Mixing
How does the knowledge of one's preference for A over B affect the probability of one's preference for C over...
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Suppose you know someone has preferences between the three pizza toppings pepperoni, olives and mushrooms. If you are told that they prefer pepperoni over olives then what is the probability that they also prefer mushrooms over olives if their preferences are transitive?
The motivation for this question comes from the following article.
https://www.nytimes.com/2008/04/08/science/08tier.html
In the article, a monkey chooses among three colors red, blue and green. It is claimed that it is likely that the monkey prefers the 3 colors roughly the same but that there is actually an order to the preferences. For example, on a scale of 1 (least preferred) to 5 (most preferred), the monkey may rate the colors a 3.9, 4.0 and a 4.1 in some order. So if you are told that the monkey prefers red to blue then there are three situations:
R > B > G
R > G > B
G > R > B
and green is preferred to blue in 2 of these 3 situations so the probability that the monkey prefers green to blue is 2/3.
But what if someone instead tells you that the monkey prefers yellow roughly the same as these three colors (let's say somewhere between 3.8 and 4.2) and that the monkey prefers blue to yellow.
You could then use the same reasoning that there are three possible situations:
B > Y > G
B > G > Y
G > B > Y
and so the monkey prefers blue over green with probability 2/3.
Each new color that the monkey prefers or does not prefer to blue appears to affect the probability that the monkey prefers blue to green. For example, when it is discovered that the monkey prefers red to blue then green becomes more probably preferred than blue. However, if it is instead discovered that yellow is less preferred to blue then blue becomes more probably preferred than green.
If there is no constraint on the scale, i.e. the color preferences can be anywhere on the real number line, and if we allow for a large enough number of colors with distinct preferences randomly distributed on the number line, then it seems it would be easy to find colors that the monkey prefers like red or doesn't prefer like yellow to blue. So then how useful is it when someone tells us that the monkey prefers red to blue or blue to yellow in determining how the monkey prefers blue to green?
It seems that the number of values there are in the range is important in determining how the new information affects the probability that the monkey prefers green to blue.
It is also possible that someone's preferences are not transitive: i.e. they prefer A to B and B to C and C to A. An example is choosing a winning side in Rock-paper-scissors.
probability conditional-probability monty-hall
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add a comment |
$begingroup$
Suppose you know someone has preferences between the three pizza toppings pepperoni, olives and mushrooms. If you are told that they prefer pepperoni over olives then what is the probability that they also prefer mushrooms over olives if their preferences are transitive?
The motivation for this question comes from the following article.
https://www.nytimes.com/2008/04/08/science/08tier.html
In the article, a monkey chooses among three colors red, blue and green. It is claimed that it is likely that the monkey prefers the 3 colors roughly the same but that there is actually an order to the preferences. For example, on a scale of 1 (least preferred) to 5 (most preferred), the monkey may rate the colors a 3.9, 4.0 and a 4.1 in some order. So if you are told that the monkey prefers red to blue then there are three situations:
R > B > G
R > G > B
G > R > B
and green is preferred to blue in 2 of these 3 situations so the probability that the monkey prefers green to blue is 2/3.
But what if someone instead tells you that the monkey prefers yellow roughly the same as these three colors (let's say somewhere between 3.8 and 4.2) and that the monkey prefers blue to yellow.
You could then use the same reasoning that there are three possible situations:
B > Y > G
B > G > Y
G > B > Y
and so the monkey prefers blue over green with probability 2/3.
Each new color that the monkey prefers or does not prefer to blue appears to affect the probability that the monkey prefers blue to green. For example, when it is discovered that the monkey prefers red to blue then green becomes more probably preferred than blue. However, if it is instead discovered that yellow is less preferred to blue then blue becomes more probably preferred than green.
If there is no constraint on the scale, i.e. the color preferences can be anywhere on the real number line, and if we allow for a large enough number of colors with distinct preferences randomly distributed on the number line, then it seems it would be easy to find colors that the monkey prefers like red or doesn't prefer like yellow to blue. So then how useful is it when someone tells us that the monkey prefers red to blue or blue to yellow in determining how the monkey prefers blue to green?
It seems that the number of values there are in the range is important in determining how the new information affects the probability that the monkey prefers green to blue.
