Calculating expected value of product of two numbersExpected value for a random variableExpected Value of a LotteryA 3 digit number is produced by randomly by rearranging the digits 1,2,3. the person receives this in money. find the expected value received?Expected Value Intermediate Counting ProblemCalculating expected value of a pareto distributionCard Game: Two Slightly Different Expected Value derivavationsCalculating the expected payoff in a 2-finger morra gameProbability that product of digits of a number is divisible by 3Expected value for a sum of two special diceHow many $4$-digit positive integers with distinct digits are there in which the sum of the first two digits equals the sum of the last two digits?
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Calculating expected value of product of two numbers
Expected value for a random variableExpected Value of a LotteryA 3 digit number is produced by randomly by rearranging the digits 1,2,3. the person receives this in money. find the expected value received?Expected Value Intermediate Counting ProblemCalculating expected value of a pareto distributionCard Game: Two Slightly Different Expected Value derivavationsCalculating the expected payoff in a 2-finger morra gameProbability that product of digits of a number is divisible by 3Expected value for a sum of two special diceHow many $4$-digit positive integers with distinct digits are there in which the sum of the first two digits equals the sum of the last two digits?
$begingroup$
Hi guys want to understand if my process on the following problem is correct or not.
Digits $1$, $2$, $3$ are randomly arranged to form a one-digit number and two-digit number. What is the expected value of the product of the two numbers?
I first knew that there are $6$ distinct ways to create a one-digit number and two-digit number based on the three available numbers ($3!$). I calculated the products of all of these distinct possibilities.
$$(1, 23): 1 × 23 = 23\
(1, 32): 1 × 32 = 32\
(2, 13): 2 × 13 = 26\
(2, 31): 2 × 31 = 62\
(3, 12): 3 × 12 = 36\
(3, 21): 3 × 21 = 63$$
So I calculated the expected value in the following way:
$$E[textProduct]=frac16×23+frac16×32+frac16×26+frac16×62+ frac16×36+frac16×63=40.33.$$
Is my idea like this correct?
probability expected-value
edited Apr 1 at 23:18
heropup
65.4k865104
asked Mar 29 at 22:28
RobinRobin
575
$endgroup$
add a comment |
$begingroup$
Hi guys want to understand if my process on the following problem is correct or not.
Digits $1$, $2$, $3$ are randomly arranged to form a one-digit number and two-digit number. What is the expected value of the product of the two numbers?
I first knew that there are $6$ distinct ways to create a one-digit number and two-digit number based on the three available numbers ($3!$). I calculated the products of all of these distinct possibilities.
$$(1, 23): 1 × 23 = 23\
(1, 32): 1 × 32 = 32\
(2, 13): 2 × 13 = 26\
(2, 31): 2 × 31 = 62\
(3, 12): 3 × 12 = 36\
(3, 21): 3 × 21 = 63$$
So I calculated the expected value in the following way:
$$E[textProduct]=frac16×23+frac16×32+frac16×26+frac16×62+ frac16×36+frac16×63=40.33.$$
Is my idea like this correct?
probability expected-value
edited Apr 1 at 23:18
heropup
65.4k865104
asked Mar 29 at 22:28
RobinRobin
575
$endgroup$
$begingroup$
Your reasoning looks solid.
$endgroup$
– lulu
Mar 29 at 22:33
1
$begingroup$
Better formatting would make it easier to follow your calculation, though. here is a good tutorial on formatting for this site.
$endgroup$
– lulu
Mar 29 at 22:34
$begingroup$
Arithmetically, it would be easier to add up all the numbers and then divide by 6.
$endgroup$
– herb steinberg
Mar 30 at 0:36
add a comment |
$begingroup$
Hi guys want to understand if my process on the following problem is correct or not.
Digits $1$, $2$, $3$ are randomly arranged to form a one-digit number and two-digit number. What is the expected value of the product of the two numbers?
I first knew that there are $6$ distinct ways to create a one-digit number and two-digit number based on the three available numbers ($3!$). I calculated the products of all of these distinct possibilities.
$$(1, 23): 1 × 23 = 23\
(1, 32): 1 × 32 = 32\
(2, 13): 2 × 13 = 26\
(2, 31): 2 × 31 = 62\
(3, 12): 3 × 12 = 36\
(3, 21): 3 × 21 = 63$$
So I calculated the expected value in the following way:
$$E[textProduct]=frac16×23+frac16×32+frac16×26+frac16×62+ frac16×36+frac16×63=40.33.$$
Is my idea like this correct?
probability expected-value
edited Apr 1 at 23:18
heropup
65.4k865104
asked Mar 29 at 22:28
RobinRobin
575
$endgroup$
Hi guys want to understand if my process on the following problem is correct or not.
