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Drawing balls from an urn without replacement: what color runs out first?
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I have $m$ white balls and $n$ black balls in an urn. I repeatedly draw balls from the urn without replacement until one color runs out. What is the probability that white will run out first?
I have the recurrence relation $p(m, n) = fracmp(m - 1, n) + np(m, n - 1)m + n$ with $p(0, n) = 1$ and $p(m, 0) = 0$, and if you guess $p(m, n) = fracnm + n$ you can show that it's a valid solution, but I have no clue how you'd derive it without the lucky guess or an intuition for why that's the solution.
probability combinatorics
asked Jul 21 at 2:47
Deifactor
133
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up vote
1
down vote
favorite
I have $m$ white balls and $n$ black balls in an urn. I repeatedly draw balls from the urn without replacement until one color runs out. What is the probability that white will run out first?
I have the recurrence relation $p(m, n) = fracmp(m - 1, n) + np(m, n - 1)m + n$ with $p(0, n) = 1$ and $p(m, 0) = 0$, and if you guess $p(m, n) = fracnm + n$ you can show that it's a valid solution, but I have no clue how you'd derive it without the lucky guess or an intuition for why that's the solution.
probability combinatorics
asked Jul 21 at 2:47
Deifactor
133
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have $m$ white balls and $n$ black balls in an urn. I repeatedly draw balls from the urn without replacement until one color runs out. What is the probability that white will run out first?
I have the recurrence relation $p(m, n) = fracmp(m - 1, n) + np(m, n - 1)m + n$ with $p(0, n) = 1$ and $p(m, 0) = 0$, and if you guess $p(m, n) = fracnm + n$ you can show that it's a valid solution, but I have no clue how you'd derive it without the lucky guess or an intuition for why that's the solution.
probability combinatorics
asked Jul 21 at 2:47
Deifactor
133
I have $m$ white balls and $n$ black balls in an urn. I repeatedly draw balls from the urn without replacement until one color runs out. What is the probability that white will run out first?
I have the recurrence relation $p(m, n) = fracmp(m - 1, n) + np(m, n - 1)m + n$ with $p(0, n) = 1$ and $p(m, 0) = 0$, and if you guess $p(m, n) = fracnm + n$ you can show that it's a valid solution, but I have no clue how you'd derive it without the lucky guess or an intuition for why that's the solution.
probability combinatorics
asked Jul 21 at 2:47
Deifactor
133
asked Jul 21 at 2:47
Deifactor
133
asked Jul 21 at 2:47
Deifactor
133
asked Jul 21 at 2:47
Deifactor
133
133
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
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accepted
Here's one way:
Consider pulling out the full sequence of all $n + m$ of them. White balls run out first if and only if the last ball in the sequence is black. The key observation now is that the probability of a given sequence happening forward is exactly the same as the probability of that sequence happening in reverse. Thus, the probability that the last ball is black is equal to the probability that the first ball is black; this probability is exactly $fracnm+n$.
answered Jul 21 at 2:56
Marcus M
8,1731847
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Here's one way:
Consider pulling out the full sequence of all $n + m$ of them. White balls run out first if and only if the last ball in the sequence is black. The key observation now is that the probability of a given sequence happening forward is exactly the same as the probability of that sequence happening in reverse. Thus, the probability that the last ball is black is equal to the probability that the first ball is black; this probability is exactly $fracnm+n$.
answered Jul 21 at 2:56
Marcus M
8,1731847
add a comment |Â
up vote
5
down vote
accepted
Here's one way:
Consider pulling out the full sequence of all $n + m$ of them. White balls run out first if and only if the last ball in the sequence is black. The key observation now is that the probability of a given sequence happening forward is exactly the same as the probability of that sequence happening in reverse. Thus, the probability that the last ball is black is equal to the probability that the first ball is black; this probability is exactly $fracnm+n$.
answered Jul 21 at 2:56
Marcus M
8,1731847
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Here's one way:
Consider pulling out the full sequence of all $n + m$ of them. White balls run out first if and only if the last ball in the sequence is black. The key observation now is that the probability of a given sequence happening forward is exactly the same as the probability of that sequence happening in reverse. Thus, the probability that the last ball is black is equal to the probability that the first ball is black; this probability is exactly $fracnm+n$.
answered Jul 21 at 2:56
Marcus M
8,1731847
Here's one way:
Consider pulling out the full sequence of all $n + m$ of them. White balls run out first if and only if the last ball in the sequence is black. The key observation now is that the probability of a given sequence happening forward is exactly the same as the probability of that sequence happening in reverse. Thus, the probability that the last ball is black is equal to the probability that the first ball is black; this probability is exactly $fracnm+n$.
answered Jul 21 at 2:56
Marcus M
8,1731847
answered Jul 21 at 2:56
Marcus M
8,1731847
answered Jul 21 at 2:56
Marcus M
8,1731847
answered Jul 21 at 2:56
Marcus M
8,1731847
8,1731847
add a comment |Â
add a comment |Â
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