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

Mellie Apr 26, 2015

#3**+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?

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Alan Apr 26, 2015

#1**+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)

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Alan Apr 26, 2015

#3**+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