## On shootout probabilities and overtime strategies

Part I: Shootout Probabilities

The Leafs have been awful in the shootout so far. In all their careers, the players on the Leafs roster are a combined 45/162. Reimer is a career 26/40.

But surely it is possible for the Leafs to win in a shootout. They've just been getting unlucky. So what is the probability?

Let's assume for the moment that every player will shoot whatever the team shoots

The probability of winning in the shootout is the probability of winning within the first 3 rounds plus the probability
of tying in the first 3 rounds times the probability of winning in one of the subsequent ones.

Let p be the team shootout % and s be the goalie's shootout save %. We have the following formula for probability of winning:

**** Warning: Math ****

3*p*(1-p)^2*s^3 + 3*p^2*(1-p)*s^3 + p^3*s^3 + 9*p^2*(1-p)*s^2*(1-s) + 3*p^3*s^2*(1-s) + 3*p^3*s*(1-s)^2

+

((1-p)^3*s^3 + p^3*(1-s)^3 + 9*p^2*(1-p)*s*(1-s)^2 + 9*p*(1-p)^2*s^2*(1-s))*p*s/((1-p)*(1-s)+p*s)

To see why this formula is true, note that to end up with a score of 1-0 after the first 3 rounds, you have to score once, miss twice, and save all 3 of the other teams shots. The probability of this happening is p (score once) times (1-p)^2 (don't score twice) times s^3 (save all 3). This corresponds to the first term in the first line. There is an extra factor of 3 in front because there are 3 different orders this can happen in.

The other terms in the first line are the probabilities of winning by scores of 2-0, 3-0, 2-1, 3-1 and 3-2 respectively.

The second line is the probability of a tie after the first 3 rounds (calculated in the same manner, broken down by the score), multiplied by p*s/((1-p)*(1-s)+p*s). This multiplication factor is the probability of eventually emerging with a win from continual shootout rounds. (Derived from geometric series).

**** End of Math ****

So we have a formula. Excellent. Using the Leaf's historical numbers, and Reimer's 0.65%, we get that the probability of the Leafs winning in the shootout is 0.4

So even if the Leafs are really 45/162 true talent shooters, and Reimer really is a 65% shootout goalie, they should still win about 40% of their shootouts. Of course, most research on shootouts shows that it's very unlikely that the Leafs' shooters will continue to shoot below average or that Tyler Bozak will continue to score on 58% of his shots. Furthermore, the probability should be increased a bit by the fact that your best shooters should go first.

If instead of using the Leafs 45/162, we use their best 3 shooters with over 10 shots (Bozak 58%, Kulemin 30%, Kessel 28%) we get that the Leafs score 28/84 and their win probability jumps to 0.48.

In reality though, almost no shooters or goalies have been consistently above or below average at shootouts over a significant number of shots, which somewhat trivializes this analysis, as if true shootout talent for everyone in the league is average, then every team's shootout win probability should be 0.5.

Part II: Overtime Strategies

There was some discussion in the FTB about overtime strategy in regards to the Leafs' awful shootout record.

Again, I will consider this question from a probabilistic point of view. Suppose a team is playing in overtime. Let p be the probability they win in overtime, q be the probability they lose in overtime, and r the probability the win in the shootout.

Then their expected points is:

2*p + q + (1-p-q)*r + (1-p-q)*(1-r) = 1 + p + r*(1-p-q)

It's the coach's job to maximize this value by correctly deploying his players. The Leafs were playing a forward pair of Tyler Bozak and Jay McClement early in OT. This would obviously reduce p as both Bozak and McClement are relatively devoid of offense. It would ostensibly reduce q due to the perceived defensive competence of Bozak and McClement (Of course, it probably didn't reduce q at all because Tyler Bozak is a bad defensive hockey player). Given the Leaf's awful shootout record this seemed like bad idea.

This kind of thing reminds us that coaches are far too risk averse when it comes to losing in overtime. For example, suppose by taking extra risks in OT (say a Kadri-Kessel, Phaneuf-Gardiner lineup), you could increase p.

Just for fun, let d be the increase in p, and suppose that it matches the increase in q. Then the change in expected points is d-2*d*r, which is clearly positive whenever r is less than 1/2. This confirms the intuitive notion that since the Leafs suck in the shootout, the should try to increase the probability they win in overtime even if it means an equal increase in the probability that they lose in overtime.

In fact, if the Leafs do indeed have a 40% chance of winning in the shootout, then they should try and increase their OT win probability even if it means a 1.5 times greater increase in their OT loss probability.

One idea that was mentioned with this in mind was pulling the goalie. Pulling the goalie would alter both p and q, however, if you get scored on while your goalie is pulled, you don't get your loser's point. This would change the expected points to:

2*p + (1-p-q)*r + (1-p-q)*(1-r) = 1 + p - q + r*(1-p-q)

This could be an increase or a decrease depending on how much p and q change. If it's late and you have an Ozone faceoff, q should hardly increase at all with your goalie pulled. If pulling your goalie increases p by d, but doesn't significantly increase q, you increase your expected points by d (even under the threat of lost loser point).

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