Two chess players play one game. If the first player wins…

Two chess players play one game. If the first player wins, this is event $A$; if the second player wins, this is event $B$. What event should be added to this set of events to obtain a complete system of events?

Step 1: What a complete system needs

• No two events can occur together (pairwise incompatible).
• At least one of them occurs in every trial (their sum is $\Omega$).

Step 2: The possible results of a chess game

A finished game of chess has exactly three possible results:

So $\Omega = \{\text{first wins},\ \text{second wins},\ \text{draw}\}$.

Step 3: Are $A$ and $B$ incompatible?

Both players cannot win the same game, so $AB=\varnothing$. ✓

Step 4: Do $A$ and $B$ cover everything?

No. If the game is drawn, neither $A$ nor $B$ occurs, so $A+B\neq\Omega$. The system is not yet complete.

Step 5: The missing event

We add

\[ \boxed{C = \overline{A+B} = \overline{A}\,\overline{B} = \text{"the game ends in a draw"}.} \]

"Neither player wins" is the same as "draw".

Step 6: Check that $A$, $B$, $C$ form a complete system

Incompatible:

\[ AB = \varnothing, \qquad AC = A\,\overline{A}\,\overline{B} = \varnothing, \qquad BC = B\,\overline{A}\,\overline{B} = \varnothing. \]

Exhaustive:

\[ A + B + C = (A+B) + \overline{A+B} = \Omega. \]

So exactly one of "first wins", "second wins", "draw" occurs in every game. ✓

Step 7: Consequence for probabilities

For a complete system, $P(A)+P(B)+P(C)=1$. For example, if the first player wins with probability $0.4$ and the second with probability $0.3$, then the probability of a draw is

\[ P(C) = 1 - 0.4 - 0.3 = 0.3. \]

Step 8: A general rule

If incompatible events $A_1,\dots,A_n$ do not yet cover $\Omega$, adding the single event $\overline{A_1+\dots+A_n}$ ("none of them occurs") always turns them into a complete system.