MHB Probability Challenge: Find Interval of Integers Drawn from Urn

AI Thread Summary
The probability challenge involves determining the likelihood that the numbers drawn from an urn of n balls form a continuous interval of integers throughout the drawing process. An allowable process requires that each subsequent ball drawn must be adjacent to the current sequence of drawn integers, represented by a string of 'L' and 'U' letters. The number of allowable sequences starting with a ball numbered r is given by the binomial coefficient (n-1 choose r-1), leading to a total of 2^(n-1) allowable processes. Since the total number of ways to draw the balls is n!, the probability of an allowable process is calculated as 2^(n-1) / n!. This probability reflects the constraints on the drawing sequence necessary for maintaining an interval of integers.
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An urn contains $n$ balls numbered $1, 2, . . . , n$. They are drawn one at a time at random until the urn is empty.
Find the probability that throughout this process the numbers on the balls which have been drawn is an interval of integers.
(That is, for $1 \leq k \leq n$, after the $k$th draw the smallest number drawn equals the largest drawn minus $k − 1$.)
 
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lfdahl said:
An urn contains $n$ balls numbered $1, 2, . . . , n$. They are drawn one at a time at random until the urn is empty.
Find the probability that throughout this process the numbers on the balls which have been drawn is an interval of integers.
(That is, for $1 \leq k \leq n$, after the $k$th draw the smallest number drawn equals the largest drawn minus $k − 1$.)
[sp]Call the process "allowable" if it satisfies that condition.

Suppose that the first ball drawn is numbered $r$. If the process is to be allowable then the number on each subsequent ball drawn must be next to either the lower end ($L$) or the upper end ($U$) of the existing consecutive run of integers. There are $n-1$ more balls to be drawn, so the process is completely specified by a string of $n-1$ letters $L$ and $U$. Also, there are exactly $r-1$ numbers less than $r$, so the string must contain $r-1$ $L$s (and $n-r$ $U$s). The number of such strings is $$n-1\choose r-1$$. Therefore there are $$n-1\choose r-1$$ allowable processes starting with $r$. So the total number of allowable processes is $$\sum_{r=1}^n{n-1\choose r-1} = 2^{n-1}.$$ The number of all ways of drawing the balls from the urn is $n!$. Therefore the probability of a process being allowable is $\dfrac{2^{n-1}}{n!}.$

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Opalg said:
[sp]Call the process "allowable" if it satisfies that condition.

Suppose that the first ball drawn is numbered $r$. If the process is to be allowable then the number on each subsequent ball drawn must be next to either the lower end ($L$) or the upper end ($U$) of the existing consecutive run of integers. There are $n-1$ more balls to be drawn, so the process is completely specified by a string of $n-1$ letters $L$ and $U$. Also, there are exactly $r-1$ numbers less than $r$, so the string must contain $r-1$ $L$s (and $n-r$ $U$s). The number of such strings is $$n-1\choose r-1$$. Therefore there are $$n-1\choose r-1$$ allowable processes starting with $r$. So the total number of allowable processes is $$\sum_{r=1}^n{n-1\choose r-1} = 2^{n-1}.$$ The number of all ways of drawing the balls from the urn is $n!$. Therefore the probability of a process being allowable is $\dfrac{2^{n-1}}{n!}.$

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Awesome - thankyou for your participation, Opalg!
 
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