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An urn has r red and w white balls that are randomly removed one at a time. Let $R_i$ be the event that the $i$th ball removed is red. Find $P(R_i)$

I started with calculating $P(R_1) = \frac{r}{r+w}$. Next, $$P(R_2) = P(R_2|R_1)\cdot P(R_1) + P(R_2|W_1)\cdot P(W_1) = $$ $$ = \frac{r-1}{r-1+w}\cdot\frac{r}{r+w}+\frac{r}{r+w-1}\cdot\frac{w}{r+w} = \frac{\left(r-1+w\right)r}{\left(r+w-1\right)\left(r+w\right)} = \frac{r}{r+w}$$

And so on. The answer is $P(R_i) = \frac{r}{r+w}$.

The textbook's answer shows that the calculations weren't necessary:

$\frac{r}{r+w}$ because each of the $r + w$ balls is equally likely to be the $i$th ball removed.

I don't see why each ball is equally likely. At any ith step we don't have $(i-1)$ balls! Could someone help me to understand this?

Yal dc
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  • you can have a look on http://math.stackexchange.com/questions/65666/finding-the-probability-that-red-ball-is-among-the-10-balls?rq=1 also. – user81411 Dec 30 '16 at 09:39

3 Answers3

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All orderings of the balls are equally likely. So the probability that the $i$-th ball is red is the same as the probability that the first ball is red.

Think of the $n=r+w$ balls as people. The probability that Charlie is the $i$-th person chosen is the same as anyone else's. So the probability that Charlie is the $i$-th person chosen is $\frac{1}{n}$.

André Nicolas
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  • I agree with the premise. But I don't see how probability of ordering relates to probability of particular ball being drawn. – Yal dc Nov 30 '13 at 15:02
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    Could you explain why having less (i-1) people doesn't reduce the denominator for ith step? – Yal dc Nov 30 '13 at 16:31
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    The Charlie's probability is the same as anyone else's only before the draw begins. After that it should change, because some people are drawn. – Yal dc Nov 30 '13 at 16:34
  • Well, it does reduce the denominator. When we have $k$ people left, one could divide by $k$, but the number of reds left would be a random variable $R_k$. If you insist on calculating the answer via conditioning, the calculations get complicated. Doable, but not a good way to solve the problem. – André Nicolas Nov 30 '13 at 16:36
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    But we don't care about Charlie's probability given that some draws have been made. The number we want is Charlies's probability of being $i$-th, period. Suppose there are $50$ people in the urn before the drawing begins. What would you say is Charlie's probability of being the $7$-th chosen? – André Nicolas Nov 30 '13 at 16:39
  • for 1 comment) I don't insist. You see, there were several similar problems that I've solved via direct computation. Had I known this nice result earlier, it would saved some time. Unfortunately, I fail to understand why the probability {ith ball is red} = {1st ball is red}. (for 2 comment) I didn't realize the "don't care" part of your comment, although I calculated this way. Thank you for stressing this. As to your question: I'm confused: though I "know" the answer is 1/50, something makes me doubt. – Yal dc Nov 30 '13 at 16:55
  • Maybe I should not have said first. I should have said the probability the $i$-th is red is the same as the probability $j$-th is red. First, last, middle, they are all the same. Last is good to sidetrack the wish to follow the drawing. Symmetries are very important in combinatorial probability. – André Nicolas Nov 30 '13 at 17:02
  • Check this, please: The sample space consists of $50!$ outcomes. Fix our dear Charlie at 7th place. We have 49 people left that gives us 49! outcomes. That's why, 1/50. – Yal dc Nov 30 '13 at 17:18
  • Applying that same reasoning to the problem in my post: the sample space is of (n+m)! outcomes. # of favorable outcomes is (n+m-1)!. But here we have Charlie's clones, so we cannot distinguish between them. $(n+m-1)! \cdot n \Rightarrow n/(n+m)$. Right? – Yal dc Nov 30 '13 at 17:22
  • That's fine. The result is clear by symmetry, but if a computation makes it clearer, good. – André Nicolas Nov 30 '13 at 17:22
  • thank you for your help and patience! – Yal dc Nov 30 '13 at 17:32
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    You are welcome. The issue becomes important when we are calculating the mean of the hypergeometric, and in a number of other situations. – André Nicolas Nov 30 '13 at 17:38
  • Why is the number of favorable outcome that $i$th ball is red $(n+m-1)!$? – user81411 Dec 31 '16 at 05:14
  • I have revisited the derivation of mean of hypergeometric distribution http://math.stackexchange.com/questions/1380460/derivation-of-mean-and-variance-of-hypergeometric-distribution. But I can't see where is the application of symmetry there? – user81411 Dec 31 '16 at 05:38
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It is important to understand this. Try to think in a way of: I draw all balls at the same time and then I give them at random the numbers $1$, $2$ etcetera. The one with number $1$ on it is considered to be the ball drawn first, etc.. Maybe that helps.

Edit

Person A and person B both are ordered to pick a ball from the urn. Habit of A: he takes the first ball touched by him. Habit of B: he touches $i-1$ distinct balls (let's say without looking at them) and after that he picks a ball distinct from the balls allready touched by him. Will B have a larger or smaller probability to pick a red ball? No. If B would have taken the touched balls out of the urn then it would be the $i$-th ball taken out that we are talking about here.

drhab
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  • Unfortunately, not much. Compare "number 1" and "number i > 1". Probability of any ball to have number 1 is $\frac1n$; where n is total number of balls. But the probability to have number i for balls left is $\frac{1}{n-(i-1)}$. Right? – Yal dc Nov 30 '13 at 15:06
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I see that it has been a bit of time, but I stumbled here from a different question. The best way to stimulate your intuition here is think in terms of cards.

Assume that you have a shuffled deck of cards with R red and W white cards. What is probability that ith card is red? This is fairly straightforward as R/(R+W).

The order of balls throws one off, in cards it is easier to see that order of initial cards does not matter.

Glorfindel
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Atul G
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