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Intuitive reason for why a succession of draws without replacement is 'fair'
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A common way of 'fairly' assigning a set of $n$ desirable items to a group of $m>n$ people is to have a bag containing, for instance, $m$ pieces of paper with $n$ of them bearing the name of an item, the others being blank, and have each person in the group pick from the bag.
Initially I wondered whether this was fair at all, in the sense that 'the order does not matter', or in other words every participant has equal probability at the start of winning an item, or the expected 'value' for each person is the same given some assignment of values to the items.
It looks like this is the case e.g. labelling the $k$th person to draw $A_k$:
$mathbbP(A_1,textsuccessful)=dfracnm=p_1$
$mathbbP(A_2,textsuccessful)= p_1fracn-1m-1+(1-p_1)fracnm-1=dfracnm=p_2;(=p_1)$
$mathbbP(A_3,textsuccessful)=p_1p_2fracn-2m-2+big((1-p_2)p_1+(1-p_1)p_2big)fracn-1m-2+(1-p_1)(1-p_2)fracnm-2=dfracnm$
etc.
I guess I am thinking of this in the wrong way, but to me it is like the algebra just magically works out. I can't quite see why intuitively this should be the case, or why one would expect this to be fair without even doing the algebra.
I tried this with a similar problem, picking letters out of a Scrabble bag to decide who starts the game (an advantage). Assigning the values $0,1,2,3dots$ to $[text ],A,B,Cdots $ it turns out the expected value for each person is the same ($11.51$ on a standard English Scrabble game), so again, this is fair.
Could anyone explain what the underlying mechanism is here? It might be completely obvious but I don't really see it.
probability intuition
asked Jul 25 at 14:02
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up vote
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A common way of 'fairly' assigning a set of $n$ desirable items to a group of $m>n$ people is to have a bag containing, for instance, $m$ pieces of paper with $n$ of them bearing the name of an item, the others being blank, and have each person in the group pick from the bag.
Initially I wondered whether this was fair at all, in the sense that 'the order does not matter', or in other words every participant has equal probability at the start of winning an item, or the expected 'value' for each person is the same given some assignment of values to the items.
It looks like this is the case e.g. labelling the $k$th person to draw $A_k$:
$mathbbP(A_1,textsuccessful)=dfracnm=p_1$
$mathbbP(A_2,textsuccessful)= p_1fracn-1m-1+(1-p_1)fracnm-1=dfracnm=p_2;(=p_1)$
$mathbbP(A_3,textsuccessful)=p_1p_2fracn-2m-2+big((1-p_2)p_1+(1-p_1)p_2big)fracn-1m-2+(1-p_1)(1-p_2)fracnm-2=dfracnm$
etc.
I guess I am thinking of this in the wrong way, but to me it is like the algebra just magically works out. I can't quite see why intuitively this should be the case, or why one would expect this to be fair without even doing the algebra.
I tried this with a similar problem, picking letters out of a Scrabble bag to decide who starts the game (an advantage). Assigning the values $0,1,2,3dots$ to $[text ],A,B,Cdots $ it turns out the expected value for each person is the same ($11.51$ on a standard English Scrabble game), so again, this is fair.
Could anyone explain what the underlying mechanism is here? It might be completely obvious but I don't really see it.
probability intuition
asked Jul 25 at 14:02
K. 622
30616
3
Instead of drawing the numbers from a bag, write them on cards, shuffle the deck, and deal it to the people. That's fair, right? You don't care whether you get the first card or the last, do you? How is it different from drawing numbers from the bag?
– saulspatz
Jul 25 at 14:12
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A common way of 'fairly' assigning a set of $n$ desirable items to a group of $m>n$ people is to have a bag containing, for instance, $m$ pieces of paper with $n$ of them bearing the name of an item, the others being blank, and have each person in the group pick from the bag.
Initially I wondered whether this was fair at all, in the sense that 'the order does not matter', or in other words every participant has equal probability at the start of winning an item, or the expected 'value' for each person is the same given some assignment of values to the items.
It looks like this is the case e.g. labelling the $k$th person to draw $A_k$:
$mathbbP(A_1,textsuccessful)=dfracnm=p_1$
$mathbbP(A_2,textsuccessful)= p_1fracn-1m-1+(1-p_1)fracnm-1=dfracnm=p_2;(=p_1)$
$mathbbP(A_3,textsuccessful)=p_1p_2fracn-2m-2+big((1-p_2)p_1+(1-p_1)p_2big)fracn-1m-2+(1-p_1)(1-p_2)fracnm-2=dfracnm$
etc.
I guess I am thinking of this in the wrong way, but to me it is like the algebra just magically works out. I can't quite see why intuitively this should be the case, or why one would expect this to be fair without even doing the algebra.
I tried this with a similar problem, picking letters out of a Scrabble bag to decide who starts the game (an advantage). Assigning the values $0,1,2,3dots$ to $[text ],A,B,Cdots $ it turns out the expected value for each person is the same ($11.51$ on a standard English Scrabble game), so again, this is fair.
Could anyone explain what the underlying mechanism is here? It might be completely obvious but I don't really see it.
probability intuition
asked Jul 25 at 14:02
K. 622
30616
A common way of 'fairly' assigning a set of $n$ desirable items to a group of $m>n$ people is to have a bag containing, for instance, $m$ pieces of paper with $n$ of them bearing the name of an item, the others being blank, and have each person in the group pick from the bag.
Initially I wondered whether this was fair at all, in the sense that 'the order does not matter', or in other words every participant has equal probability at the start of winning an item, or the expected 'value' for each person is the same given some assignment of values to the items.
