Random Incidence Paradox Question
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10 friends have a collection of 1000 candies, 15 have collection of 1300 candies, and 5 have collection of 600 candies. What is the average number of candies your friends have? If you mix all these coins and randomly pick one out, what is the expected size of collection that candy's owner has.
For the first part of question, I calculated as : (10*1000 / 10) +(15*1300 / 15)+ (5*600 / 5) which turns out to be 1083. However, I am confused about second part of question. If total candies are 32500 and the probability that you pick it out from first part of friends : (10*1000 / 32500). And similarly for the rest, (15*1300 / 32500), (5*600 / 32500). Then you multiply probability with count in that group. Final equation becomes:
0.3*10*1000 + 0.6*15*1300 + 5*600*0.092 = 14976. But I am not really sure about my answer
probability expected-value
asked Nov 15 at 15:39
puffles
669
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up vote
0
down vote
favorite
10 friends have a collection of 1000 candies, 15 have collection of 1300 candies, and 5 have collection of 600 candies. What is the average number of candies your friends have? If you mix all these coins and randomly pick one out, what is the expected size of collection that candy's owner has.
For the first part of question, I calculated as : (10*1000 / 10) +(15*1300 / 15)+ (5*600 / 5) which turns out to be 1083. However, I am confused about second part of question. If total candies are 32500 and the probability that you pick it out from first part of friends : (10*1000 / 32500). And similarly for the rest, (15*1300 / 32500), (5*600 / 32500). Then you multiply probability with count in that group. Final equation becomes:
0.3*10*1000 + 0.6*15*1300 + 5*600*0.092 = 14976. But I am not really sure about my answer
probability expected-value
asked Nov 15 at 15:39
puffles
669
1
Note: The expected size can't be greater than all the possible sizes.
– lulu
Nov 15 at 15:40
So instead of 0.3*10*1000, I should just take 0.3*1000 ? Since that is the size and the probability for that size
– puffles
Nov 15 at 15:45
I can't really follow your calculation. There are only three possible collection sizes. You shouldn't be multiplying by the number of times each occurs (that factor has already been used in computing the probabilities).
– lulu
Nov 15 at 15:46
Yes, that is correct. Only it isn't exactly $.3$...it's not a good idea to round probabilities to the extent that they no longer sum to $1$. Since everything in sight is rational, I'd just keep it in fractional form until the end.
– lulu
Nov 15 at 15:47
The first question is the mean of the distribution given by the 30 numbers {1000,...,1000,1300,...,1300,600,...,600}. The second is the mean of the "size bias" distribution of the first distribution, that is the distribution in which, in effect, each group is weighted by the size of the group. The mean comes out to be the sum of the squares of the original list divided by the sum of the list. The latter mean is bigger. Similarly, compare the average class size in a college, computed (1) from the teacher point of view, and (2) from the students' point of view. Same computations.
– Ned
Nov 18 at 22:52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
10 friends have a collection of 1000 candies, 15 have collection of 1300 candies, and 5 have collection of 600 candies. What is the average number of candies your friends have? If you mix all these coins and randomly pick one out, what is the expected size of collection that candy's owner has.
For the first part of question, I calculated as : (10*1000 / 10) +(15*1300 / 15)+ (5*600 / 5) which turns out to be 1083. However, I am confused about second part of question. If total candies are 32500 and the probability that you pick it out from first part of friends : (10*1000 / 32500). And similarly for the rest, (15*1300 / 32500), (5*600 / 32500). Then you multiply probability with count in that group. Final equation becomes:
0.3*10*1000 + 0.6*15*1300 + 5*600*0.092 = 14976. But I am not really sure about my answer
probability expected-value
asked Nov 15 at 15:39
puffles
669
10 friends have a collection of 1000 candies, 15 have collection of 1300 candies, and 5 have collection of 600 candies. What is the average number of candies your friends have? If you mix all these coins and randomly pick one out, what is the expected size of collection that candy's owner has.
For the first part of question, I calculated as : (10*1000 / 10) +(15*1300 / 15)+ (5*600 / 5) which turns out to be 1083. However, I am confused about second part of question. If total candies are 32500 and the probability that you pick it out from first part of friends : (10*1000 / 32500). And similarly for the rest, (15*1300 / 32500), (5*600 / 32500). Then you multiply probability with count in that group. Final equation becomes:
0.3*10*1000 + 0.6*15*1300 + 5*600*0.092 = 14976. But I am not really sure about my answer
probability expected-value
probability expected-value
asked Nov 15 at 15:39
puffles
669
asked Nov 15 at 15:39
puffles
669
asked Nov 15 at 15:39
puffles
669
asked Nov 15 at 15:39
puffles
669
asked Nov 15 at 15:39
puffles
669
669
1
Note: The expected size can't be greater than all the possible sizes.
