When Replacement doesn't matter (drawing marbles)
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A bag contains black and white marbles $(X>0, Y>0)$. We are asked the probability of successfully drawing a black marbles and b white marbles $(a>0,b>0)$ in exactly $a+b$ draws. For what values $X,Y,a,b$ does drawing with or without replacement not change the probability of success? In other words, in what specific situations does replacement not matter? What is the smallest number of marbles in the bag to have a solution?
I've generalized the equations for the chances of success:
$$textw/o replacement = cfracbinomXabinomYbbinomX+Ya+b\
textw/ replacement = cfracbinoma+baX^aY^b(X+Y)^a+b$$
I've tested these equations with small values $1,1,1,1, 2,2,1,1,2,2,2,1$, etc. that are easily checked by hand, and they appear to hold true. What I have found is that drawing without replacement does not always give an advantage over drawing with, so I assume by IVT, as $n$ approaches infinity there will be some integer set where the probabilities intersect. The problem is I don't know how to proceed from here in finding a Magic Set other than with trial and error. A discovered answer would be easy to check, but I wouldn't know how to potentially find one other than brute force.
probability combinatorics
edited Aug 7 at 21:02
Davide Morgante
2,070221
asked Aug 7 at 20:32
Calibur
63
 |Â
show 1 more comment
up vote
1
down vote
favorite
A bag contains black and white marbles $(X>0, Y>0)$. We are asked the probability of successfully drawing a black marbles and b white marbles $(a>0,b>0)$ in exactly $a+b$ draws. For what values $X,Y,a,b$ does drawing with or without replacement not change the probability of success? In other words, in what specific situations does replacement not matter? What is the smallest number of marbles in the bag to have a solution?
I've generalized the equations for the chances of success:
$$textw/o replacement = cfracbinomXabinomYbbinomX+Ya+b\
textw/ replacement = cfracbinoma+baX^aY^b(X+Y)^a+b$$
I've tested these equations with small values $1,1,1,1, 2,2,1,1,2,2,2,1$, etc. that are easily checked by hand, and they appear to hold true. What I have found is that drawing without replacement does not always give an advantage over drawing with, so I assume by IVT, as $n$ approaches infinity there will be some integer set where the probabilities intersect. The problem is I don't know how to proceed from here in finding a Magic Set other than with trial and error. A discovered answer would be easy to check, but I wouldn't know how to potentially find one other than brute force.
probability combinatorics
edited Aug 7 at 21:02
Davide Morgante
2,070221
asked Aug 7 at 20:32
Calibur
63
"What is the smallest number of marbles in the bag to have a solution?" One marble in the bag (or even zero) is the smallest. Generally, drawing one marble (or zero) will give the same result with our without replacement. But that's boring.
– Arthur
Aug 7 at 20:37
There must be at least $Xge a, Yge b$ marbles in the bag.
– Doug M
Aug 7 at 20:42
IVT only applies to continuous functions, and marbles are discrete objects. The specific case where replacement does not matter is the case that only one marble is chosen.
– Doug M
Aug 7 at 20:48
Yes. For replacement to be a factor, a+b must be >= 2, and both colors must exists in the bag.
– Calibur
Aug 7 at 20:50
As the number of marbles approaches infinity, the discreet addition of marbles becomes comparatively continuous. With the Sock problem from MindYourDecisions (youtube.com/watch?v=GuqcnrtdUCc), this logic explains that some sock drawer exists such that the chance of pulling of matching pair is 50%, but not that any previous sock drawer can be added to to attain this property.
– Calibur
Aug 7 at 20:56
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A bag contains black and white marbles $(X>0, Y>0)$. We are asked the probability of successfully drawing a black marbles and b white marbles $(a>0,b>0)$ in exactly $a+b$ draws. For what values $X,Y,a,b$ does drawing with or without replacement not change the probability of success? In other words, in what specific situations does replacement not matter? What is the smallest number of marbles in the bag to have a solution?
I've generalized the equations for the chances of success:
$$textw/o replacement = cfracbinomXabinomYbbinomX+Ya+b\
textw/ replacement = cfracbinoma+baX^aY^b(X+Y)^a+b$$
I've tested these equations with small values $1,1,1,1, 2,2,1,1,2,2,2,1$, etc. that are easily checked by hand, and they appear to hold true. What I have found is that drawing without replacement does not always give an advantage over drawing with, so I assume by IVT, as $n$ approaches infinity there will be some integer set where the probabilities intersect. The problem is I don't know how to proceed from here in finding a Magic Set other than with trial and error. A discovered answer would be easy to check, but I wouldn't know how to potentially find one other than brute force.
probability combinatorics
edited Aug 7 at 21:02
Davide Morgante
2,070221
asked Aug 7 at 20:32
Calibur
63
A bag contains black and white marbles $(X>0, Y>0)$. We are asked the probability of successfully drawing a black marbles and b white marbles $(a>0,b>0)$ in exactly $a+b$ draws. For what values $X,Y,a,b$ does drawing with or without replacement not change the probability of success? In other words, in what specific situations does replacement not matter? What is the smallest number of marbles in the bag to have a solution?
