Loser Keeps Bieber. We Stood No Chance :(

Suppose there is $t$ minutes left and Team Canada is winning by $1$ . If they are defending, then the score will not change with probability $0.96^t.$ If the score is tied, then we both teams start attacking again and we can assume that, similar to the overtime situation, that there is a $50\%$ chance of either team winning. So with this defensive strategy the overall probability of Team Canada winning is $0.96^t+\frac{1}{2}(1- 0.96^t)=\frac{1+0.96^t}{2}$

Now suppose that with $t$ minutes left Team Canada is winning by $1$ and it keeps attacking until the end. The probability that at the end of the game Team Canada is up by $1$ is $p_0=\sum \limits_{i=0}^{\frac{t}{2}} 0.9^{t-2i}\cdot 0.05^{2i} \cdot \frac{t!}{i!(t-i)!} \cdot\frac{(t-i)!}{i!(t-2i)!}$ (Here $i$ is the number of times each team scored. We assume that in one minute at most one goal is scored. This assumption is questionable, but should not change the strategy much).

The probability that at the end of the game the score is tied is, with the same assumptions, $p_1=\sum \limits_{i=0}^{\frac{t}{2}} 0.9^{t-2i-1}\cdot 0.05^{2i+1} \cdot \frac{t!}{i!(t-i)!} \cdot\frac{(t-i)!}{(i+1)!(t-2i-1)!}$ Because of the natural symmetry of the problem (the two teams are presumed generally equal) the probability of Team Canada winning is $\frac{1}{2}p_1+p_0+\frac{1}{2}(1-p_0)$ (representing the probability of it winning in overtime, in regulation by $1$ and in regulation by more than $1$ respectively).

These probabilities are hard to calculate by hand, so a computer was used to figure out the time $t$ when the defensive strategy becomes more beneficial than the offensive one. As expected, for smaller $t$ the defensive strategy is better than the offensive, while for larger $t$ the offensive strategy is preferred. The "break even" point is at about $10$ minutes (with about $83\%$ probability of Team Canada winning).

This seems to indicate that the best strategy is to start defending when there is $10$ minutes left. However, this is not true! Indeed, at $10$ minutes left the defensive strategy is only marginally better than the purely offensive one. On the other hand, if there is $5$ minutes left, and Team Canada is winning, then the defensive strategy is much better for it than the offensive one. Therefore, with $10$ minutes left in the game the "attack until $5$ minutes left" strategy is better than both purely offensive and purely defensive strategies. So this strategy is the closest to the optimal one for team Canada. (Many thanks to Skylar Saveland, whose brilliant Python simulation led to this correction. The simulation does confirm that the "attack until $5$ minutes left" strategy is close to the best possible).

Note that this is just an approximation to the optimal strategy (as the question indicates). For example, we do not discuss when the team should start playing defensively if it is up by $2$ or more. We also do not take into account the possibility of a mixed strategy with protecting the later lead. Most importantly, we disregard that Team USA is also capable of changing strategy and, in reality, will attack all out when they are trailing by $1$ late in the game.

$$ What a 'lucky' guess!? LOL

Instead of using binomial method, I prefer using intuitive method. As what I know about ice hockey game, this game lasts for 60 minutes and it is divided into 3 periods with each period lasting for 20 minutes. Hockey game has similar characteristic with football (soccer), which is the game will end with small difference goal between two team in average. Since US team as strong as Canada team, so if team Canada leads the game, let say 1 goal, with only one period left, the best strategy for team Canada from the given options is clearly, by using common sense: continue attacking throughout. If winning with minutes or less left, play defensively . But it just only a 'guess' and I don't wanna look so cocky but if the question is multiple choices and is based on real life problem, you almost could answer the problem by only using intuitive thinking. The same logic also I used to answer problem Surviving The Titanic - Part 2 .

Anyway, who is Bieber?? One thing that I've learnt from my dad is, "as a true gentleman, don't ever tease a girl". That advice is always I keep in my mind. So please don't be so mean to her. #Serious_Face #LOL

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

If probability of scoring a goal in 1 min is p then in any n min. Interval probability of: No goal scored: p0= (1-p)^n;

1 goal scored: p1= C(n,1) p (1-p)^n-1;

2 scored: p2=C(n,2) p^2 (1-p)^n-2; ....... n scored: pn=C(n,n)*p^n

Expectation value of number of goals scored:

= 0 p0+1 p1+2 p2+.....+n pn =np (Please calculate, I could not edit the equations here)

Similar formulation is valid for expected number of goals conceded. Now we consider the 3 attacking strategies where Canada attacks first upto T minutes before completion and then defends. Expected goals scored = 0.05(20-T) Expected goals conceded = 0.05(20-T)+0.04T So expected goal deficit in 20 min play is 0.04T. Obviously this deficit has to be minimized. So among the 3 the best strategy is to attack upto 5 min before extra time. (expected deficit 0.2 goals) Now we consider the defensive strategy where team defends till score is level and then attacks. Here in every case they will concede a goal already or play defensively upto the end. So the expected goal deficit here will be between 0.8 and 1 (for 20 min defensive play expectation value will be 0.04x20 = 0.8). This is obviously the worst strategy of all four. So best strategy: Attack for 15 minutes and defend rest 5.