Tonight’s Panthers/Capitals game lasted an improbable 20 rounds, with the teams recording the following round by round outcomes (X for misses, 0 for goals)

Washington: X X X O X X O X X O O X X X X X O X X X

Florida: X X X O X X O X X O O X X X X X O X X O

Estimating the likelihood of a shootout reaching 20 rounds is actually fairly straightforward, provided that you make some reasonable assumptions.

**Assumption 1:** The save percentage for each goalie is 67%.

For starters, the overall NHL save rate on shootouts is about 67%. And while I have made the point that not all NHL goalies are created equal in terms of stopping shootout attempts, in this game, both Florida goalie Roberto Luongo (67.3% save rate) and Washington counterpart Braden Holtby (66.0%) are nearly identical to the league average. As a result, 67% seems like a safe bet.

**Assumption 2: **Independent trials.

I don’t know for sure that each shootout attempt is independent of other ones, but I have have no reason not to believe in independence in this example.

Of course, you could argue that goalies get ‘hot.’ However, given that there were several goals mixed into the Panthers/Capitals string of attempts, it doesn’t seem like Luongo or Holtby were riding a hot-hand/hot-glove.

Under **Assumption 1 **and **Assumption 2,** there are two parts of the shootout that we need to consider. We start with the Best-of-3, because NHL shootouts are only extended to extra rounds when the teams are tied after three rounds. Using a 67% save rate and the binomial distribution, the probability that a shootout continues onwards, and past the first three rounds, is about 34%. Let’s call this *Pr(TiedAfter3).*

Next, we want to consider the probability that the shootout extended in each successive round. This is just the probability of either both goalies making the save (0.67*0.67) or allowing a goal (0.33*0.33), which adds to about 0.56 and is labelled as *Pr(Extends)*.

So, the probability of a shootout lasting four rounds or more is just a function of *Pr(TiedAfter3)*, *Pr(Extends)*, and (1-*Pr(Extends)*).

For example, the probability a shootout is decided in 6 rounds is:

*Pr(TiedAfter3) * Pr(Extends)* Pr(Extends)*(1-Pr(Extends)) = 0.05.*

For 20 rounds, we just have lots of extensions to plug in.

Pr(TiedAfter3) * Pr(Extends)^16*(1-Pr(Extends)) = about **1.5**** in 100,000. **

So, I estimate the probability of the shootout lasting exactly 20 rounds to be about 1.5 in 100,000.

Similarly, about 4** out of every 100,000 shootouts would last 20 rounds or greater.**

With about 13% of NHL games coming down to the shootout (or 160 games per season), we could expect a shootout to last 20 rounds or longer roughly once every 200 years.

**But what if your assumptions are invalid?**

Glad you asked.

NHL rules stipulate that shooters may be used twice only after all players on each team have been given one attempt. Because this rule forces coaches to use their weaker offensive players, its reasonable to believe that even though Holtby and Luongo started the shootout at a 67% save rate, by the 18th shooter (the last on the bench), that rate was closer to 90%.

Allowing for (i) the 4th shooter to score with a 33% success rate and the 18th shooter a 10% rate, (ii) a constant decline in offensive ability between shooters No. 4 and No. 18, and (iii) assuming shooters No. 19 and No. 20 are back to 33% rate, we get much more reasonable probabilities as follows:

Probability of the shootout lasting exactly 20 rounds: 4** in 10,000**

Under this version, about **once every decade** will we see a shootout lasting 20 rounds or more.

In the future, it would be interesting to compare how these probabilities compare to the observed proportions of shootouts that last a given round.