Lightbulb Circuit Combinations

An electrician arranges 12 light bulbs in a circle and connects each pair of bulbs with a wire. A pair of bulbs refers to any two bulbs that are directly connected by a wire. Each wire is independently active with a probability of $\frac{1}{2}$ . A group of light bulbs (ranging from 2 to 12) forms a complete circuit if there is an active wire between every possible pair of bulbs in that group. What is the expected number of complete circuits that can be formed with exactly 4 light bulbs?

Answer

First, we count the number of ways to choose a subset of 4 light bulbs from 12:

\binom{12}{4} = 495.

Label these potential circuits/combinations from 1 to 495. Let

X_1, X_2, \dots, X_{495}

be indicators for each circuit, where $$ X_i = 1 $$ if the $i^\text{th}$ circuit is complete and 0 otherwise. The total number of 4-bulb circuits is

T = X_1 + X_2 + \cdots + X_{495}.

By linearity of expectation,

E[T] = 495 \times E[X_1].

A circuit is complete if every one of the 6 wires between the 4 bulbs is active (there are 6 pairs without repetition and without order in a set of 4), which occurs with probability

\left(\frac{1}{2}\right)^6 = \frac{1}{64}.

Thus,

E[X_1] = \frac{1}{64}.

Therefore, the expected number of complete 4-bulb circuits is

E[T] = 495 \times \frac{1}{64} \approx 7.73.