Expected Value for uniform random permutations using random variables
Question: Let $n geq 2$ be an integer. You are given n beer bottles $B_1, B_2,...B_n$ and one cider bottle $C$. Consider a uniformly random permutation of these $n + 1$ bottles. The positions in this permutation are numbered as $1, 2, 3,...,n + 1$.
Define the random variable X to be:
X= the position of the leftmost beer bottle.
What is the expected value E(X) of the random variable X?
Answer: $frac{(n+2)}{(n+1)}$
Attempt:
Total ways to arrange bottles: (n+1)! ways
The position of the leftmost beer: 1 way
E(X) = $$ sum_{i=0}^{n+1} i*Pr(X = position, of, the, leftmost, beer)$$
E(X) = $$ sum_{i=0}^{n+1} = 1*frac{{n+1 choose 1}*1}{(n+1)!} $$
$$ sum_{i=0}^{n+1} = frac{1}{(n-1)!} $$
Can't seem to simplify it to get the correct answer.
probability probability-theory discrete-mathematics random-variables expected-value
edited Nov 27 '18 at 20:14
greedoid
38.2k114795
asked Nov 27 '18 at 20:03
Toby
1577
add a comment |
Question: Let $n geq 2$ be an integer. You are given n beer bottles $B_1, B_2,...B_n$ and one cider bottle $C$. Consider a uniformly random permutation of these $n + 1$ bottles. The positions in this permutation are numbered as $1, 2, 3,...,n + 1$.
Define the random variable X to be:
X= the position of the leftmost beer bottle.
What is the expected value E(X) of the random variable X?
Answer: $frac{(n+2)}{(n+1)}$
Attempt:
Total ways to arrange bottles: (n+1)! ways
The position of the leftmost beer: 1 way
E(X) = $$ sum_{i=0}^{n+1} i*Pr(X = position, of, the, leftmost, beer)$$
E(X) = $$ sum_{i=0}^{n+1} = 1*frac{{n+1 choose 1}*1}{(n+1)!} $$
$$ sum_{i=0}^{n+1} = frac{1}{(n-1)!} $$
Can't seem to simplify it to get the correct answer.
probability probability-theory discrete-mathematics random-variables expected-value
edited Nov 27 '18 at 20:14
greedoid
38.2k114795
asked Nov 27 '18 at 20:03
Toby
1577
Please be extremely careful in writing summations. Remember that the expectation of a random variable $X$ is summing all products of the form $i cdot mathrm{Pr}(X = i).$ Adjust this in edits.
– Sean Roberson
Nov 27 '18 at 20:08
add a comment |
Question: Let $n geq 2$ be an integer. You are given n beer bottles $B_1, B_2,...B_n$ and one cider bottle $C$. Consider a uniformly random permutation of these $n + 1$ bottles. The positions in this permutation are numbered as $1, 2, 3,...,n + 1$.
Define the random variable X to be:
X= the position of the leftmost beer bottle.
What is the expected value E(X) of the random variable X?
Answer: $frac{(n+2)}{(n+1)}$
Attempt:
Total ways to arrange bottles: (n+1)! ways
The position of the leftmost beer: 1 way
E(X) = $$ sum_{i=0}^{n+1} i*Pr(X = position, of, the, leftmost, beer)$$
E(X) = $$ sum_{i=0}^{n+1} = 1*frac{{n+1 choose 1}*1}{(n+1)!} $$
$$ sum_{i=0}^{n+1} = frac{1}{(n-1)!} $$
Can't seem to simplify it to get the correct answer.
probability probability-theory discrete-mathematics random-variables expected-value
edited Nov 27 '18 at 20:14
greedoid
38.2k114795
asked Nov 27 '18 at 20:03
Toby
1577
Question: Let $n geq 2$ be an integer. You are given n beer bottles $B_1, B_2,...B_n$ and one cider bottle $C$. Consider a uniformly random permutation of these $n + 1$ bottles. The positions in this permutation are numbered as $1, 2, 3,...,n + 1$.
Define the random variable X to be:
X= the position of the leftmost beer bottle.
What is the expected value E(X) of the random variable X?
Answer: $frac{(n+2)}{(n+1)}$
Attempt:
Total ways to arrange bottles: (n+1)! ways
The position of the leftmost beer: 1 way
E(X) = $$ sum_{i=0}^{n+1} i*Pr(X = position, of, the, leftmost, beer)$$
E(X) = $$ sum_{i=0}^{n+1} = 1*frac{{n+1 choose 1}*1}{(n+1)!} $$
$$ sum_{i=0}^{n+1} = frac{1}{(n-1)!} $$
Can't seem to simplify it to get the correct answer.
probability probability-theory discrete-mathematics random-variables expected-value
probability probability-theory discrete-mathematics random-variables expected-value
edited Nov 27 '18 at 20:14
greedoid
38.2k114795
asked Nov 27 '18 at 20:03
Toby
1577
edited Nov 27 '18 at 20:14
greedoid
38.2k114795
asked Nov 27 '18 at 20:03
Toby
1577
edited Nov 27 '18 at 20:14
greedoid
38.2k114795
edited Nov 27 '18 at 20:14
greedoid
38.2k114795
edited Nov 27 '18 at 20:14
greedoid
38.2k114795
38.2k114795
asked Nov 27 '18 at 20:03
Toby
1577
asked Nov 27 '18 at 20:03
Toby
1577
asked Nov 27 '18 at 20:03
Toby
1577
1577
Please be extremely careful in writing summations. Remember that the expectation of a random variable $X$ is summing all products of the form $i cdot mathrm{Pr}(X = i).$ Adjust this in edits.
