OThe first round of the NHL playoffs ended earlier this week. There were 8 7-game series. Two were decided in 4 games, 2 more in 5 games, and the other 4 in 6 games. It got me thinking about the expectation. So I did some probabilities in my head on the way home. Then on paper and a program at home.

Assume Team A has a probability, **p**, of winning any game. Team B will win with probability **(1-p)**.

**Win in 4:**

It is the probability of winning 4 straight.

Team A = p^4

Team B = (1-p)^4

If p = 0.5 (equally matched teams), then 1 in 8 series will end in 4 games (0.5^4 + 0.5^4).

**Win in 5:**

This is the probability of winning 4 of 5 games. But team must win last game. The prob of winning 4 games is p^4. The probability of losing a game is (1-p).

The answer (for Team A) is not = p^4 * (1-p) because the loss can occur in any of the first four games. The previous equation is the probability of winning 4 of 5 games. But there are 4 ways to win 4 of 5 games (cannot win the first 4 because the series is over before 5th game). More precisely, it is 4 choose 3 or combination(4, 3).

Combination(m, n) = m!/(n!*(m-n)!)

Team A = C(4, 3) * p^4 * (1-p)

Team B = C(4, 3) * p * (1-p)^4

If p = 0.5, then 1 in 4 series end in 5 games (4*p^5 + 4*p^5 = 8*0.5^5 = 2^3/2^5 = 1/4).

**Win in 6:**

In a 6-game series, the losing team has 2 wins. The winning team wins the last game but the first 3 wins can happen in any of the first 5 games.

We can generalize the formulae from above as

Team A = C(5, 3) * p^4 * (1-p)^2

Team B = C(5, 3) * p^2 * (1-p)^4

If p = 0.5, then 5 out of 16 series end in 6 games (10*p^6 + 10*p^6 = 20/64 = 5/16).

**Win in 7:**

In a 7-game series, losing team has 3 wins. Winning team has 3 wins in the first 6 games.

Team A = C(6, 3) * p^4 * (1-p)^3

Team B = C(6, 3) * p^3 * (1-p)^4

If p = 0.5, then 5 out of 16 series end in 7 games (20*p^7 + 20*p^7 = 40/128 = 5/16).

**Check your math:**

Always check your math: 1/8 + 1/4 + 5/16 + 5/16 = (2+4+5+5)/16 = 16/16. Adds up to 1. Check.

**Program:**

Naturally, I wrote a python method that computes this given the length of the series (bestof) and the probability of Team A winning any game (prob).

def analytical(bestof, prob): A = prob B = 1.0 - prob # prob(b) = 1 - prob(a) wins = bestof//2 + 1 aprob = [0.0] * wins bprob = [0.0] * wins for i in range(wins): aprob[i] = m_choose_n(i+wins-1, wins-1) * pow(A, wins) * pow(B, i) bprob[i] = m_choose_n(i+wins-1, wins-1) * pow(B, wins) * pow(A, i) return aprob, bprob

Some results:

Best of 7; p=0.5

```
4: 0.0625 0.0625 = 0.1250
5: 0.1250 0.1250 = 0.2500
6: 0.1562 0.1562 = 0.3125
7: 0.1562 0.1562 = 0.3125
0.5000 0.5000
```

Best of 5; p=0.5

```
3: 0.1250 0.1250 = 0.2500
4: 0.1875 0.1875 = 0.3750
5: 0.1875 0.1875 = 0.3750
0.5000 0.5000
```

Best of 7; p=0.6

```
4: 0.1296 0.0256 = 0.1552
5: 0.2074 0.0614 = 0.2688
6: 0.2074 0.0922 = 0.2995
7: 0.1659 0.1106 = 0.2765
0.7102 0.2898
```

Best of 7; p=0.75

```
4: 0.3164 0.0039 = 0.3203
5: 0.3164 0.0117 = 0.3281
6: 0.1978 0.0220 = 0.2197
7: 0.0989 0.0330 = 0.1318
0.9294 0.0706
```