Michael's expected value

Firstly, note that there are ${10 \choose 3} = 120$ ways of choosing $3$ distinct numbers from the geometric series. Now to find the expected value of the sum, we need to sum each possible combination and divide the total by $120$ (because the numbers are selected uniformly at random). So the sum is: $\frac{\left (\frac{10}{9} + \frac{20}{9} + \frac{40}{9}\right ) + \left (\frac{10}{9} + \frac{20}{9} + \frac{80}{9}\right ) + \dots}{120}$ Now, notice that each term will appear ${9 \choose 2} = 36$ times in the sum, so the sum can be simplified to: $\frac{36 \cdot \left (\frac{10}{9} + \frac{20}{9} + \frac{40}{9} + \dots + \frac{5120}{9} \right )}{120}$ Now taking out factors we further simplify the sum to: $\frac{10 \cdot 36 \cdot (1 + 2 + 4 + \dots + 512)}{9 \cdot 120} = \frac{1 + 2 + 4 + \dots + 512}{3}$ Now we will evaluate the geometric series $1 + 2 + 4 + \dots + 512$ . This is equal to: $S_{10} = \frac{1 \cdot (2^{10} - 1)}{2 - 1} = 2^{10} - 1 = 1023$ Therefore, the expected value is: $\frac{1023}{3} = \fbox{341}$

• The sum of the series is $a_1 \dfrac{q^n - 1}{q - 1} = \dfrac{10}{9} \dfrac{2^10 - 1}{2 - 1} = 10 \dfrac{341}{3}$ .
• The average of the 10 numbers is then $\dfrac{341}{3}$ .
• So on average, the sum of three numbers would be three times this number $\Rightarrow \boxed{341}$ .

Let us calculate the result for: 1, 2, 4, ... 512

We will multiply this result by 10 / 9 at the end.

With any of these numbers selected, we can select any 2 other numbers from the rest in 9 * 8/2 = 36 ways Adding all the cases for all the numbers will give us: (1+2+4+8+...+512) * 36 = 1023 * 36

The total number of ways of selecting 3 nums is of course the number of ways of selecting 3 things from 10, which is 120

Hence, result = 1023 * 36 / 120

Multiplying this result by 10 / 9 we get , final result = 1023 * 36 / 120 * 10 / 9 = 341

The number of ways of picking three numbers from ten is $10\choose3$ $=120$ If we fix a certain number of the geometric series- the number of ways we could choose two other is $9\choose2$ $= 36$ So if we pick a number from the series for our three number sum it could appear in $36$ sums.

Now, our $Expected Sum$ $=\frac{Sum of All Possible Sums}{Number of All Possible Sums}$

If you think about the $Sum of All Possible Sums$ its really just $36\times$ the sum of all the numbers in the geometric series. Since in adding all of the possible sums each member of the geometric series will appear $36$ times. We can factor out the $36$ to get a sum of a geometric series. So assuming the reader is aware of the sum of a geometric series we get:

$Expected Sum$ $=\frac{Sum of All Possible Sums}{Number of All Possible Sums} =\frac{36(\frac{10}{9})(2^{10}-1)}{120}=\boxed{341}$

all ideas are about symmetric.

let $x_1.x_2,x_3$ are the numbers that choose from the set of $a_i,i=1,2,3,...,10$ .

easily to find the number of ways to write $(x_1,x_2,x_3)$ that is $\binom{10}{3}=120$ . when you list $(x_1,x_2,x_3)$ for each $x_1$ you will find $\binom{9}{2}=36$ ways to choose $x_2,x_3$ .

so the expected value that you want will be

$\sum_{i=1}^{10}[\frac{1}{\binom{10}{3}}\cdot \binom{9}{2} \cdot a_i] = 341$

The key to solving this problem is to write out the expected value as an expression, and then cleverly simplify the expression.

Note that the expected value of the sum is

$(\frac{1}{\binom{10}{3}})(\frac{10}{9}+\frac{20}{9}+\frac{40}{9}) + (\frac{1}{\binom{10}{3}})(\frac{10}{9}+\frac{20}{9}+\frac{80}{9}) + (\frac{1}{\binom{10}{3}})(\frac{10}{9}+\frac{20}{9}+\frac{160}{9})...$

where the sum continues for all possible triplets of elements. If we expand and collect, we get

$(\frac{1}{\binom{10}{3}})(\frac{10}{9}+\frac{10}{9}+\frac{10}{9} \ldots \frac{10}{9} + \frac{20}{9}+\frac{20}{9}+\frac{20}{9} \ldots \frac{20}{9}+\frac{40}{9}+\frac{40}{9}+\frac{40}{9} \ldots \frac{40}{9}...\frac{5120}{9}+\frac{5120}{9}+\frac{5120}{9} \ldots \frac{5120}{9})$

How many times does each term above repeat? Each term is used in a triplet with two other distinct elements of the sequence. There are $\binom{9}{2}$ ways to choose the other two elements, so each term must appear $\binom{9}{2}$ times. The simplification of the above quantity is now trivial (and is left to the reader), leaving a final answer of $\boxed{341}$ .

