In the box there are $3$ coins. The probabilities of getting heads while tossing are equal to:
- $\frac{1}{2}$ on the first one,
- $\frac{1}{3}$ on the second one,
- $\frac{1}{4}$ on the third one.
A man chose randomly $1$ of those $3$ coins and tossed it until getting heads. Then chose randomly another one and similarly - tossed it until getting heads. What's the probability that in the box the only coin that is left is the symmetric one if we know that the man tossed exactly $n$ times?
I thought about counting the possibilities for $3 \cdot 2$ pairs of chosen coins in both orders and then the pair that we need in both orders. The first of two expressions would be put in the denominator and the other in the numerator. However, when I started with that approach I got to a point where calculating that all got too complicated. That's why I got to think that there must be more clever way to approach that problem. Any help would be much appreciated.
Edit:
Using the hint given by @lulu in the comments, I got something like this:
$$\mathbb{P} \left( \text{coins were: } \ \frac{1}{3} \text{ and } \frac{1}{4} \ \Big| \ \text{there were n toses} \right) =$$ $$ = \frac{\mathbb{P}(\text{there were n toses} \ \Big| \ \text{2 coins were:} \ \frac{1}{3} \text{ and } \frac{1}{4}) \cdot \mathbb{P}(\text{coins were: } \ \frac{1}{3} \text{ and } \frac{1}{4})}{\mathbb{P}(\text{there were n toses})}$$
$\mathbb{P}(\text{there were n toses}) = $
- $\mathbb{P} \left( \text{there were n toses} \ \Big| \ \text{2 coins chosen were those of prob. } \ \frac{1}{2} \text{ and } \frac{1}{3} \right) +$
- $\mathbb{P} \left( \text{there were n toses} \ \Big| \ \text{2 coins chosen were those of prob. } \ \frac{1}{3} \text{ and } \frac{1}{4} \right) +$
- $\mathbb{P} \left( \text{there were n toses} \ \Big| \ \text{2 coins chosen were those of prob. } \ \frac{1}{2} \text{ and } \frac{1}{4} \right)$
We get:
($k$ is the number of failed tosses on one coin and $n-k-2$ are failed tosses on another)
$1. = \displaystyle \sum_{k=0}^{n-1} \left( \frac{1}{2} \right)^k \frac{1}{2} \left( \frac{2}{3} \right)^{n-k-2} \frac{1}{3}$
$2. = \displaystyle \sum_{k=0}^{n-1} \left( \frac{2}{3} \right)^k \frac{1}{3} \left( \frac{3}{4} \right)^{n-k-2} \frac{1}{4}$
$3. = \displaystyle \sum_{k=0}^{n-1} \left( \frac{1}{2} \right)^k \frac{1}{2} \left( \frac{3}{4} \right)^{n-k-2} \frac{1}{4}$
It looks all good, but now, how do I express $1, 2, 3$ in terms of $n$, losing $k$ in the process?