Calculating statement execution likelihood

Derek Jones from The Shape of Code

In the following code, how often will the variable b be incremented, compared to a?

If we assume that the variables x and y have values drawn from the same distribution, then the condition (x < y) will be true 50% of the time (ignoring the situation where both values are equal), i.e., b will be incremented half as often as a.

a++;
if (x < y)
   {
   b++;
   if (x < z)
      {
      c++;
      }
   }

If the value of z is drawn from the same distribution as x and y, how often will c be incremented compared to a?

The test (x < y) reduces the possible values that x can take, which means that in the comparison (x < z), the value of x is no longer drawn from the same distribution as z.

Since we are assuming that z and y are drawn from the same distribution, there is a 50% chance that (z < y).

If we assume that (z < y), then the values of x and z are drawn from the same distribution, and in this case there is a 50% change that (x < z) is true.

Combining these two cases, we deduce that, given the statement a++; is executed, there is a 25% probability that the statement c++; is executed.

If the condition (x < z) is replaced by (x > z), the expected probability remains unchanged.

If the values of x, y, and z are not drawn from the same distribution, things get complicated.

Let's assume that the probability of particular values of x and y occurring are alpha e^{-sx} and beta e^{-ty}, respectively. The constants alpha and beta are needed to ensure that both probabilities sum to one; the exponents s and t control the distribution of values. What is the probability that (x < y) is true?

Probability theory tells us that P(A < B) = int{-infty}{+infty} f_B(x) F_A(x) dx, where: f_B is the probability distribution function for B (in this case: beta e^{-tx}), and F_A the the cumulative probability distribution for A (in this case: alpha(1-e^{-sx})).

Doing the maths gives the probability of (x < y) being true as: {alpha beta s}/{s+t}.

The (x < z) case can be similarly derived, and combining everything is just a matter of getting the algebra right; it is left as an exercise to the reader :-)