The approach I used to solve this problem is the following.

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Definitions:

We first note that the $n^{th}$ term of a geometric sequence is given by $a_n = a_1 r^{n-1}$ . We then define the set $S$ containing the ten-term geometric sequence where $a_1 = \frac{10}{9}$ and $r = 2$ .

$\ \ \ \ \ \ \ \ \ S = \{\frac{10}{9} \times 2^{n-1}\} _{n=1}^{10}$

Furthermore, we define the set $T$ containing all the triples formed from $S$ .

$\ \ \ \ \ \ \ \ \ T = \{ s \in S\ \ \ |\ \ \ \ |s| = 3\}$

The number of these triples is equal to $|T| = \ _{10}C_{3} = 120$ . The set of the sum of these triples can then be defined as $\sigma = \{\sum\limits_{l = 1}^{|s|} s_{n,l} \}_{s_n \in T}$ .

Here, we should note that a triple $s_n$ has the form $\frac{10}{9}(2^i, 2^j, 2^k)_n$ where $i\neq j\neq k$ and $i,j,k \in [0,9]$ .

We can then get the solution by solving the mean value of $\sigma$ .

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Calculation of mean value of $\sigma$ :

The mean value of $\sigma$ is defined as $\bar{\sigma} = \frac{\sum\limits_{i=1}^{|\sigma|} \sigma_i}{|\sigma|}$ , where $|\sigma| = |T| = 120$ . With this, we're left with the task of calculating the value of $\sum\limits_{i=1}^{|\sigma|} \sigma_i$ .

We can simplify the summation by noting that in the entire collection of triples, a specific number is repeated 36 times. This is found by the fact that a certain number in the set $S$ will be paired with any other two numbers in $\ _{9}C_{2} = 36$ ways.

Since we are solving for the sum of all the triples, we find that the value of the summation is equal to $36 \times \frac{10}{9} \times \sum\limits_{n=1}^{10} 2^{n-1}$ .

Combining the numerator and the denominator, we get

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bar{\sigma} = \frac{36 \times \frac{10}{9} \times \sum\limits_{n=1}^{10} 2^{n-1}}{120}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bar{\sigma} = \frac{36 \times \frac{10}{9} \times 1023}{120}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bar{\sigma} = 341$

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Answer: $\bar{\sigma} = 341$

Though this question may appear to be quite complicated, it is actually pretty simple.

The numbers in the sequence are $\frac{10}{9}, \frac{20}{9}, \frac{40}{9}, \ldots \frac{5120}{9}$ We know this because the first term is $\frac{10}{9}$ and the common ration is 2.

The sum of these terms is equal to $\frac{10+20+40+...+2560+5120}{9}=\frac{10230}{9}$ by using the fact that

$\qquad \qquad \qquad \quad \quad \quad 2^0+2^1+\ldots + 2^n = 2^{n+1}$

However, we do not want to add all ten; we just want to choose three of the ten. So, we have

$\qquad\qquad \qquad\qquad \frac{3}{10}\cdot\frac{10230}{9}=\frac{3\cdot10230}{10\cdot 9}=\frac{1023}{3}=\boxed{341}$

We will choose 3 numbers from 10 numbers in 10C3 = 120 ways. Since in each choice we choose 3 numbers we are actually choosing (3*120) = 360 numbers. Thus each number is chosen (360 / 10) = 36 times

So we multiply 36 to each of the numbers 10/9,20/9,.....,5120/9 and get 40,80,...,20480.

Adding this (40 + 80 + .... + 20480) we get 40920 and since this is the sum of all the numbers taken in all possible groups of three dividing this by 120 gives us the answer as 341

First of all, I consider the sum of all possible groups of three numbers. To do that, I calculate the number of times that each member appears in this sum. We select a random term. We can select the other two terms in $\binom{9}{2}=36$ . So this sum is: $36\cdot \frac{10}{9} \cdot {1+2+4+...+512}=40 \cdot 1023$ since the sum of the first $n$ powers of two is $2^{n+1}-1$ . To obtain the average value, we have to divide this sum by the number of possible terns, which is $\binom{10}{3}=120$ . So $\frac{40 \cdot 1023 }{120}=341$ which is the answer.

The sequence runs (10/9) * 1, (10/9) * 2, (10/9) * 4, ... (10/9) * 512. Now each term will be added into the sum in (9 choose 2) triples , since we choose the other two distinct numbers from the remaining nine. Then 10/9 (1+2+4+... +512) * (9 c 2) is the total sum. Also, 1/(10 c 3) is the probability that each triple will occur. So we multiply the two together to get the expected value. Using the identity 1+2+4...+ 2^n = 2^(n+1) -1, we can evaluate this expression quickly and get the answer 341.