It is also possible that someone's preferences are not transitive: i.e. they prefer A to B and B to C and C to A. An example is choosing a winning side in Rock-paper-scissors.
probability conditional-probability monty-hall
$endgroup$
$begingroup$
Interestingly, the assumption that preference in general is transitive is incorrect
$endgroup$
– Zubin Mukerjee
Jan 23 at 6:13
add a comment |
$begingroup$
Suppose you know someone has preferences between the three pizza toppings pepperoni, olives and mushrooms. If you are told that they prefer pepperoni over olives then what is the probability that they also prefer mushrooms over olives if their preferences are transitive?
The motivation for this question comes from the following article.
https://www.nytimes.com/2008/04/08/science/08tier.html
In the article, a monkey chooses among three colors red, blue and green. It is claimed that it is likely that the monkey prefers the 3 colors roughly the same but that there is actually an order to the preferences. For example, on a scale of 1 (least preferred) to 5 (most preferred), the monkey may rate the colors a 3.9, 4.0 and a 4.1 in some order. So if you are told that the monkey prefers red to blue then there are three situations:
R > B > G
R > G > B
G > R > B
and green is preferred to blue in 2 of these 3 situations so the probability that the monkey prefers green to blue is 2/3.
But what if someone instead tells you that the monkey prefers yellow roughly the same as these three colors (let's say somewhere between 3.8 and 4.2) and that the monkey prefers blue to yellow.
You could then use the same reasoning that there are three possible situations:
B > Y > G
B > G > Y
G > B > Y
and so the monkey prefers blue over green with probability 2/3.
Each new color that the monkey prefers or does not prefer to blue appears to affect the probability that the monkey prefers blue to green. For example, when it is discovered that the monkey prefers red to blue then green becomes more probably preferred than blue. However, if it is instead discovered that yellow is less preferred to blue then blue becomes more probably preferred than green.
If there is no constraint on the scale, i.e. the color preferences can be anywhere on the real number line, and if we allow for a large enough number of colors with distinct preferences randomly distributed on the number line, then it seems it would be easy to find colors that the monkey prefers like red or doesn't prefer like yellow to blue. So then how useful is it when someone tells us that the monkey prefers red to blue or blue to yellow in determining how the monkey prefers blue to green?
It seems that the number of values there are in the range is important in determining how the new information affects the probability that the monkey prefers green to blue.
It is also possible that someone's preferences are not transitive: i.e. they prefer A to B and B to C and C to A. An example is choosing a winning side in Rock-paper-scissors.
probability conditional-probability monty-hall
$endgroup$
Suppose you know someone has preferences between the three pizza toppings pepperoni, olives and mushrooms. If you are told that they prefer pepperoni over olives then what is the probability that they also prefer mushrooms over olives if their preferences are transitive?
The motivation for this question comes from the following article.
https://www.nytimes.com/2008/04/08/science/08tier.html
In the article, a monkey chooses among three colors red, blue and green. It is claimed that it is likely that the monkey prefers the 3 colors roughly the same but that there is actually an order to the preferences. For example, on a scale of 1 (least preferred) to 5 (most preferred), the monkey may rate the colors a 3.9, 4.0 and a 4.1 in some order. So if you are told that the monkey prefers red to blue then there are three situations:
R > B > G
R > G > B
G > R > B
and green is preferred to blue in 2 of these 3 situations so the probability that the monkey prefers green to blue is 2/3.
But what if someone instead tells you that the monkey prefers yellow roughly the same as these three colors (let's say somewhere between 3.8 and 4.2) and that the monkey prefers blue to yellow.
You could then use the same reasoning that there are three possible situations:
B > Y > G
B > G > Y
G > B > Y
and so the monkey prefers blue over green with probability 2/3.
Each new color that the monkey prefers or does not prefer to blue appears to affect the probability that the monkey prefers blue to green. For example, when it is discovered that the monkey prefers red to blue then green becomes more probably preferred than blue. However, if it is instead discovered that yellow is less preferred to blue then blue becomes more probably preferred than green.
If there is no constraint on the scale, i.e. the color preferences can be anywhere on the real number line, and if we allow for a large enough number of colors with distinct preferences randomly distributed on the number line, then it seems it would be easy to find colors that the monkey prefers like red or doesn't prefer like yellow to blue. So then how useful is it when someone tells us that the monkey prefers red to blue or blue to yellow in determining how the monkey prefers blue to green?
