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Assume we have $n$ red and $n$ green balls in a box. What is the probability that blindfolded, you will pick a red ball on the third pick, if you learn that at least one red ball was picked on the first two picks, assuming the first two balls were not replaced?

My thoughts:

Using Bayes theorem, it is easy to show that if we consecutively pick balls blindfolded without replacement, the probability that the $i$th pick is say, red, is $\dfrac 12 \; (\star)$ for any $i.$ (seems counter-intuitive in the first place!).

Then, let $A$ be the event of picking red in the third pick and $B$ be the event of picking at least one red in the first and second picks. What we want is $P(A|B)$ which by Bayes is $\dfrac{P(B|A)P(A)}{P(B)}.$ Using $(\star),\; P(A) = 1/2.$ Also $P(B)=\dfrac 12 + \dfrac 12 - \dfrac 14=\dfrac 34.$

For $P(B|A),$ I think, it's easier to compute $P(B^c|A)$ where $B^c$ is the complement of $B$ i.e. the event that the first and second picks are green. If I define $D$ to be the event that the first pick is green and $E$ the second pick is green, then I have found no reason to believe that $D|A$ and $E|A$ are independent and so $P(B^c|A) \neq P(D|A) P(E|A).$ This is where I got stuck!

Ehsan M. Kermani
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    I gave a different way to understand this kind of problem, on a similar question at http://math.stackexchange.com/q/1043245/21820. – user21820 Dec 09 '14 at 07:09
  • So if you don't mind using that method to this problem it becomes very easy. =) – user21820 Dec 09 '14 at 07:12
  • Dear @user21820, thank you, that totally works. It is indeed elegant. Though, using Bayes is a mystery to me now. – Ehsan M. Kermani Dec 09 '14 at 07:30
  • I don't recommend using Bayes' rule unless the problem itself invites such considerations. In general always remember that probability can be defined as $#(solutions)/#(possibilities)$ and sometimes all the symmetries will immediately become apparent. For another example try finding for a shuffled standard deck the expected number of consecutive pairs of cards of the same suit, where overlapping pairs are counted as separate pairs. – user21820 Dec 09 '14 at 08:09
  • @user21820, That's easy using the indicator random variables for suited consecutive pairs and linearity of expectation. If we don't count (king, ace) as a pair, we'll have $48$ suited pairs and each expectation equals $\frac{2!51!}{52!},$ so the expected number is $48 \frac{2!51!}{52!}= \frac{48}{26} \simeq 1.84$ – Ehsan M. Kermani Dec 09 '14 at 09:06
  • Yes that is the method (even though my question was slightly misinterpreted). I guess it wasn't a very good example, but counting permutations helps to motivate an equivalent discrete solution. – user21820 Dec 09 '14 at 09:51

2 Answers2

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The blindfolded is just supposed to be the same as a random selection. What is the chance the first two picks were red? Without the information that there was at least one red ball in the first two picks, it is $\frac 12 \cdot \frac {n-1}{2n-1}$. The chance of one of each is $\frac n{2n-1}$, again without knowing that there was at least one red in the first two. We need to scale these up based on the new information, so the chance of two red is $\frac 12 \cdot \frac {n-1}{2n-1}\cdot \frac{4n-2}{3n-1}$. Similarly the chance the first two picks are opposite is $\frac n{2n-1}\cdot \frac{4n-2}{3n-1}.$ After the first two picks, either there are $n-1$ of each color and the chance the third is red is $\frac 12$, or there are $n-2$ red ones and $n$ green ones and the chance of the third is red is $\frac {n-2}{2n}$

Ross Millikan
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There is another way to apply Bayes's Theorem. Recall that $P(B|A)P(A) = P(B \cap A).$ Therefore $$ P(A \mid B) = \frac{P(B \cap A)}{P(B)}. $$

The method suggested by this answer to a similar question (cited in a comment on your question), applied to your question, would be to count permutations of the balls in which at least one of the first two is red and the third is red, and divide by the number of permutations in which at least one of the first two is red.

To use Bayes's Theorem, you might let $A$ be the event that the third ball is red, and $B$ be the event that at least one of the first two balls is red. Then the statement of Bayes's Theorem above starts to look a lot like the permutation-counting approach, except that it has (for example) $P(B)$ in place of "number of permutations that satisfy $B$."

In fact, one way to compute $P(B)$ would be simply to find the number of permutations that satisfy $B$ and divide by the total number of permutations. If we apply the same method to compute $P(B \cap A)$, then in $\frac{P(B \cap A)}{P(B)}$ we can multiply both the numerator and denominator by the total number of permutations, and find that

$$ P(A \mid B) = \frac{\mbox{number of permutations with initial RRR, RGR, or GRR}} {\mbox{number of permutations with initial RR, RG, or GR}} $$

which is the same as the permutation-counting method.

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David K
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