Math expectations
- Thread starter jac213
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pka
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If \(\displaystyle X_k\) is the outcome on the kth roll then \(\displaystyle \sum\limits_{k = 1}^{10} {X_k }\) is the overall sum.jac213 said:
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!
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.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** Consider a one-die game.
. . \(\displaystyle \begin{array}{c|c} \text{Win} & \text{Prob}\\ \hline \\[-3mm] \$1 & \frac{1}{6} \\ \\[-3mm] \$2 & \frac{1}{6} \\ \\[-3mm] \$3 & \frac{1}{6} \\ \\[-3mm] \$4 & \frac{1}{6} \\ \\[-3mm] \$5 & \frac{1}{6} \\ \\[-3mm] \$6 & \frac{1}{6} \end{array}\)
\(\displaystyle \text{Then: }\:E \;=\;1\left(\tfrac{1}{6}\right) + 2\left(\tfrac{1}{6}\right) + 3\left(\tfrac{1}{6}\right) + 4\left(\tfrac{1}{6}\right) + 5\left(\tfrac{1}{6}\right) + 6\left(\tfrac{1}{6}\right) \;=\;\frac{21}{6}\)
. . . . \(\displaystyle \text{Therefore: }\;E \;=\;\$3.50\qquad \left(\frac{1+6}{2} = 3.5\right)\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** Consider a two-dice game.
. . \(\displaystyle \begin{array}{c|c}\text{Win} & \text{Prob} \\ \hline \\[-3mm] 2 & \frac{1}{36} \\ \\[-3mm] 3 & \frac{2}{36} \\ \\[-3mm] 4 & \frac{3}{36} \\ \\[-3mm] 5 & \frac{4}{36} \\ \\[-3mm] 6 & \frac{5}{36} \\ \\[-3mm] 7 & \frac{6}{\36} \end{array} \qquad\qquad\begin{array}{c|c}\text{Win} & \text{Prob} \\ \hline \\[-3mm] 8 & \frac{5}{36} \\ \\[-3mm] 9 & \frac{4}{36} \\ \\[-3mm] 10 & \frac{3}{36} \\ \\[-3mm] 11 & \frac{2}{36} \\ \\[-3mm] 12 & \frac{1}{36} \end{array}\)
\(\displaystyle \text{Then: }\;E \;=\;2\left(\tfrac{1}{6}\right) + 3\left(\tfrac{2}{36}\right) + 4\left(\tfrac{3}{36}\right) + \hdots 10\left(\tfrac{3}{36}\right) + 11\left(\tfrac{2}{36}\right) + 12\left(\tfrac{1}{36}\right) \;=\;\frac{252}{36}\)
. . \(\displaystyle \text{Therefore: }\;E \;=\;\$7 \qquad\left(\frac{2+12}{2} = 7\right)\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** Consider a 3-dice game.
. . \(\displaystyle \begin{array}{c|c} \text{Win} & \text{Prob} \\ \hline \\[-3mm] 3 & \frac{1}{216} \\ \\[-3mm] 4 & \frac{3}{216} \\ \\[-3mm] 5 & \frac{6}{216}\\ \\[-3mm] 6 & \frac{10}{216}\\ \\[-3mm] 7 & \frac{15}{216}\\ \\[-3mm] 8 & \frac{21}{216}\\ \\[-3mm] 9 & \frac{25}{216}\\ \\[-3mm] 10 & \frac{27}{216} \end{array} \qquad\begin{array}{c|c}\text{Win} & {Prob} \\ \hline \\[-3mm] 11 & \frac{27}{216}\\ \\[-3mm] 12 & \frac{25}{216}\\ \\[-3mm] 13 & \frac{21}{216}\\ \\[-3mm] 14 & \frac{15}{216}\\ \\[-3mm] 15 & \frac{10}{216}\\ \\[-3mm] 16 & \frac{6}{216}\\ \\[-3mm] 17 & \frac{3}{216}\\ \\[-3mm] 18 & \frac{1}{216} \end{array}\)
\(\displaystyle \text{Then: }\;E \;=\;3\left(\tfrac{1}{216}\right) + 4\left(\tfrac{3}{216}\right) + 5\left(\tfrac{6}{216}\right) + \hdots + 16\left(\tfrac{6}{216}\right) + 17\left(\tfrac{3}{216}\right) + 18\left(\tfrac{1}{216}\right)\;=\;\frac{2268}{216}\)
. . \(\displaystyle \text{Therefore: } \;E \;=\;\$10.50\qquad \left(\frac{3+18}{2} = 10.5\right)\)
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.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** Consider a one-die game.
