2. Coins vs qubits

Basic Probability

A fair coin is flipped 10 times. Let's call $p$ the probability that all the coin flips turn heads. Now, consider a qubit in an equal superposition state $(|0\rangle + |1\rangle)/\sqrt{2}$ which is being measured 10 times without preparation. Let's call $q$ the probability that all the measurements result in $|0\rangle$. What is the ratio $p/q$?

Hint

Think about what happens if a qubit is measured, and then measured again without preparing the same initial state.

Solution

Coin flips

For the classical case of 10 coin flips the probability of 10 heads is $$ p = (\frac{1}{2})^{10} $$

Qubits

For a quantum bit the first measurement collapses the state vector and the consequent 9 measurements result in the same value. $$ q = \frac{1}{2} $$

Therefore, $p/q = (1/2)^9 = 1/512$.