It seems that the number of values there are in the range is important in determining how the new information affects the probability that the monkey prefers green to blue.
It is also possible that someone's preferences are not transitive: i.e. they prefer A to B and B to C and C to A. An example is choosing a winning side in Rock-paper-scissors.
probability conditional-probability monty-hall
probability conditional-probability monty-hall
edited Jan 23 at 6:04
asked Jan 23 at 3:25
user637421
$begingroup$
Interestingly, the assumption that preference in general is transitive is incorrect
$endgroup$
– Zubin Mukerjee
Jan 23 at 6:13
add a comment |
$begingroup$
Interestingly, the assumption that preference in general is transitive is incorrect
$endgroup$
– Zubin Mukerjee
Jan 23 at 6:13
$begingroup$
Interestingly, the assumption that preference in general is transitive is incorrect
$endgroup$
– Zubin Mukerjee
Jan 23 at 6:13
$begingroup$
Interestingly, the assumption that preference in general is transitive is incorrect
$endgroup$
– Zubin Mukerjee
Jan 23 at 6:13
add a comment |
1 Answer
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oldest
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$begingroup$
If you assume there is a strict ordering, then as a prior, every permutation is equally likely. Each known preference eliminates all permutations inconsistent with that preference, while having no impact on the relative weights of the others. So, eliminate all eliminated permutations then count what proportion of them fit the new question you're asking.
So in the case of three things, it's intially six cases:
ABC
ACB
BAC
BCA
CAB
CBA
A>B leaves three cases:
ABC
ACB
CAB
Among these, C>B two thirds of the time.
answered Jan 23 at 5:57
Steven IrrgangSteven Irrgang
71126
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
If you assume there is a strict ordering, then as a prior, every permutation is equally likely. Each known preference eliminates all permutations inconsistent with that preference, while having no impact on the relative weights of the others. So, eliminate all eliminated permutations then count what proportion of them fit the new question you're asking.
So in the case of three things, it's intially six cases:
ABC
ACB
BAC
BCA
CAB
CBA
A>B leaves three cases:
ABC
ACB
CAB
Among these, C>B two thirds of the time.
answered Jan 23 at 5:57
Steven IrrgangSteven Irrgang
71126
$endgroup$
add a comment |
$begingroup$
If you assume there is a strict ordering, then as a prior, every permutation is equally likely. Each known preference eliminates all permutations inconsistent with that preference, while having no impact on the relative weights of the others. So, eliminate all eliminated permutations then count what proportion of them fit the new question you're asking.
So in the case of three things, it's intially six cases:
ABC
ACB
BAC
BCA
CAB
CBA
A>B leaves three cases:
ABC
ACB
CAB
Among these, C>B two thirds of the time.
answered Jan 23 at 5:57
Steven IrrgangSteven Irrgang
71126
$endgroup$
add a comment |
$begingroup$
If you assume there is a strict ordering, then as a prior, every permutation is equally likely. Each known preference eliminates all permutations inconsistent with that preference, while having no impact on the relative weights of the others. So, eliminate all eliminated permutations then count what proportion of them fit the new question you're asking.
So in the case of three things, it's intially six cases:
ABC
ACB
BAC
BCA
CAB
CBA
A>B leaves three cases:
ABC
ACB
CAB
Among these, C>B two thirds of the time.
answered Jan 23 at 5:57
Steven IrrgangSteven Irrgang
71126
$endgroup$
If you assume there is a strict ordering, then as a prior, every permutation is equally likely. Each known preference eliminates all permutations inconsistent with that preference, while having no impact on the relative weights of the others. So, eliminate all eliminated permutations then count what proportion of them fit the new question you're asking.
So in the case of three things, it's intially six cases:
ABC
ACB
BAC
BCA
CAB
CBA
A>B leaves three cases:
ABC
ACB
CAB
Among these, C>B two thirds of the time.
answered Jan 23 at 5:57
Steven IrrgangSteven Irrgang
71126
answered Jan 23 at 5:57
Steven IrrgangSteven Irrgang
71126
answered Jan 23 at 5:57
Steven IrrgangSteven Irrgang
71126
answered Jan 23 at 5:57
Steven IrrgangSteven Irrgang
71126
71126
add a comment |
add a comment |
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$begingroup$
Interestingly, the assumption that preference in general is transitive is incorrect
$endgroup$
– Zubin Mukerjee
Jan 23 at 6:13