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An eight-sided die has its faces numbered from 1 to 8. What is the expected value of the roll of the die?

 Apr 26, 2015

Best Answer 

 #3
avatar+33266 
+10

There is a probability of 1/8 that the number 1 will show.  

There is a probability of 1/8 that the number 2 will show.

 and so on until you get to:

There is a probability of 1/8 that the number 8 will show.

 

The expected value is the sum of (probability that the number will show)*(the number) or in this case:

(1/8)*1 + (1/8)*2 + (1/8)*3 + ... +(1/8)*8

or

(1/8)*(1  2 + 3 + 4 + 5 + 6 + 7 + 8) = (1/8)*36 = 4.5

 

We can use a shortcut to sum the numbers 1 through 8.  It is given by 8*(8+1)/2.

 

Does this help?

 Apr 26, 2015
 #1
avatar+33266 
+10

Each side has a probability of 1/8 of showing (assuming a fair die).  The expected value is one eighth of the sum of the numbers from 1 to 8 or (1/8)*8*9/2 = 9/2 = 4.5

 

(The sum of the integers from 1 to n is n(n+1)/2)

 Apr 26, 2015
 #2
avatar+1821 
+5

I don't understand. Sorry Alan.

 Apr 26, 2015
 #3
avatar+33266 
+10
Best Answer

There is a probability of 1/8 that the number 1 will show.  

There is a probability of 1/8 that the number 2 will show.

 and so on until you get to:

There is a probability of 1/8 that the number 8 will show.

 

The expected value is the sum of (probability that the number will show)*(the number) or in this case:

(1/8)*1 + (1/8)*2 + (1/8)*3 + ... +(1/8)*8

or

(1/8)*(1  2 + 3 + 4 + 5 + 6 + 7 + 8) = (1/8)*36 = 4.5

 

We can use a shortcut to sum the numbers 1 through 8.  It is given by 8*(8+1)/2.

 

Does this help?

Alan Apr 26, 2015
 #4
avatar+1821 
+5

Yup! I understand now! Sorry for all the trouble!

 Apr 26, 2015

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