It looks like this is the case e.g. labelling the $k$th person to draw $A_k$:
$mathbbP(A_1,textsuccessful)=dfracnm=p_1$
$mathbbP(A_2,textsuccessful)= p_1fracn-1m-1+(1-p_1)fracnm-1=dfracnm=p_2;(=p_1)$
$mathbbP(A_3,textsuccessful)=p_1p_2fracn-2m-2+big((1-p_2)p_1+(1-p_1)p_2big)fracn-1m-2+(1-p_1)(1-p_2)fracnm-2=dfracnm$
etc.
I guess I am thinking of this in the wrong way, but to me it is like the algebra just magically works out. I can't quite see why intuitively this should be the case, or why one would expect this to be fair without even doing the algebra.
I tried this with a similar problem, picking letters out of a Scrabble bag to decide who starts the game (an advantage). Assigning the values $0,1,2,3dots$ to $[text ],A,B,Cdots $ it turns out the expected value for each person is the same ($11.51$ on a standard English Scrabble game), so again, this is fair.
Could anyone explain what the underlying mechanism is here? It might be completely obvious but I don't really see it.
probability intuition
asked Jul 25 at 14:02
K. 622
30616
asked Jul 25 at 14:02
K. 622
30616
asked Jul 25 at 14:02
K. 622
30616
asked Jul 25 at 14:02
K. 622
30616
30616
3
Instead of drawing the numbers from a bag, write them on cards, shuffle the deck, and deal it to the people. That's fair, right? You don't care whether you get the first card or the last, do you? How is it different from drawing numbers from the bag?
– saulspatz
Jul 25 at 14:12
add a comment |Â
3
Instead of drawing the numbers from a bag, write them on cards, shuffle the deck, and deal it to the people. That's fair, right? You don't care whether you get the first card or the last, do you? How is it different from drawing numbers from the bag?
– saulspatz
Jul 25 at 14:12
3
3
Instead of drawing the numbers from a bag, write them on cards, shuffle the deck, and deal it to the people. That's fair, right? You don't care whether you get the first card or the last, do you? How is it different from drawing numbers from the bag?
– saulspatz
Jul 25 at 14:12
Instead of drawing the numbers from a bag, write them on cards, shuffle the deck, and deal it to the people. That's fair, right? You don't care whether you get the first card or the last, do you? How is it different from drawing numbers from the bag?
– saulspatz
Jul 25 at 14:12
add a comment |Â
1 Answer
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up vote
0
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Short answer: 'Intuitively': everyone has an equal chance at each of the $m$ slips of paper.
Longer answer: If a later person, say X, knows in advance what happened with the earlier draws, the probability of X getting a desirable item may change depending on how many of the desirable items were picked earlier. But the change in probability could go up or down depending on whether the earlier drawers were less or more lucky than average.
It might help to think of everyone drawing a slip of paper, but not looking until all have drawn. Then everyone looks at their paper simultaneously.
answered Jul 25 at 14:14
paw88789
28.1k12248
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Short answer: 'Intuitively': everyone has an equal chance at each of the $m$ slips of paper.
Longer answer: If a later person, say X, knows in advance what happened with the earlier draws, the probability of X getting a desirable item may change depending on how many of the desirable items were picked earlier. But the change in probability could go up or down depending on whether the earlier drawers were less or more lucky than average.
It might help to think of everyone drawing a slip of paper, but not looking until all have drawn. Then everyone looks at their paper simultaneously.
answered Jul 25 at 14:14
paw88789
28.1k12248
add a comment |Â
up vote
0
down vote
Short answer: 'Intuitively': everyone has an equal chance at each of the $m$ slips of paper.
Longer answer: If a later person, say X, knows in advance what happened with the earlier draws, the probability of X getting a desirable item may change depending on how many of the desirable items were picked earlier. But the change in probability could go up or down depending on whether the earlier drawers were less or more lucky than average.
It might help to think of everyone drawing a slip of paper, but not looking until all have drawn. Then everyone looks at their paper simultaneously.
answered Jul 25 at 14:14
paw88789
28.1k12248
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Short answer: 'Intuitively': everyone has an equal chance at each of the $m$ slips of paper.
Longer answer: If a later person, say X, knows in advance what happened with the earlier draws, the probability of X getting a desirable item may change depending on how many of the desirable items were picked earlier. But the change in probability could go up or down depending on whether the earlier drawers were less or more lucky than average.
It might help to think of everyone drawing a slip of paper, but not looking until all have drawn. Then everyone looks at their paper simultaneously.
answered Jul 25 at 14:14
paw88789
28.1k12248
Short answer: 'Intuitively': everyone has an equal chance at each of the $m$ slips of paper.
Longer answer: If a later person, say X, knows in advance what happened with the earlier draws, the probability of X getting a desirable item may change depending on how many of the desirable items were picked earlier. But the change in probability could go up or down depending on whether the earlier drawers were less or more lucky than average.
It might help to think of everyone drawing a slip of paper, but not looking until all have drawn. Then everyone looks at their paper simultaneously.
answered Jul 25 at 14:14
paw88789
28.1k12248
answered Jul 25 at 14:14
paw88789
28.1k12248
answered Jul 25 at 14:14
paw88789
28.1k12248
answered Jul 25 at 14:14
paw88789
28.1k12248
28.1k12248
add a comment |Â
add a comment |Â
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3
Instead of drawing the numbers from a bag, write them on cards, shuffle the deck, and deal it to the people. That's fair, right? You don't care whether you get the first card or the last, do you? How is it different from drawing numbers from the bag?
– saulspatz
Jul 25 at 14:12