How can quantum computing win 97% of times in coin flipping experiment?

Question

I'm new to this field of science. I'm curious about how quantum computing can win 97% of times in a coin flipping experiment?

Refer this link: Ted Talk by Shohini Ghose

To give an idea about how this coin experiment works:

1. Quantum Computer plays a move but it is not revealed to the Opponent[Human].
2. Opponent[Human] plays a move and it is also not revealed to the Quantum Computer.
3. Finally Quantum Computer plays a move.
4. Results are shown. If its heads, then Quantum Computer wins. Else, Opponent[Human] wins.

Here, playing a move refers to "Flipping the coin". The video talks of superposition and how it can recover heads every time in the final move. This made me think about 2 possibilities:

1. Quantum computer is tracking something low level in hardware. So, it knows every time, what did the opponent played.
2. Superposition and the third state is just a way, to not consider Opponent's Move (i.e. Ignoring Opponent's move). So, it is actually all the moves of Quantum Computer. So, it knows how to win. If this is the case, then actually there is no randomness or uncertainty added in the game by Opponent.

Just same as how the magician does the coin flipping trick. He practices and controls the flipping power of his thumb, so he knows every time he flips the coin, whether its going to be heads or tails.

Answer 1

We translate this game into conventional QC terminology as follows:

• The coin is a single qbit state $|\psi\rangle=\begin{bmatrix}\alpha \\ \beta\end{bmatrix}$ where $\alpha, \beta \in \mathbb{C}$ and $|\alpha|^2 + |\beta|^2 = 1$
• "Flipping" the coin is application of the bit-flip operator $X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
• The "heads" state is defined as $|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and "tails" as $|1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$
• The quantum computer "plays" by applying the Hadamard operator $H = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \end{bmatrix}$

The game is conducted as follows:

1. The coin is initialized to the $|0\rangle$ "heads" state.
2. The computer plays, applying the Hadamard operator to the coin (operators are applied using matrix multiplication). The coin enters the $|+\rangle = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$ state.
3. The human plays, choosing whether to flip the coin (apply the $X$ operator). However, since the $X$ operator just flips the state vector upside down, $X$ has no effect; $X|+\rangle = |+\rangle$. No matter what, the state is $|+\rangle$ after the human plays.
4. The computer plays, applying the Hadamard operator again, taking the coin to the $|0\rangle$ "heads" state.

Thus the computer always wins, except for $3\%$ of the time when errors cause it to not behave according to this model.

The example deals with the basic quantum states $|0\rangle$, $|1\rangle$, $|+\rangle$, and $|-\rangle$, plus the common ways of moving between them with the $X$, $H$, and $Z$ gates. This is the same material you learn about in IBM's Hello Quantum game.

This isn't a great example to use to illustrate how a quantum computer works (since it only makes sense if you already know how a quantum computer works), but I guess the speaker only had ten minutes.