Two Interesting, Short Probability Questions

2019-02-28

If you have $N$ coins, and $x$ of them are special in the sense that they have head both front and back. And you pick one up randomly and flip it $k$ times and get 10 heads, what is the probability that the coin you picked is one of the special ones?

Solution: Say $A$ is the event that you get 10 heads, $B$ is the event you picked one of the special coins. Then, by Law of Total Probability, $P (A) = P (A | B) \cdot P (B) + P (A | B^{C}) \cdot P (B^{C}) = 1 \cdot \frac{x}{N} + (\frac{1}{2})^{k} \cdot \frac{N - x}{N}$ Now, from the Bayes Theorem, $P (B | A) = \frac{P (A, B)}{P (A)} = \frac{\frac{1}{N}}{1 \cdot \frac{x}{N} + (\frac{1}{2})^{k} \cdot \frac{N - x}{N}} = \frac{1}{x + \frac{N - 1}{2^{k}}}$ Say for $N = 100$ , $x = 10$ , and $k = 2$ , the probability will be about $2.9 %$ .

Say you have $N$ ropes, or lines, or spaghetti. Each time you pick two ends and tie them up together. What is the expected number of loops?

Solution: This one is for bored people. So for each round, after you picked up two ends and connect them, they either form a circle, or they do not. In either case the total number of ropes available decreases by $1$ . Say in a round we have $i$ ropes left, then let $R_{i}$ be a Bernoulli r.v. denoting if a loop is formed, by Law of Total Probability, $E (R_{i}) = \sum_{j} P (picked up rope j) \cdot P (forming a loop) = \sum_{j = 1}^{2 i} \frac{1}{2 i} \cdot \frac{1}{2 i - 1} = \frac{1}{2 i - 1}$ Summing over all $N$ rounds, and let $R$ be the r.v. of total number of loops formed. $E (R) = E (\sum_{i} R_{i}) = \sum_{i} E (R_{i}) = \sum_{i = 1}^{N} \frac{1}{2 i - 1}$ Second step is true since expectation is a linear operator. So for $N = 100$ , this is about py> expected_n_of_loops = lambda N: sum(1.0/(2*i - 1) for i in range(1,N)) py> expected_n_of_loops(100) # 3.2793170636734916
Note the generator expression, without of a list or a tuple, in the sum() function.