Problem: An urn contains 3 red and 7 black balls. Player A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn)

I have the solution: $$\Bbb{P}(A)=\frac{3\cdot 9!+6\cdot 3 \cdot 7!+ 7\cdot6\cdot5\cdot4\cdot3\cdot5!+7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot3\cdot3!}{10!}$$ But i have no idea how they came up with this. Can someone explain this? Thank you.


Imagine that the balls have secret identifying numbers. and imagine that we draw out all the balls, whether someone has already won or not.

There are $10!$ ways to draw out the balls, all equally likely.

Player A could win on the first draw. How many sequences correspond to this? The winning red ball could be any of the $3$, and then the rest of the balls could come in any one of $9!$ possible orders, for a total of $(3)(9!)$.

Or else Player A could win on the third draw. That happens if the first ball is black, the second is black, the third is red, and the rest are arranged in any order. The first ball can then be chosen in $7$ ways, and for each choice the second can be chosen in $6$ ways. Then we have $3$ ways for the third, since it has to be red, and then $7!$ for the rest, for a total of $(7)(6)(3)(7!)$. This should be the second entry in the numerator. It isn't, presumably a typo. No wonder you are puzzled.

Or else Player $A$ could win on the fifth draw. The same reasoning gives $(7)(6)(5)(4)(3)(5!)$ ways. That's correctly written down.

Finally, there could be a win on the $7$-th draw. The number of sequences that give this is correctly written down.

  • $\begingroup$ Great! I get it now. Thank you. $\endgroup$ – Kes Nov 28 '12 at 20:17
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    $\begingroup$ There are better ways to do the problem. Probability of win on first is $\frac{3}{10}$. Probability of win on third is $\frac{7}{10}\frac{6}{9}\frac{3}{8}$. And so on. Add up. $\endgroup$ – André Nicolas Nov 28 '12 at 20:22

Well, the second entry on top is wrong (there should be another factor of $7$), so that could be the source of your confusion. Let's assume it's just a typo, and see how they might have gone about it.

There are $4$ mutually exclusive possibilities to consider for ways Player A can draw the red ball:

(i) The first ball drawn is red. Probability of this is $\dfrac3{10}$

(ii) The first two balls drawn are black and the third one is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac38=\dfrac{7\cdot 6\cdot 3}{10\cdot 9\cdot 8}$.

(iii) The first four balls drawn are black and the fifth one is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac58\cdot\dfrac47\cdot\dfrac36=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3}{10\cdot 9\cdot 8\cdot 7\cdot 6}$.

(iv) The first six balls drawn are black and the seventh is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac58\cdot\dfrac47\cdot\dfrac36\cdot\dfrac25\cdot\dfrac34=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}$.

[Do you see why these probabilities are correct?]

Since these events are mutually exclusive, the probability that Player A gets the red ball is simply the sum of these probabilities. The least common denominator is clearly $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4=\frac{10!}{3!}$. It seems your book chose to use $10!$ as the common denominator, instead (probably to reduce the length of the answer). Rewriting the probabilities with $10!$ as denominator, we therefore have $$\begin{align}\Bbb P(A) &= \frac{3\cdot9!}{10!}+\frac{7\cdot 6\cdot 3\cdot 7!}{10!}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 5!}{10!}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3\cdot 3!}{10!}\\ &= \frac{3\cdot9!+7\cdot 6\cdot 3\cdot 7!+7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 5!+7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3\cdot 3!}{10!}.\end{align}$$

  • $\begingroup$ Thank you, this helped me a lot $\endgroup$ – Kes Nov 28 '12 at 20:36
  • $\begingroup$ Thanks for the explanation & I understand that this is the brute-force-ish approach; do you think this can be generalized for any number of red and black balls? $\endgroup$ – KartikKannapur Jun 3 '17 at 19:23
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    $\begingroup$ @KartikKannapur: It is absolutely generalizable, but this way of displaying/calculating the probability will obviously get uglier with more balls. $\endgroup$ – Cameron Buie Jun 3 '17 at 20:13
  • $\begingroup$ I'd recommend seeking a closed form, instead. $\endgroup$ – Cameron Buie Jun 3 '17 at 20:17

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