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Odds vs. Probability

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Date: 12/06/96 at 12:06:47
From: Jessica Peace
Subject: Calculating odds

Here is the question:  Assume four individuals are each drawing two
numbers from a box that contains eight numbers.  The numbers in the
box are assigned in numerical order, 1-8.  Also assume that the four
individuals are identified in alphabetical order: A, B, C, and D.

Individual 'A' wants to draw numbers 2 & 7; individuals B, C, and D
have no preference as to which two numbers they pull from the box.
What are the odds, or how do I write the formula to determine the
odds, that individual 'A' will draw the two numbers that he wants?

Also, what will be the odds if the numbers in the box are grouped by
lots of two; i.e. 2 and 7, 1 and 8, 3 and 6, and 4 and 5?  Would this
mean the odds would be 4:1 that Individual 'A' would draw the lot of
2 and 7?  How would the order of drawing from the box affect the odds
in both scenarios?

Thank you.
Jessica
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Date: 12/06/96 at 15:46:41
From: Doctor Wilkinson
Subject: Re: calculating odds

First of all, it's important to know how the term "odds" is used, as
opposed to "probability".

If I toss a coin, it may come down heads or tails. The two
possibilities are equally likely (assuming the coin is fair, etc).

The probability of it coming down heads is 1/2, because that is one
of the two equally likely possibilities.

The term "odds" is used usually with "against" and is used to compare
the unfavorable with the favorable possibilities (instead of the
favorable possibilities with all the possibilities), so the odds
against the coin's coming down heads are 1:1.

Similarly, if you roll a die, the probability of a 3 is 1/6, since
there are six possibilities, one of which is a 3.  The odds against a
3 are 5:1, since there are 5 unfavorable possibilities and one
favorable.  The probability of throwing either a 2 or a 3 is 2/6 or
1/3, and the odds against are 2:1.

It is easy to translate from probability to odds, but they are not the
same thing.

Now that I've confused you thoroughly, let's look at your problem.

The other players (B, C, and D) really have nothing to do with it.
A picks two numbers out of eight, and there is only one outcome that
he wants.  So the key to the problem is to figure out how many
possible pairs of numbers there are.  Then the probability of getting
a 2 and a 7 will just be 1 divided by the total number of
possibilities, and the odds against will be the ratio of the number of
unfavorable possiblities to 1.

So how many possibilities are there altogether?  If you don't know how
to count them, let me know.

For the second part of the problem, you have only 4 possibilities, one
favorable and 3 unfavorable, so the probability of winning is 1/4, and
the odds are 3:1 against.

-Doctor Wilkinson,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Permutations and Combinations
High School Probability

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