I've generalized the equations for the chances of success:
$$textw/o replacement = cfracbinomXabinomYbbinomX+Ya+b\
textw/ replacement = cfracbinoma+baX^aY^b(X+Y)^a+b$$
I've tested these equations with small values $1,1,1,1, 2,2,1,1,2,2,2,1$, etc. that are easily checked by hand, and they appear to hold true. What I have found is that drawing without replacement does not always give an advantage over drawing with, so I assume by IVT, as $n$ approaches infinity there will be some integer set where the probabilities intersect. The problem is I don't know how to proceed from here in finding a Magic Set other than with trial and error. A discovered answer would be easy to check, but I wouldn't know how to potentially find one other than brute force.
probability combinatorics
edited Aug 7 at 21:02
Davide Morgante
2,070221
asked Aug 7 at 20:32
Calibur
63
edited Aug 7 at 21:02
Davide Morgante
2,070221
edited Aug 7 at 21:02
Davide Morgante
2,070221
edited Aug 7 at 21:02
Davide Morgante
2,070221
2,070221
asked Aug 7 at 20:32
Calibur
63
asked Aug 7 at 20:32
Calibur
63
asked Aug 7 at 20:32
Calibur
63
63
"What is the smallest number of marbles in the bag to have a solution?" One marble in the bag (or even zero) is the smallest. Generally, drawing one marble (or zero) will give the same result with our without replacement. But that's boring.
– Arthur
Aug 7 at 20:37
There must be at least $Xge a, Yge b$ marbles in the bag.
– Doug M
Aug 7 at 20:42
IVT only applies to continuous functions, and marbles are discrete objects. The specific case where replacement does not matter is the case that only one marble is chosen.
– Doug M
Aug 7 at 20:48
Yes. For replacement to be a factor, a+b must be >= 2, and both colors must exists in the bag.
– Calibur
Aug 7 at 20:50
As the number of marbles approaches infinity, the discreet addition of marbles becomes comparatively continuous. With the Sock problem from MindYourDecisions (youtube.com/watch?v=GuqcnrtdUCc), this logic explains that some sock drawer exists such that the chance of pulling of matching pair is 50%, but not that any previous sock drawer can be added to to attain this property.
– Calibur
Aug 7 at 20:56
 |Â
show 1 more comment
"What is the smallest number of marbles in the bag to have a solution?" One marble in the bag (or even zero) is the smallest. Generally, drawing one marble (or zero) will give the same result with our without replacement. But that's boring.
– Arthur
Aug 7 at 20:37
There must be at least $Xge a, Yge b$ marbles in the bag.
– Doug M
Aug 7 at 20:42
IVT only applies to continuous functions, and marbles are discrete objects. The specific case where replacement does not matter is the case that only one marble is chosen.
– Doug M
Aug 7 at 20:48
Yes. For replacement to be a factor, a+b must be >= 2, and both colors must exists in the bag.
– Calibur
Aug 7 at 20:50
As the number of marbles approaches infinity, the discreet addition of marbles becomes comparatively continuous. With the Sock problem from MindYourDecisions (youtube.com/watch?v=GuqcnrtdUCc), this logic explains that some sock drawer exists such that the chance of pulling of matching pair is 50%, but not that any previous sock drawer can be added to to attain this property.
– Calibur
Aug 7 at 20:56
"What is the smallest number of marbles in the bag to have a solution?" One marble in the bag (or even zero) is the smallest. Generally, drawing one marble (or zero) will give the same result with our without replacement. But that's boring.
– Arthur
Aug 7 at 20:37
"What is the smallest number of marbles in the bag to have a solution?" One marble in the bag (or even zero) is the smallest. Generally, drawing one marble (or zero) will give the same result with our without replacement. But that's boring.
– Arthur
Aug 7 at 20:37
There must be at least $Xge a, Yge b$ marbles in the bag.
– Doug M
Aug 7 at 20:42
There must be at least $Xge a, Yge b$ marbles in the bag.
– Doug M
Aug 7 at 20:42
IVT only applies to continuous functions, and marbles are discrete objects. The specific case where replacement does not matter is the case that only one marble is chosen.
– Doug M
Aug 7 at 20:48
IVT only applies to continuous functions, and marbles are discrete objects. The specific case where replacement does not matter is the case that only one marble is chosen.
– Doug M
Aug 7 at 20:48
Yes. For replacement to be a factor, a+b must be >= 2, and both colors must exists in the bag.
– Calibur
Aug 7 at 20:50
Yes. For replacement to be a factor, a+b must be >= 2, and both colors must exists in the bag.
– Calibur
Aug 7 at 20:50
As the number of marbles approaches infinity, the discreet addition of marbles becomes comparatively continuous. With the Sock problem from MindYourDecisions (youtube.com/watch?v=GuqcnrtdUCc), this logic explains that some sock drawer exists such that the chance of pulling of matching pair is 50%, but not that any previous sock drawer can be added to to attain this property.
– Calibur
Aug 7 at 20:56
As the number of marbles approaches infinity, the discreet addition of marbles becomes comparatively continuous. With the Sock problem from MindYourDecisions (youtube.com/watch?v=GuqcnrtdUCc), this logic explains that some sock drawer exists such that the chance of pulling of matching pair is 50%, but not that any previous sock drawer can be added to to attain this property.
– Calibur
Aug 7 at 20:56
 |Â
show 1 more comment
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"What is the smallest number of marbles in the bag to have a solution?" One marble in the bag (or even zero) is the smallest. Generally, drawing one marble (or zero) will give the same result with our without replacement. But that's boring.
– Arthur
Aug 7 at 20:37
There must be at least $Xge a, Yge b$ marbles in the bag.
– Doug M
Aug 7 at 20:42
IVT only applies to continuous functions, and marbles are discrete objects. The specific case where replacement does not matter is the case that only one marble is chosen.
– Doug M
Aug 7 at 20:48
Yes. For replacement to be a factor, a+b must be >= 2, and both colors must exists in the bag.
– Calibur
Aug 7 at 20:50
As the number of marbles approaches infinity, the discreet addition of marbles becomes comparatively continuous. With the Sock problem from MindYourDecisions (youtube.com/watch?v=GuqcnrtdUCc), this logic explains that some sock drawer exists such that the chance of pulling of matching pair is 50%, but not that any previous sock drawer can be added to to attain this property.
– Calibur
Aug 7 at 20:56