(A First Course in Probability, Sheldon Ross) Example 5d: An urn contains $n$ balls, one of which is special. If $k$ of these balls are withdrawn one at a time, with each selection being highly equally likely to be any of the balls that remain at that time, what is the probability that the special case is chosen?

Let $A_i$ be the event that the special ball is the $i$th ball chosen.

Two approaches are suggested to determine $P(A_i)$.

$\underline{\text{Approach }1}$

Total number of ways to select $k$ balls from the urn (with ordering) = $\frac{n!}{(n-k)!}$

Total number of ways to select $k$ balls with the $i$th ball as the special ball (with ordering) = $\frac{(n-1)!}{(n-k)!}$

Hence, $P(A_i)= \frac{(n-1)!}{(n-k)!} \bigg/ \frac{n!}{(n-k)!} =\frac{(n-1)!}{n!}=\frac{1}{n}$.

$\underline{\text{Approach }2}$

Since each of the $n$ balls is equally likely to be the $i$th ball chosen, it follows that $P(A_i)=\frac{1}{n}$.

$\underline{\text{My question}}$

I tried the example and my workings agree with Approach $1$. The sample space of the experiment is the set of $k$-tuples such that each entry corresponds to one of the $n$ balls and no two entries are the same. Any element in the event has the form $(x_1,\ldots,x_{i-1}, \text{special ball} , x_{i+1},\ldots,x_k)$.

But I do not understand Approach $2$. It seems like a completely different experiment with its sample space. I would describe the experiment as :

An urn contains $n$ balls, one of which is special. One ball is chosen at random from the urn. The sample space of the new experiment is $\{ball 1, ball 2,\ldots, ball n\}$ and the event is $\{special ball\}$.

Since $A_i$ is an event of the sample space proposed in the problem, shouldn't it be a subset of the sample space? For instance, $A_1 = \{(special ball, x_2,\ldots,x_k) : x_i \in \{n \text{ balls in urn}\}\text{ for }i=2,\ldots,k\}$. I don't understand how this is reduced to the boxed problem above.

  • $\begingroup$ Say you had the balls numbered $1$ to $n$ instead. Would you agree that the probabilities of drawing any given number $k \in \{1, 2,\ldots, n\}$ on the $i^{th}$ draw are equal? $\endgroup$ – dxiv Jan 11 '18 at 2:35
  • $\begingroup$ Then that's what approach 2 says: one of the numbers $k$ is designated to be "special", but that doesn't change its probability of being drawn on the $i^{th}$ draw, which remains at $1/n\,$. $\endgroup$ – dxiv Jan 11 '18 at 3:42
  • $\begingroup$ No, I only rephrased the problem to an equivalent of approach 2 which hopefully makes it more clear what the reasoning behind it is. $\endgroup$ – dxiv Jan 11 '18 at 3:56
  • $\begingroup$ I already explained it as better as I could in the first two comments. Guess I don't understand where the difficulty lies with recognizing that "one special ball among $n$ balls" is the same thing as "one special number $k$ in $\{1, 2,\ldots,n\}$". $\endgroup$ – dxiv Jan 11 '18 at 4:09
  • $\begingroup$ @dxiv In the original question in the book, it was already assumed that the balls in the urn are distinguishable. Thanks for your comments, but i don't get what you are saying. $\endgroup$ – yh016 Jan 11 '18 at 4:27

Perhaps you agree that it doesn't matter what $i$ is; the probability is the same whether $i$ is $1$ or $5$. Then we can take $i$ to be $1$. We can then ignore all subsequent drawings -- and our sample space and experiment is now exactly how you described it in your take on Approach #2.

  • $\begingroup$ Why can we ignore the subsequent drawings? This reminds me of a similar problem related to the Gambler's Ruin. Let $T_{i}(k)$ be the probability of hitting state $i$ without hitting state $0$, starting at state $k$. Then $T_{i}(i-2) = T_{i-1}(i-2) \times T_{i}(i-1)$, which we ignore the subsequent games played after hitting state $i$ as well. $\endgroup$ – yh016 Jan 11 '18 at 3:18
  • $\begingroup$ @yh016 Essential is that $P(A_i)=P(A_1)$ for every $i$ (as made clear in the answer). If you understand that (do you?) then in order to find $P(A_i)$ it is enough to focus on the first drawing of a ball, because it is enough to find $P(A_1)$. In that sense it can be said that subsequent drawings are irrelevant. This approach is IMHO much more elegant than the other. $\endgroup$ – drhab Jan 11 '18 at 8:35
  • $\begingroup$ Yup, I get that $P(A_i)=P(A_1)$, and so it suffice to find $P(A_1)$ instead of $P(A_2)$ etc. To find $P(A_1)$, I don't understand why we can consider the sample space of $\{ball 1,ball 2,\ldots, ball n\}$, instead of all set of k-tuples such that each entry corresponds to one of the n balls and no two entries are the same, as proposed in the problem. I'm missing something and I can't figure it out. Sorry. :( $\endgroup$ – yh016 Jan 11 '18 at 10:53
  • $\begingroup$ I have included more explanation on the part whereby I have much difficulty understanding in the original post. $\endgroup$ – yh016 Jan 11 '18 at 11:33
  • $\begingroup$ @yh016 As soon as we draw the first ball, we know whether we have drawn the special ball first or not. Just as we can ignore (I.e. exclude from our definition of experiment) whether we take a bathroom break because it will not affect whether we draw the special ball first, we can ignore subsequent drawings because they will not affect whether we draw the special ball first. $\endgroup$ – BallBoy Jan 11 '18 at 11:47

Approach #2 is best understood as follows. It has the same sample space as Approach #1. However, the argumentation takes advantage of symmetry. We have $P(A_i)=\frac{\#(\text{The ith ball drawn is the special ball})}{\#(\text{Full sample space})}$. But by symmetry, $P(A_i)$ = $P(\text{ball #2 is drawn ith)} = P(\text{ball #3 is drawn ith}) = \dots$, and the sum of all these probabilities is $1$, because the union of all sets consisting of outcomes where the $i$th ball drawn is any particular ball is the full sample space. Then we have $nP(A_i) = 1$, so $P(A_i) = \frac{1}{n}$.

  • $\begingroup$ Thanks for your clear explanation. I'd wait for different answers before choosing the most helpful answer. $\endgroup$ – yh016 Jan 11 '18 at 2:54
  • $\begingroup$ @yh016 You're welcome! There are often multiple ways to look at these types of problems. I just posted a different one in case it's more intuitive to you. $\endgroup$ – BallBoy Jan 11 '18 at 3:07

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