# What's the probability that the only coin left is fair given that exactly $n$ tosses were required to obtain heads with each of the first two coins?

In the box there are $$3$$ coins. The probabilities of getting heads while tossing are equal to:

1. $$\frac{1}{2}$$ on the first one,
2. $$\frac{1}{3}$$ on the second one,
3. $$\frac{1}{4}$$ 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:

$$\mathbb{P} \left( \text{coins were: } \ \frac{1}{3} \text{ and } \frac{1}{4} \ \Big| \ \text{there were n toses} \right) =$$ $$= \frac{\mathbb{P}(\text{there were n toses} \ \Big| \ \text{2 coins were:} \ \frac{1}{3} \text{ and } \frac{1}{4}) \cdot \mathbb{P}(\text{coins were: } \ \frac{1}{3} \text{ and } \frac{1}{4})}{\mathbb{P}(\text{there were n toses})}$$

$$\mathbb{P}(\text{there were n toses}) =$$

1. $$\mathbb{P} \left( \text{there were n toses} \ \Big| \ \text{2 coins chosen were those of prob. } \ \frac{1}{2} \text{ and } \frac{1}{3} \right) +$$
2. $$\mathbb{P} \left( \text{there were n toses} \ \Big| \ \text{2 coins chosen were those of prob. } \ \frac{1}{3} \text{ and } \frac{1}{4} \right) +$$
3. $$\mathbb{P} \left( \text{there were n toses} \ \Big| \ \text{2 coins chosen were those of prob. } \ \frac{1}{2} \text{ and } \frac{1}{4} \right)$$

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?

• Bayes Theorem is good here. Note that the order of the first two choices doesn't matter, so there are only three cases. Just compute the probability that it takes exactly $n$ tosses for each of the three cases and apply Bayes.
– lulu
Commented Jun 5, 2023 at 18:11
• @lulu I edited my question uning your hint but now I have a problem with series of probability. Commented Jun 6, 2023 at 0:29
• Hello. There is a small typo in "($k$ is the number of failed tosses on one coin and $n−n−2$ are failed toses on another)". I should be $n-k-2$. Commented Jun 6, 2023 at 2:33
• @RenatoFernandes yes, you are right, I edited my question. Thank you! Commented Jun 6, 2023 at 2:37
• tip: $\displaystyle \sum_{k=0}^{n-1} \left( \frac{1}{2} \right)^k \left( \frac{2}{3} \right)^{n-k-2} = \sum_{k=0}^{n-1} \left( \frac{1}{2} \right)^k \left( \frac{2}{3} \right)^{-k} \left( \frac{2}{3} \right)^{n-2} =\left( \frac{2}{3} \right)^{n-2}\; \sum_{k=0}^{n-1} \left( \frac{3}{4} \right)^k$ Commented Jun 6, 2023 at 3:17