1.1 Probabilistic bits

Probabilities are used to quantify the likelihood of events – the more likely an event is, the larger its probability. An event that occurs with certainty has probability 11 (indeed, it has a 100% chance of happing) while an event that never occurs has probability 0.

As an example, we can think of tossing a coin. If you toss a coin and cover it with your hands without looking at it, there are two possible events – either the coin shows “heads” or it shows “tails”. For a fair coin, the two events are equally likely to occur, so we assign a probability of 12=0.50\frac{1}{2}=0.50 or 50%50\% to both (this is represented by [Uncaptioned image] in Fig. 1.1). However, the coin could be biased and more likely to land on one side than the other. Depending on how biased it is, we can imagine a whole spectrum of possibilities: one extremely biased coin might always show “heads” while another might always show “tails” (see [Uncaptioned image] and [Uncaptioned image] on the left and right side of Fig. 1.1). The first coin shows “tails” with probability 0 (since it always shows “heads”) while the second shows “tails” with probability 11.


Figure 1.1: A probabilistic bit describing the state of a random donkey coin.

Since we don’t want to worry about designing coins of different shapes and sizes, it is helpful to somehow abstract the information represented by a coin toss. We can do this by associating the outcomes “heads” and “tails” with the bit values 0 and 11, respectively. The coin toss can then be described by two probabilities: the probability p0p_{0} that it is equal to 0 (“heads”), and the probability p1p_{1} that it is equal to 1 (“tails”). Such a bit that attains its two possible values with some probabilities is called a probabilistic bit. Note that the two probabilities p0,p10p_{0},p_{1}\geq 0 must necessarily add to one: p0+p1=1p_{0}+p_{1}=1. For example, if the coin is fair then p0=p1=12p_{0}=p_{1}=\frac{1}{2}, as explained above.

Did you notice that it would be enough to specify just one of the probabilities p0p_{0} or p1p_{1} since either of them can be obtained from the other? For clarity we will always specify both and write them down as a vector:

p=(p0p1).p=\begin{pmatrix}p_{0}\\ p_{1}\end{pmatrix}. (1.1)

We call this vector the probability distribution or the state of the probabilistic bit. This vector notation is convenient not only for collecting all probabilities together in a nice table, but it will also provide a geometrical way to visualize a probabilistic bit. Moreover, it will help us to see the analogy between probabilistic and quantum bits.

We will call the states associated to 0 or “heads” and to 11 or “tails” deterministic, since in this case there is no uncertainty – is it fully determined which side of the coin is up. In Eq. 1.1, they correspond to probability distributions with p0=1p_{0}=1, p1=0p_{1}=0 and p0=0p_{0}=0, p1=1p_{1}=1, respectively. Since they will be used very often, it is convenient to introduce a shorthand notation for them:

[0]=(10),[1]=(01).\displaystyle[0]=\begin{pmatrix}1\\ 0\end{pmatrix},\qquad[1]=\begin{pmatrix}0\\ 1\end{pmatrix}. (1.6)

This notation might look confusing at first, but you can also think of [0][0] and [1][1] as [Uncaptioned image] and [Uncaptioned image] or as [heads][\text{heads}] and [tails][\text{tails}], if you wish.

These two states form a basis of all states, meaning that we can express any other state of a probabilistic bit as a linear combination of them:

(p0p1)=p0(10)+p1(01)=p0[0]+p1[1].\begin{pmatrix}p_{0}\\ p_{1}\end{pmatrix}=p_{0}\begin{pmatrix}1\\ 0\end{pmatrix}+p_{1}\begin{pmatrix}0\\ 1\end{pmatrix}=p_{0}[0]+p_{1}[1]. (1.7)

Every state is a two-dimensional vector, which we can visualize in a two-dimensional coordinate system. Then the possible states of a probabilistic bit is precisely the line segment connecting the two points [0][0] and [1][1] which correspond to the two deterministic states of the bit (see Fig. 1.2).

Figure 1.2: The blue line segment corresponds to the states of a probabilistic bit.

Exercise 1.1 (Understanding the blue line segment).

According to Fig. 1.2, the possible states of a probabilistic bit form a line segment. Take some time to investigate this. See if you can answer the following questions:

  1. 1.

    Why do the states of a probabilistic bit all lie on a line?

  2. 2.

    Why does this line end at the coordinate axes and does not go further?

  3. 3.

    Which point on the line segment corresponds to a fair coin?

Solution.
  1. 1.

    Since one of the two events has to occur, the two probabilities necessarily add to 1. This means that p0+p1=1p_{0}+p_{1}=1. If you write this as p1=1p0p_{1}=1-p_{0} you recognize this as the equation of a line with slope minus one.

  2. 2.

    If the line would go further one of the probabilities would become negative. Since probabilities cannot be negative, we must require that p00p_{0}\geq 0 and p10p_{1}\geq 0, which is equivalent to saying that the line segment must end at the coordinate axes.

  3. 3.

    This is the midpoint, where p0=p1=12p_{0}=p_{1}=\frac{1}{2}.