Posted on June 16, 2026
Monthly Mindbenders (June 2026). A Markov chain problem with a surprising invariant.
Your favorite team is in the Stanley Cup Finals, a best-of-7 series (first to win 4 games). Against this opponent, your team wins each game with probability \(\frac{1}{3}\), unless your team is strictly behind in the series (fewer wins than losses so far), in which case your team wins with probability \(\frac{2}{3}\).
What is the probability that your team wins the series?
Simpler version: What if it were best-of-3 (first to win 2 games)?
Extra challenge: What if it were best-of-77 (first to win 39 games)?
All three variants have the same answer: \(\dfrac{1}{3}\).
In fact, the probability is \(\frac{1}{3}\) for any best-of-\((2n-1)\) series, regardless of \(n\).
Represent each state as (your wins, opponent wins). Your team is behind when your wins are strictly fewer than your losses, so it wins the next game with probability \(\frac{2}{3}\); otherwise with probability \(\frac{1}{3}\).
Working backward from the terminal states:
| State | Status | P(win series) |
|---|---|---|
| (1,1) | Tied | \(\frac{1}{3}\) |
| (1,0) | Ahead | \(\frac{1}{3}(1)+\frac{2}{3}\cdot\frac{1}{3}=\frac{5}{9}\) |
| (0,1) | Behind | \(\frac{2}{3}\cdot\frac{1}{3}+\frac{1}{3}(0)=\frac{2}{9}\) |
| (0,0) | Tied | \(\frac{1}{3}\cdot\frac{5}{9}+\frac{2}{3}\cdot\frac{2}{9}=\frac{9}{27}=\frac{1}{3}\) |
The same backward induction over all 15 non-terminal states. The tied states (shaded green) all equal \(\frac{1}{3}\):
| State | Status | P(win series) |
|---|---|---|
| (3,3) | Tied | \(\frac{1}{3}\) |
| (3,2) | Ahead | \(\frac{5}{9}\) |
| (2,3) | Behind | \(\frac{2}{9}\) |
| (3,1) | Ahead | \(\frac{19}{27}\) |
| (2,2) | Tied | \(\frac{1}{3}\) |
| (1,3) | Behind | \(\frac{4}{27}\) |
| (3,0) | Ahead | \(\frac{65}{81}\) |
| (2,1) | Ahead | \(\frac{37}{81}\) |
| (1,2) | Behind | \(\frac{22}{81}\) |
| (0,3) | Behind | \(\frac{8}{81}\) |
| (2,0) | Ahead | \(\frac{139}{243}\) |
| (1,1) | Tied | \(\frac{1}{3}\) |
| (0,2) | Behind | \(\frac{52}{243}\) |
| (1,0) | Ahead | \(\frac{301}{729}\) |
| (0,1) | Behind | \(\frac{214}{729}\) |
| (0,0) | Tied | \(\frac{1}{3}\) |
Every tied state has probability exactly \(\frac{1}{3}\). Since the series starts at \((0,0)\), the answer is \(\frac{1}{3}\).
The pattern is no coincidence. There is an elegant invariant that explains all three variants at once.
Claim. For any state \((a, b)\) with \(a \geq b\) (your team not behind), \[ P(a,b) + 2\,P(b,a) = 1. \]
This implies that every tied state has probability \(\frac{1}{3}\): setting \(a=b=k\) gives \(3\,P(k,k)=1\).
Proof by backward induction.
Base case. When \(a = n\) (your team just won the series), \(P(n,b)=1\) and \(P(b,n)=0\), so \(1 + 2(0) = 1\). ✓
Inductive step for tied states \((k,k)\). Let \(u = P(k+1,k)\) and \(v = P(k,k+1)\). Since \(k+1 > k\), the identity gives \(u + 2v = 1\), so \(u = 1 - 2v\). Then
\[ P(k,k) = \tfrac{1}{3}\,u + \tfrac{2}{3}\,v = \tfrac{1}{3}(1-2v) + \tfrac{2}{3}\,v = \tfrac{1}{3}. \]
The \(v\) terms cancel exactly because the two transition probabilities (\(\frac{1}{3}\) and \(\frac{2}{3}\)) are precisely the weights needed. ✓
Inductive step for ahead states \((a,b)\) with \(a > b\). Your team is not behind, so it wins the next game with probability \(\frac{1}{3}\); your opponent is not behind from its perspective (it is behind), so from \((b,a)\) your team wins with probability \(\frac{2}{3}\).
\[ P(a,b) = \tfrac{1}{3}P(a+1,b) + \tfrac{2}{3}P(a,b+1), \]
\[ P(b,a) = \tfrac{2}{3}P(b+1,a) + \tfrac{1}{3}P(b,a+1). \]
Adding \(P(a,b) + 2P(b,a)\):
\[ = \tfrac{1}{3}P(a{+}1,b) + \tfrac{2}{3}P(b,a{+}1) + \tfrac{2}{3}P(a,b{+}1) + \tfrac{4}{3}P(b{+}1,a). \]
By induction applied to \((a,b+1)\) (with \(a \geq b+1\)):
\[ P(a,b+1)+2P(b+1,a)=1 \implies \tfrac{2}{3}P(a,b{+}1)+\tfrac{4}{3}P(b{+}1,a) = \tfrac{2}{3}. \]
So the sum becomes:
\[ P(a,b)+2P(b,a) = \tfrac{1}{3}P(a{+}1,b)+\tfrac{2}{3}P(b,a{+}1) + \tfrac{2}{3}. \]
By induction applied to \((a+1, b)\) (with \(a+1 > b\)):
\[ P(a+1,b)+2P(b,a+1)=1 \implies \tfrac{1}{3}P(a{+}1,b)+\tfrac{2}{3}P(b,a{+}1) = \tfrac{1}{3}. \]
Therefore \(P(a,b)+2P(b,a) = \frac{1}{3}+\frac{2}{3} = 1\). ✓
For any best-of-\((2n-1)\) series — best-of-3, best-of-7, best-of-77, or any odd length — the probability is the same:
\[ P(\text{win series}) = \frac{1}{3}. \]
The "rubber band" effect (winning with \(\frac{2}{3}\) when behind) never manages to push the overall winning probability above the base rate of \(\frac{1}{3}\). It merely keeps the game competitive enough to preserve that exact value.