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 $(| 0 ⟩ + | 1 ⟩) / \sqrt{2}$ which is being measured 10 times without preparation. Let's call $q$ the probability that all the measurements result in $| 0 ⟩$ . 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, $p / q = (1 / 2)^{9} = 1 / 512$ .