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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?

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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 $\frac{1}{n}$.

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  • $\begingroup$ I agree with the premise. But I don't see how probability of ordering relates to probability of particular ball being drawn. $\endgroup$ – Yal dc Nov 30 '13 at 15:02
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    $\begingroup$ Could you explain why having less (i-1) people doesn't reduce the denominator for ith step? $\endgroup$ – Yal dc Nov 30 '13 at 16:31
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    $\begingroup$ 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. $\endgroup$ – Yal dc Nov 30 '13 at 16:34
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Nov 30 '13 at 16:36
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    $\begingroup$ 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? $\endgroup$ – André Nicolas Nov 30 '13 at 16:39
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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.

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  • $\begingroup$ 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? $\endgroup$ – 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.

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