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 and , respectively. The constants and are needed to ensure that both probabilities sum to one; the exponents and control the distribution of values. What is the probability that `(x < y)`

is true?

Probability theory tells us that , where: is the probability distribution function for (in this case: ), and the the cumulative probability distribution for (in this case: ).

Doing the maths gives the probability of `(x < y)`

being true as: .

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 :-)