# Understanding different approaches for the problem

(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 If $k$ of these balls are withdrawn one at a time, with each selection being highly equally likely to be any of the balls that remain at that time, what is the probability that the special case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.

• 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? – dxiv Jan 11 '18 at 2:35
• 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 $1/n\,$. – dxiv Jan 11 '18 at 3:42
• No, I only rephrased the problem to an equivalent of approach 2 which hopefully makes it more clear what the reasoning behind it is. – dxiv Jan 11 '18 at 3:56
• 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 $\{1, 2,\ldots,n\}$". – dxiv Jan 11 '18 at 4:09
• @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. – 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.
• 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. – yh016 Jan 11 '18 at 3:18
• @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. – drhab Jan 11 '18 at 8:35
• 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. :( – yh016 Jan 11 '18 at 10:53
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 $P(A_i) = \frac{1}{n}$.