Digits $1$, $2$, $3$ are randomly arranged to form a one-digit number and two-digit number. What is the expected value of the product of the two numbers?
I first knew that there are $6$ distinct ways to create a one-digit number and two-digit number based on the three available numbers ($3!$). I calculated the products of all of these distinct possibilities.
$$(1, 23): 1 × 23 = 23\
(1, 32): 1 × 32 = 32\
(2, 13): 2 × 13 = 26\
(2, 31): 2 × 31 = 62\
(3, 12): 3 × 12 = 36\
(3, 21): 3 × 21 = 63$$
So I calculated the expected value in the following way:
$$E[textProduct]=frac16×23+frac16×32+frac16×26+frac16×62+ frac16×36+frac16×63=40.33.$$
Is my idea like this correct?
probability expected-value
probability expected-value
edited Apr 1 at 23:18
heropup
65.4k865104
asked Mar 29 at 22:28
RobinRobin
575
edited Apr 1 at 23:18
heropup
65.4k865104
asked Mar 29 at 22:28
RobinRobin
575
edited Apr 1 at 23:18
heropup
65.4k865104
edited Apr 1 at 23:18
heropup
65.4k865104
edited Apr 1 at 23:18
heropup
65.4k865104
65.4k865104
asked Mar 29 at 22:28
RobinRobin
575
asked Mar 29 at 22:28
RobinRobin
575
asked Mar 29 at 22:28
RobinRobin
575
575
$begingroup$
Your reasoning looks solid.
$endgroup$
– lulu
Mar 29 at 22:33
1
$begingroup$
Better formatting would make it easier to follow your calculation, though. here is a good tutorial on formatting for this site.
$endgroup$
– lulu
Mar 29 at 22:34
$begingroup$
Arithmetically, it would be easier to add up all the numbers and then divide by 6.
$endgroup$
– herb steinberg
Mar 30 at 0:36
add a comment |
$begingroup$
Your reasoning looks solid.
$endgroup$
– lulu
Mar 29 at 22:33
1
$begingroup$
Better formatting would make it easier to follow your calculation, though. here is a good tutorial on formatting for this site.
$endgroup$
– lulu
Mar 29 at 22:34
$begingroup$
Arithmetically, it would be easier to add up all the numbers and then divide by 6.
$endgroup$
– herb steinberg
Mar 30 at 0:36
$begingroup$
Your reasoning looks solid.
$endgroup$
– lulu
Mar 29 at 22:33
$begingroup$
Your reasoning looks solid.
$endgroup$
– lulu
Mar 29 at 22:33
1
1
$begingroup$
Better formatting would make it easier to follow your calculation, though. here is a good tutorial on formatting for this site.
$endgroup$
– lulu
Mar 29 at 22:34
$begingroup$
Better formatting would make it easier to follow your calculation, though. here is a good tutorial on formatting for this site.
$endgroup$
– lulu
Mar 29 at 22:34
$begingroup$
Arithmetically, it would be easier to add up all the numbers and then divide by 6.
$endgroup$
– herb steinberg
Mar 30 at 0:36
$begingroup$
Arithmetically, it would be easier to add up all the numbers and then divide by 6.
$endgroup$
– herb steinberg
Mar 30 at 0:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A more formal way to perform this computation is to observe that if $(a,b,c)$ represents a permutation of the set $1,2,3$, then take $f(a,b,c) = (10a + b)c$ as the product formed. Then we observe $$f(a,b,c) + f(b,c,a) + f(c,a,b) = (10a + b)c + (10b + c)a + (10c + a)b = 11(ab + bc + ca),$$ and because this expression is symmetric in $a, b, c$, the sum of the other three permutations is also equal to this value. Therefore, the expectation is simply $$frac226(ab+bc+ca)$$ and without loss of generality we may take $a = 1$, $b = 2$, $c = 3$ to obtain $$frac226(2 + 6 + 3) = frac1213.$$
It is worth noting that this solution applies to any choice of three distinct decimal digits $1$ through $9$.
