Math expectations

jac213

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Jun 6, 2009
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I'm a little confused on the following problem:

A fair die is rolled 10 times. Find the math expectation of the numerical sum of the outcomes
 
jac213 said:
A fair die is rolled 10 times. Find the math expectation of the numerical sum of the outcomes
Are you asking for the sum of the outcomes on each of the ten rolls?
If so, the least value is 10: ten ones.
The greatest sum is 60: ten sixes.

If this is not what you mean, then please clarify.
 
I know for one roll, the chance of getting any one number is 1/6. What I'm confused about is the numerical sum of the outcomes part. Not exactly sure what that means.
 
jac213 said:
I know for one roll, the chance of getting any one number is 1/6. What I'm confused about is the numerical sum of the outcomes part. Not exactly sure what that means.
Well, that does not answer the question I posted to you.
What sum are you asking about?
Is it the sum of all ten outcomes?
 
jac213 said:
It is asking for the sum of 10 outcomes
If \(\displaystyle X_k\) is the outcome on the kth roll then \(\displaystyle \sum\limits_{k = 1}^{10} {X_k }\) is the overall sum.
What do you know about expected value?
Does \(\displaystyle E\left( {\sum\limits_{k = 1}^{10} {X_k } } \right) = \sum\limits_{k = 1}^{10} {E\left( {X_k } \right)} ?\).
 
Hello, jac213!

A fair die is rolled 10 times.
Find the math expectation of the numerical sum of the outcomes

I believe the expected value is the average of the smallest possible sum and largest possible sum, **
. . but I have not seen a proof of this. .(Perhaps someone here can devise one.)

With 10 rolls, the smallest sum is 10 and the largest sum is 60.
. . Hence, the expected value is 35.


To illustrate this value, consider this proposition.
You are to roll ten dice and will receive (in the dollars) the total of the ten dice.
What is your expectation?

With no fees or penalties, you will show a profit, of course.
Playing this game repeatedly, you can expect to win an average of $35 per game.


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** Consider a one-die game.

. . \(\displaystyle \begin{array}{c|c} \text{Win} & \text{Prob}\\ \hline \\[-3mm] \$1 & \frac{1}{6} \\ \\[-3mm] \$2 & \frac{1}{6} \\ \\[-3mm] \$3 & \frac{1}{6} \\ \\[-3mm] \$4 & \frac{1}{6} \\ \\[-3mm] \$5 & \frac{1}{6} \\ \\[-3mm] \$6 & \frac{1}{6} \end{array}\)

\(\displaystyle \text{Then: }\:E \;=\;1\left(\tfrac{1}{6}\right) + 2\left(\tfrac{1}{6}\right) + 3\left(\tfrac{1}{6}\right) + 4\left(\tfrac{1}{6}\right) + 5\left(\tfrac{1}{6}\right) + 6\left(\tfrac{1}{6}\right) \;=\;\frac{21}{6}\)

. . . . \(\displaystyle \text{Therefore: }\;E \;=\;\$3.50\qquad \left(\frac{1+6}{2} = 3.5\right)\)

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** Consider a two-dice game.

. . \(\displaystyle \begin{array}{c|c}\text{Win} & \text{Prob} \\ \hline \\[-3mm] 2 & \frac{1}{36} \\ \\[-3mm] 3 & \frac{2}{36} \\ \\[-3mm] 4 & \frac{3}{36} \\ \\[-3mm] 5 & \frac{4}{36} \\ \\[-3mm] 6 & \frac{5}{36} \\ \\[-3mm] 7 & \frac{6}{\36} \end{array} \qquad\qquad\begin{array}{c|c}\text{Win} & \text{Prob} \\ \hline \\[-3mm] 8 & \frac{5}{36} \\ \\[-3mm] 9 & \frac{4}{36} \\ \\[-3mm] 10 & \frac{3}{36} \\ \\[-3mm] 11 & \frac{2}{36} \\ \\[-3mm] 12 & \frac{1}{36} \end{array}\)

\(\displaystyle \text{Then: }\;E \;=\;2\left(\tfrac{1}{6}\right) + 3\left(\tfrac{2}{36}\right) + 4\left(\tfrac{3}{36}\right) + \hdots 10\left(\tfrac{3}{36}\right) + 11\left(\tfrac{2}{36}\right) + 12\left(\tfrac{1}{36}\right) \;=\;\frac{252}{36}\)

. . \(\displaystyle \text{Therefore: }\;E \;=\;\$7 \qquad\left(\frac{2+12}{2} = 7\right)\)


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** Consider a 3-dice game.

. . \(\displaystyle \begin{array}{c|c} \text{Win} & \text{Prob} \\ \hline \\[-3mm] 3 & \frac{1}{216} \\ \\[-3mm] 4 & \frac{3}{216} \\ \\[-3mm] 5 & \frac{6}{216}\\ \\[-3mm] 6 & \frac{10}{216}\\ \\[-3mm] 7 & \frac{15}{216}\\ \\[-3mm] 8 & \frac{21}{216}\\ \\[-3mm] 9 & \frac{25}{216}\\ \\[-3mm] 10 & \frac{27}{216} \end{array} \qquad\begin{array}{c|c}\text{Win} & {Prob} \\ \hline \\[-3mm] 11 & \frac{27}{216}\\ \\[-3mm] 12 & \frac{25}{216}\\ \\[-3mm] 13 & \frac{21}{216}\\ \\[-3mm] 14 & \frac{15}{216}\\ \\[-3mm] 15 & \frac{10}{216}\\ \\[-3mm] 16 & \frac{6}{216}\\ \\[-3mm] 17 & \frac{3}{216}\\ \\[-3mm] 18 & \frac{1}{216} \end{array}\)

\(\displaystyle \text{Then: }\;E \;=\;3\left(\tfrac{1}{216}\right) + 4\left(\tfrac{3}{216}\right) + 5\left(\tfrac{6}{216}\right) + \hdots + 16\left(\tfrac{6}{216}\right) + 17\left(\tfrac{3}{216}\right) + 18\left(\tfrac{1}{216}\right)\;=\;\frac{2268}{216}\)

. . \(\displaystyle \text{Therefore: } \;E \;=\;\$10.50\qquad \left(\frac{3+18}{2} = 10.5\right)\)

 
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