland Australia

It's nice to know these equations and the odds of throwing certain numbers of dice, drawing a certain card etc. Very well researched hub , Eugene. However under the heading "Probability of an Event" it says; "There are two types of probability, empirical and empirical."(should the second one be "classical"?)

eugbug 10 months ago from Ireland Author

Thanks Jodah and well spotted! That's what I get for racing through the proof reading!

Digital MD 10 months ago

Thanks for sharing and reiterating the basic mathematics we learn in our early years of schooling! Actually, this topic is very useful in real life even if you engange in a field which does not deal much on numbers such as mine. I agree with Jodah, well-researched hub!

eugbug 10 months ago from Ireland Author

Thanks LM, I learned this stuff in school over 30 years ago, but it was refreshing to revisit it!

Larry Rankin 10 months ago from Oklahoma

Wonderful insight into odds.

eugbug 10 months ago from Ireland Author

Thanks Larry!

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