– lulu
Nov 15 at 15:40
So instead of 0.3*10*1000, I should just take 0.3*1000 ? Since that is the size and the probability for that size
– puffles
Nov 15 at 15:45
I can't really follow your calculation. There are only three possible collection sizes. You shouldn't be multiplying by the number of times each occurs (that factor has already been used in computing the probabilities).
– lulu
Nov 15 at 15:46
Yes, that is correct. Only it isn't exactly $.3$...it's not a good idea to round probabilities to the extent that they no longer sum to $1$. Since everything in sight is rational, I'd just keep it in fractional form until the end.
– lulu
Nov 15 at 15:47
The first question is the mean of the distribution given by the 30 numbers {1000,...,1000,1300,...,1300,600,...,600}. The second is the mean of the "size bias" distribution of the first distribution, that is the distribution in which, in effect, each group is weighted by the size of the group. The mean comes out to be the sum of the squares of the original list divided by the sum of the list. The latter mean is bigger. Similarly, compare the average class size in a college, computed (1) from the teacher point of view, and (2) from the students' point of view. Same computations.
– Ned
Nov 18 at 22:52
add a comment |
1
Note: The expected size can't be greater than all the possible sizes.
– lulu
Nov 15 at 15:40
So instead of 0.3*10*1000, I should just take 0.3*1000 ? Since that is the size and the probability for that size
– puffles
Nov 15 at 15:45
I can't really follow your calculation. There are only three possible collection sizes. You shouldn't be multiplying by the number of times each occurs (that factor has already been used in computing the probabilities).
– lulu
Nov 15 at 15:46
Yes, that is correct. Only it isn't exactly $.3$...it's not a good idea to round probabilities to the extent that they no longer sum to $1$. Since everything in sight is rational, I'd just keep it in fractional form until the end.
– lulu
Nov 15 at 15:47
The first question is the mean of the distribution given by the 30 numbers {1000,...,1000,1300,...,1300,600,...,600}. The second is the mean of the "size bias" distribution of the first distribution, that is the distribution in which, in effect, each group is weighted by the size of the group. The mean comes out to be the sum of the squares of the original list divided by the sum of the list. The latter mean is bigger. Similarly, compare the average class size in a college, computed (1) from the teacher point of view, and (2) from the students' point of view. Same computations.
– Ned
Nov 18 at 22:52
1
1
Note: The expected size can't be greater than all the possible sizes.
– lulu
Nov 15 at 15:40
Note: The expected size can't be greater than all the possible sizes.
– lulu
Nov 15 at 15:40
So instead of 0.3*10*1000, I should just take 0.3*1000 ? Since that is the size and the probability for that size
– puffles
Nov 15 at 15:45
So instead of 0.3*10*1000, I should just take 0.3*1000 ? Since that is the size and the probability for that size
– puffles
Nov 15 at 15:45
I can't really follow your calculation. There are only three possible collection sizes. You shouldn't be multiplying by the number of times each occurs (that factor has already been used in computing the probabilities).
– lulu
Nov 15 at 15:46
I can't really follow your calculation. There are only three possible collection sizes. You shouldn't be multiplying by the number of times each occurs (that factor has already been used in computing the probabilities).
– lulu
Nov 15 at 15:46
Yes, that is correct. Only it isn't exactly $.3$...it's not a good idea to round probabilities to the extent that they no longer sum to $1$. Since everything in sight is rational, I'd just keep it in fractional form until the end.
– lulu
Nov 15 at 15:47
Yes, that is correct. Only it isn't exactly $.3$...it's not a good idea to round probabilities to the extent that they no longer sum to $1$. Since everything in sight is rational, I'd just keep it in fractional form until the end.
– lulu
Nov 15 at 15:47
The first question is the mean of the distribution given by the 30 numbers {1000,...,1000,1300,...,1300,600,...,600}. The second is the mean of the "size bias" distribution of the first distribution, that is the distribution in which, in effect, each group is weighted by the size of the group. The mean comes out to be the sum of the squares of the original list divided by the sum of the list. The latter mean is bigger. Similarly, compare the average class size in a college, computed (1) from the teacher point of view, and (2) from the students' point of view. Same computations.
