The same coin twice

Four probability questions. Try each one before revealing the solution.

Two coins

You receive 2 fair coins, each with a 50% probability of landing on heads. You flip them both. What is the probability of both coins landing on heads?

Solution

The outcomes of both coins are independent, giving a final probability of $\frac{1}{4}$.

One coin, twice

You receive a single fair coin, with a 50% probability of landing on heads. You flip it twice. What is the probability of both flips landing on heads?

Solution

The outcomes of both flips are independent, giving a final probability of $\frac{1}{4}$.

Two random coins

You receive 2 unfair coins. Each coin's probability of landing on heads is chosen uniformly from the interval $[0,1]$ (the coins might have different probabilities). You flip both coins. What is the probability of both coins landing on heads?

Solution

The probability of the first coin landing on heads can be calculated by averaging the probabilities out of the uniform distribution:

$${\int }_0^{1}pdp=\frac{1}{2}$$

The second coin can be calculated the same way for a probability of $\frac{1}{2}$. In total, since the coins are independent, the final probability is $\frac{1}{4}$.

One random coin, twice

You receive a single unfair coin, whose probability of landing on heads is chosen uniformly from the interval $[0,1]$. You flip the coin twice. What is the probability of both flips landing on heads?

Solution

A naive approach might be to treat this as equivalent to the last problem, with a probability of $\frac{1}{4}$, since the flips are independently sampled from the unfair coin. However, this is incorrect!

The first flip has a probability of $\frac{1}{2}$, calculated the same way as the previous problem. However, the first flip updates the distribution of the second flip. Intuitively, if you see the coin land on heads, it makes you suspect that the coin is biased in favor of heads, making it more likely that the second flip also lands on heads. (In the extreme case, if you flip a coin and get heads 100 times in a row, you would be concerned that something is wrong with the coin, leading you to believe that the 101st flip would also land on heads.)

The final probability is calculated by averaging the probabilities of both flips landing on heads out of a single sample of the uniform distribution:

$${\int }_0^1{p}^{2}dp=\frac{1}{3}$$