. . \(\displaystyle \begin{array}{c|c} \text{Win} & \text{Prob}\\ \hline \\[-3mm] \$1 & \frac{1}{6} \\ \\[-3mm] \$2 & \frac{1}{6} \\ \\[-3mm] \$3 & \frac{1}{6} \\ \\[-3mm] \$4 & \frac{1}{6} \\ \\[-3mm] \$5 & \frac{1}{6} \\ \\[-3mm] \$6 & \frac{1}{6} \end{array}\)
\(\displaystyle \text{Then: }\:E \;=\;1\left(\tfrac{1}{6}\right) + 2\left(\tfrac{1}{6}\right) + 3\left(\tfrac{1}{6}\right) + 4\left(\tfrac{1}{6}\right) + 5\left(\tfrac{1}{6}\right) + 6\left(\tfrac{1}{6}\right) \;=\;\frac{21}{6}\)
. . . . \(\displaystyle \text{Therefore: }\;E \;=\;\$3.50\qquad \left(\frac{1+6}{2} = 3.5\right)\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** Consider a two-dice game.
. . \(\displaystyle \begin{array}{c|c}\text{Win} & \text{Prob} \\ \hline \\[-3mm] 2 & \frac{1}{36} \\ \\[-3mm] 3 & \frac{2}{36} \\ \\[-3mm] 4 & \frac{3}{36} \\ \\[-3mm] 5 & \frac{4}{36} \\ \\[-3mm] 6 & \frac{5}{36} \\ \\[-3mm] 7 & \frac{6}{\36} \end{array} \qquad\qquad\begin{array}{c|c}\text{Win} & \text{Prob} \\ \hline \\[-3mm] 8 & \frac{5}{36} \\ \\[-3mm] 9 & \frac{4}{36} \\ \\[-3mm] 10 & \frac{3}{36} \\ \\[-3mm] 11 & \frac{2}{36} \\ \\[-3mm] 12 & \frac{1}{36} \end{array}\)
\(\displaystyle \text{Then: }\;E \;=\;2\left(\tfrac{1}{6}\right) + 3\left(\tfrac{2}{36}\right) + 4\left(\tfrac{3}{36}\right) + \hdots 10\left(\tfrac{3}{36}\right) + 11\left(\tfrac{2}{36}\right) + 12\left(\tfrac{1}{36}\right) \;=\;\frac{252}{36}\)
. . \(\displaystyle \text{Therefore: }\;E \;=\;\$7 \qquad\left(\frac{2+12}{2} = 7\right)\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** Consider a 3-dice game.
. . \(\displaystyle \begin{array}{c|c} \text{Win} & \text{Prob} \\ \hline \\[-3mm] 3 & \frac{1}{216} \\ \\[-3mm] 4 & \frac{3}{216} \\ \\[-3mm] 5 & \frac{6}{216}\\ \\[-3mm] 6 & \frac{10}{216}\\ \\[-3mm] 7 & \frac{15}{216}\\ \\[-3mm] 8 & \frac{21}{216}\\ \\[-3mm] 9 & \frac{25}{216}\\ \\[-3mm] 10 & \frac{27}{216} \end{array} \qquad\begin{array}{c|c}\text{Win} & {Prob} \\ \hline \\[-3mm] 11 & \frac{27}{216}\\ \\[-3mm] 12 & \frac{25}{216}\\ \\[-3mm] 13 & \frac{21}{216}\\ \\[-3mm] 14 & \frac{15}{216}\\ \\[-3mm] 15 & \frac{10}{216}\\ \\[-3mm] 16 & \frac{6}{216}\\ \\[-3mm] 17 & \frac{3}{216}\\ \\[-3mm] 18 & \frac{1}{216} \end{array}\)
\(\displaystyle \text{Then: }\;E \;=\;3\left(\tfrac{1}{216}\right) + 4\left(\tfrac{3}{216}\right) + 5\left(\tfrac{6}{216}\right) + \hdots + 16\left(\tfrac{6}{216}\right) + 17\left(\tfrac{3}{216}\right) + 18\left(\tfrac{1}{216}\right)\;=\;\frac{2268}{216}\)
. . \(\displaystyle \text{Therefore: } \;E \;=\;\$10.50\qquad \left(\frac{3+18}{2} = 10.5\right)\)