– Ned
Nov 18 at 22:52
The first question is the mean of the distribution given by the 30 numbers {1000,...,1000,1300,...,1300,600,...,600}. The second is the mean of the "size bias" distribution of the first distribution, that is the distribution in which, in effect, each group is weighted by the size of the group. The mean comes out to be the sum of the squares of the original list divided by the sum of the list. The latter mean is bigger. Similarly, compare the average class size in a college, computed (1) from the teacher point of view, and (2) from the students' point of view. Same computations.
– Ned
Nov 18 at 22:52
add a comment |
1 Answer
1
active
oldest
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up vote
0
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Your answer for the first part is correct. However, for the second part we want to know how many coins the owner has. Note the following:
10000/32500 candies belong to the owner who has 1000 candies
19500/32500 candies belong to the owner who has 1300 candies
3000/32500 candies belong to the owner who has 600 candies
The above values are all probabilities so just multiply them with the number of candies the owner has. For an intuition note that there is 10000/32500 chance that the candy you choose belongs to the owner who has 1000 candies. Just use this to calculate the expected value.
answered Nov 18 at 22:00
helloworld
477
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Your answer for the first part is correct. However, for the second part we want to know how many coins the owner has. Note the following:
10000/32500 candies belong to the owner who has 1000 candies
19500/32500 candies belong to the owner who has 1300 candies
3000/32500 candies belong to the owner who has 600 candies
The above values are all probabilities so just multiply them with the number of candies the owner has. For an intuition note that there is 10000/32500 chance that the candy you choose belongs to the owner who has 1000 candies. Just use this to calculate the expected value.
answered Nov 18 at 22:00
helloworld
477
add a comment |
up vote
0
down vote
accepted
Your answer for the first part is correct. However, for the second part we want to know how many coins the owner has. Note the following:
10000/32500 candies belong to the owner who has 1000 candies
19500/32500 candies belong to the owner who has 1300 candies
3000/32500 candies belong to the owner who has 600 candies
The above values are all probabilities so just multiply them with the number of candies the owner has. For an intuition note that there is 10000/32500 chance that the candy you choose belongs to the owner who has 1000 candies. Just use this to calculate the expected value.
answered Nov 18 at 22:00
helloworld
477
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Your answer for the first part is correct. However, for the second part we want to know how many coins the owner has. Note the following:
10000/32500 candies belong to the owner who has 1000 candies
19500/32500 candies belong to the owner who has 1300 candies
3000/32500 candies belong to the owner who has 600 candies
The above values are all probabilities so just multiply them with the number of candies the owner has. For an intuition note that there is 10000/32500 chance that the candy you choose belongs to the owner who has 1000 candies. Just use this to calculate the expected value.
answered Nov 18 at 22:00
helloworld
477
Your answer for the first part is correct. However, for the second part we want to know how many coins the owner has. Note the following:
10000/32500 candies belong to the owner who has 1000 candies
19500/32500 candies belong to the owner who has 1300 candies
3000/32500 candies belong to the owner who has 600 candies
The above values are all probabilities so just multiply them with the number of candies the owner has. For an intuition note that there is 10000/32500 chance that the candy you choose belongs to the owner who has 1000 candies. Just use this to calculate the expected value.
answered Nov 18 at 22:00
helloworld
477
answered Nov 18 at 22:00
helloworld
477
answered Nov 18 at 22:00
helloworld
477
answered Nov 18 at 22:00
helloworld
477
477
add a comment |
add a comment |
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1
Note: The expected size can't be greater than all the possible sizes.
– lulu
Nov 15 at 15:40
So instead of 0.3*10*1000, I should just take 0.3*1000 ? Since that is the size and the probability for that size
– puffles
Nov 15 at 15:45
I can't really follow your calculation. There are only three possible collection sizes. You shouldn't be multiplying by the number of times each occurs (that factor has already been used in computing the probabilities).
– lulu
Nov 15 at 15:46
Yes, that is correct. Only it isn't exactly $.3$...it's not a good idea to round probabilities to the extent that they no longer sum to $1$. Since everything in sight is rational, I'd just keep it in fractional form until the end.
– lulu
Nov 15 at 15:47
The first question is the mean of the distribution given by the 30 numbers {1000,...,1000,1300,...,1300,600,...,600}. The second is the mean of the "size bias" distribution of the first distribution, that is the distribution in which, in effect, each group is weighted by the size of the group. The mean comes out to be the sum of the squares of the original list divided by the sum of the list. The latter mean is bigger. Similarly, compare the average class size in a college, computed (1) from the teacher point of view, and (2) from the students' point of view. Same computations.
– Ned
Nov 18 at 22:52