– Sean Roberson
Nov 27 '18 at 20:08
add a comment |
Please be extremely careful in writing summations. Remember that the expectation of a random variable $X$ is summing all products of the form $i cdot mathrm{Pr}(X = i).$ Adjust this in edits.
– Sean Roberson
Nov 27 '18 at 20:08
Please be extremely careful in writing summations. Remember that the expectation of a random variable $X$ is summing all products of the form $i cdot mathrm{Pr}(X = i).$ Adjust this in edits.
– Sean Roberson
Nov 27 '18 at 20:08
Please be extremely careful in writing summations. Remember that the expectation of a random variable $X$ is summing all products of the form $i cdot mathrm{Pr}(X = i).$ Adjust this in edits.
– Sean Roberson
Nov 27 '18 at 20:08
add a comment |
1 Answer
1
active
oldest
votes
Left most beer bottle can be on position 1 or 2. If it is on position 1 then cider bootle can be on any position between 2 and n+1, so we have $n$ possibilities for cider bootle among all $n+1$ positions. Else if cider bottle is on position 1 and leftmost bottle is on position 2. So we have: $$E(X) = 1cdot P(X=1)+2cdot P(X=2) $$
$$ = {nover n+1}+2{1over n+1}$$
answered Nov 27 '18 at 20:08
greedoid
38.2k114795
why are you basing it off the cider bottle's probability of positioning?
– Toby
Nov 27 '18 at 20:24
3
All the permutations depend only on the cider bottle. It uniqely determine each permuatiton.
– greedoid
Nov 27 '18 at 20:27
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Left most beer bottle can be on position 1 or 2. If it is on position 1 then cider bootle can be on any position between 2 and n+1, so we have $n$ possibilities for cider bootle among all $n+1$ positions. Else if cider bottle is on position 1 and leftmost bottle is on position 2. So we have: $$E(X) = 1cdot P(X=1)+2cdot P(X=2) $$
$$ = {nover n+1}+2{1over n+1}$$
answered Nov 27 '18 at 20:08
greedoid
38.2k114795
why are you basing it off the cider bottle's probability of positioning?
– Toby
Nov 27 '18 at 20:24
3
All the permutations depend only on the cider bottle. It uniqely determine each permuatiton.
– greedoid
Nov 27 '18 at 20:27
add a comment |
Left most beer bottle can be on position 1 or 2. If it is on position 1 then cider bootle can be on any position between 2 and n+1, so we have $n$ possibilities for cider bootle among all $n+1$ positions. Else if cider bottle is on position 1 and leftmost bottle is on position 2. So we have: $$E(X) = 1cdot P(X=1)+2cdot P(X=2) $$
$$ = {nover n+1}+2{1over n+1}$$
answered Nov 27 '18 at 20:08
greedoid
38.2k114795
why are you basing it off the cider bottle's probability of positioning?
– Toby
Nov 27 '18 at 20:24
3
All the permutations depend only on the cider bottle. It uniqely determine each permuatiton.
– greedoid
Nov 27 '18 at 20:27
add a comment |
Left most beer bottle can be on position 1 or 2. If it is on position 1 then cider bootle can be on any position between 2 and n+1, so we have $n$ possibilities for cider bootle among all $n+1$ positions. Else if cider bottle is on position 1 and leftmost bottle is on position 2. So we have: $$E(X) = 1cdot P(X=1)+2cdot P(X=2) $$
$$ = {nover n+1}+2{1over n+1}$$
answered Nov 27 '18 at 20:08
greedoid
38.2k114795
Left most beer bottle can be on position 1 or 2. If it is on position 1 then cider bootle can be on any position between 2 and n+1, so we have $n$ possibilities for cider bootle among all $n+1$ positions. Else if cider bottle is on position 1 and leftmost bottle is on position 2. So we have: $$E(X) = 1cdot P(X=1)+2cdot P(X=2) $$
$$ = {nover n+1}+2{1over n+1}$$
answered Nov 27 '18 at 20:08
greedoid
38.2k114795
answered Nov 27 '18 at 20:08
greedoid
38.2k114795
answered Nov 27 '18 at 20:08
greedoid
38.2k114795
answered Nov 27 '18 at 20:08
greedoid
38.2k114795
38.2k114795
why are you basing it off the cider bottle's probability of positioning?
– Toby
Nov 27 '18 at 20:24
3
All the permutations depend only on the cider bottle. It uniqely determine each permuatiton.
– greedoid
Nov 27 '18 at 20:27
add a comment |
why are you basing it off the cider bottle's probability of positioning?
– Toby
Nov 27 '18 at 20:24
3
All the permutations depend only on the cider bottle. It uniqely determine each permuatiton.
– greedoid
Nov 27 '18 at 20:27
why are you basing it off the cider bottle's probability of positioning?
– Toby
Nov 27 '18 at 20:24
why are you basing it off the cider bottle's probability of positioning?
– Toby
Nov 27 '18 at 20:24
3
3
All the permutations depend only on the cider bottle. It uniqely determine each permuatiton.
– greedoid
Nov 27 '18 at 20:27
All the permutations depend only on the cider bottle. It uniqely determine each permuatiton.
– greedoid
Nov 27 '18 at 20:27
add a comment |
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Please be extremely careful in writing summations. Remember that the expectation of a random variable $X$ is summing all products of the form $i cdot mathrm{Pr}(X = i).$ Adjust this in edits.
– Sean Roberson
Nov 27 '18 at 20:08