Two-leg knockout probability: trees and penalties

In every two-leg knockout tie, fans ask the same question: what are my team’s chances of going through? Two-leg ties are a great probability lab, because they combine two events in sequence, a tiebreak and a penalty shootout. Here we use two hypothetical teams, A and B, with made-up probabilities. The goal is not to predict anything, but to show the math that any forecasting model relies on.

Each match has three possible results for Team A: win (W), draw (D) or loss (L). Over two matches the sample space has 3 × 3 = 9 paths: WW, WD, WL, DW, DD, DL, LW, LD and LL. This is exactly a tree diagram: the first level has three branches, and each branch splits into three more.

To get the probability of one specific path we use the product rule. If the two matches are independent (the first result does not change the odds of the second), the probability that both events happen is the product of their probabilities.

Independence is a simplification. In reality, a team that wins the first leg may play the second more defensively, which changes the odds. Richer models use conditional probability, $P(B \mid A)$, instead of $P(B)$. For learning the structure, the independent model is the right starting point.

One more detail: under the current rules the tie is decided by aggregate goal difference, and a level aggregate goes to penalties. To avoid modeling exact scores, assume every win in this example is by one goal. Then WL and LW cancel out, DD does too, and those three paths go to penalties.

Suppose Team A’s probabilities are as follows. First leg, at home: win 0.5, draw 0.3, loss 0.2. Second leg, away: win 0.3, draw 0.3, loss 0.4. In each match the three probabilities add up to 1, as they must. Also suppose that, if it goes to penalties, each team has a 50% chance of winning.

Check. Compute penalties directly instead of using the complement: WL gives 0.5 × 0.4 = 0.20, LW gives 0.2 × 0.3 = 0.06 and DD gives 0.3 × 0.3 = 0.09. The sum is 0.35, matching step 4. And the three groups together add up to 0.39 + 0.26 + 0.35 = 1, confirming no path was missed or double-counted. By the same logic, Team B advances with probability 1 − 0.565 = 0.435.

Notice how much penalties matter: if A had a 55% shootout edge instead of 50%, the total would rise to 0.39 + 0.35 × 0.55 = 0.5825. A small edge in an event that occurs in over a third of scenarios moves the final answer.

The shootout itself can be modeled too. Suppose every taker scores with probability 0.75 and kicks are independent. What is the chance a team scores all of its first five? By the product rule it is 0.75 to the fifth power, about 0.237. So the chance of missing at least one is the complementary event:

The complement is the right tool whenever you see the words “at least one”. Computing “exactly one miss, or exactly two, or…” directly would mean adding five separate terms, each with its own combinations. With the complement it is a single calculation: 1 − 0.237 ≈ 0.763. In this model, a team misses at least one of its five kicks in about 76% of shootouts, which helps explain why shootouts rarely end 5–5.

• Adding instead of multiplying. The chance of winning the first leg and the second is 0.5 × 0.3 = 0.15, not 0.5 + 0.3. “And” between independent events means product; “or” between non-overlapping paths means sum.
• Missing paths. Listing only WW, DD and LL drops six of the nine paths. Draw the full tree before calculating.
• Treating a level aggregate as a loss. Paths that go to penalties still carry a chance of qualifying; ignoring them underrates both teams.
• Assuming independence silently. It is a modeling assumption, not a fact. If the first leg changes how the second is played, use conditional probability.
• Confusing probability with certainty. A team with a 56.5% chance still goes out in about 43.5% of scenarios.