answered Mar 30 at 0:36
heropupheropup
65.4k865104
$endgroup$
add a comment |
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1 Answer
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$begingroup$
A more formal way to perform this computation is to observe that if $(a,b,c)$ represents a permutation of the set $1,2,3$, then take $f(a,b,c) = (10a + b)c$ as the product formed. Then we observe $$f(a,b,c) + f(b,c,a) + f(c,a,b) = (10a + b)c + (10b + c)a + (10c + a)b = 11(ab + bc + ca),$$ and because this expression is symmetric in $a, b, c$, the sum of the other three permutations is also equal to this value. Therefore, the expectation is simply $$frac226(ab+bc+ca)$$ and without loss of generality we may take $a = 1$, $b = 2$, $c = 3$ to obtain $$frac226(2 + 6 + 3) = frac1213.$$
It is worth noting that this solution applies to any choice of three distinct decimal digits $1$ through $9$.
answered Mar 30 at 0:36
heropupheropup
65.4k865104
$endgroup$
add a comment |
$begingroup$
A more formal way to perform this computation is to observe that if $(a,b,c)$ represents a permutation of the set $1,2,3$, then take $f(a,b,c) = (10a + b)c$ as the product formed. Then we observe $$f(a,b,c) + f(b,c,a) + f(c,a,b) = (10a + b)c + (10b + c)a + (10c + a)b = 11(ab + bc + ca),$$ and because this expression is symmetric in $a, b, c$, the sum of the other three permutations is also equal to this value. Therefore, the expectation is simply $$frac226(ab+bc+ca)$$ and without loss of generality we may take $a = 1$, $b = 2$, $c = 3$ to obtain $$frac226(2 + 6 + 3) = frac1213.$$
It is worth noting that this solution applies to any choice of three distinct decimal digits $1$ through $9$.
answered Mar 30 at 0:36
heropupheropup
65.4k865104
$endgroup$
add a comment |
$begingroup$
A more formal way to perform this computation is to observe that if $(a,b,c)$ represents a permutation of the set $1,2,3$, then take $f(a,b,c) = (10a + b)c$ as the product formed. Then we observe $$f(a,b,c) + f(b,c,a) + f(c,a,b) = (10a + b)c + (10b + c)a + (10c + a)b = 11(ab + bc + ca),$$ and because this expression is symmetric in $a, b, c$, the sum of the other three permutations is also equal to this value. Therefore, the expectation is simply $$frac226(ab+bc+ca)$$ and without loss of generality we may take $a = 1$, $b = 2$, $c = 3$ to obtain $$frac226(2 + 6 + 3) = frac1213.$$
It is worth noting that this solution applies to any choice of three distinct decimal digits $1$ through $9$.
answered Mar 30 at 0:36
heropupheropup
65.4k865104
$endgroup$
A more formal way to perform this computation is to observe that if $(a,b,c)$ represents a permutation of the set $1,2,3$, then take $f(a,b,c) = (10a + b)c$ as the product formed. Then we observe $$f(a,b,c) + f(b,c,a) + f(c,a,b) = (10a + b)c + (10b + c)a + (10c + a)b = 11(ab + bc + ca),$$ and because this expression is symmetric in $a, b, c$, the sum of the other three permutations is also equal to this value. Therefore, the expectation is simply $$frac226(ab+bc+ca)$$ and without loss of generality we may take $a = 1$, $b = 2$, $c = 3$ to obtain $$frac226(2 + 6 + 3) = frac1213.$$
It is worth noting that this solution applies to any choice of three distinct decimal digits $1$ through $9$.
answered Mar 30 at 0:36
heropupheropup
65.4k865104
answered Mar 30 at 0:36
heropupheropup
65.4k865104
answered Mar 30 at 0:36
heropupheropup
65.4k865104
answered Mar 30 at 0:36
heropupheropup
65.4k865104
65.4k865104
add a comment |
add a comment |
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$begingroup$
Your reasoning looks solid.
$endgroup$
– lulu
Mar 29 at 22:33
1
$begingroup$
Better formatting would make it easier to follow your calculation, though. here is a good tutorial on formatting for this site.
$endgroup$
– lulu
Mar 29 at 22:34
$begingroup$
Arithmetically, it would be easier to add up all the numbers and then divide by 6.
$endgroup$
– herb steinberg
Mar 30 at 0:36