Can we qualitatively predict the strategy of the German and US teams in today's World Cup soccer match? In today's World Cup soccer match between Germany and the US, both teams only need a draw to advance to the next round. There's been speculation about possible collusion, especially given the friendly relationship between the two coaches, but let's assume that both teams will follow their professional ethos and will maximize their own chances of getting to the next round (not of winning this particular game) without trying to make a deal with the other team.
This article claims that there's no reason to worry that the game will be boring, and both teams will play to win, until at some point late in the game, if the score is even, they may decide to go for a draw. That makes some intuitive sense; so I was wondering: Can we provide mathematical support for this qualitative prediction about the optimal strategies of the teams?
For simplicity, let's assume that the teams advance if they win or draw and don't advance if they lose (which is not quite true, since if they lose the result will depend also on the result of the other game in the group, in different ways for the two teams); and also ignore the incentive to win that arises because that's likely to lead to a weaker opponent in the next round.
 A: I tried to deal with the complete problem analytically, but what I came up with  seemed so involved and unenlightening that I propose a further simplification: Let's assume that after a goal is scored, the scoring team is likely enough to win by adopting a conservative approach that we can neglect the probability of the other team equalizing.
Then the game reduces to trying to score a goal or make it to the final whistle before the other team scores a goal. At each point in the game, each team can choose among various approaches (in game-theoretical terms the "actions"), some more conservative and some more risky. Given a fixed approach of their opponents, each of the team's approaches can be characterized by a rate $\lambda_1$ at which they score goals and a rate $\lambda_2$ at which their opponents score goals. In switching from a more conservative to a more risky approach, they will increase both $\lambda_1$ and $\lambda_2$. (If one increases while the other decreases, that means that one of the approaches is dominated by the other and need not be considered.)
Now the game is a simple continuous-time Markov chain with transition rates $\lambda_1$ and $\lambda_2$ from an initial tied state to two final winning states, and the probability distribution after time $t$ is
$$
\pmatrix{
c_1\left(1-\mathrm e^{-\lambda t}\right)\\
\mathrm e^{-\lambda t}\\
c_2\left(1-\mathrm e^{-\lambda t}\right)
}
$$
with $\lambda=\lambda_1+\lambda_2$ and $c_i=\lambda_i/\lambda$ (where the order of the entries is win, draw, lose). Thus, if time $t$ remains to be played and the team adopts the same approach for this entire time, its probability of advancing to the next round will be
$$
p=c_1\left(1-\mathrm e^{-\lambda t}\right)+\mathrm e^{-\lambda t}\;,
$$
the sum of the first two entries. Let's analyze this for large and small $t$. For $t\gg1/\lambda$, we can neglect the exponential terms and have $p\approx c_1$, i.e. the probability approaches the proportion of goals scored by the team. For $t\ll1/\lambda$, we can expand the exponential terms to first order in $t$ and get $p\approx1-(1-c_1)\lambda t=1-c_2\lambda t=1-\lambda_2 t$, so the probability decreases in proportion to the rate at which the opponents score goals.
Both limiting results make sense – if the game goes on forever, eventually someone is going to score a goal, and you want to make sure that it's likely to be you; if the game lasts only a short time, it's quadratically unlikely that both teams would shoot a goal during that time, so it doesn't pay to invest in shooting one first and you just want to minimize the chances of the other team shooting one.
To summarize, in the short term you want to minimize $\lambda_2$, and in the long term you want to maximize $c_1=\lambda_1/(\lambda_1+\lambda_2)$ (or equivalently minimize $c_2$). Clearly the most conservative approach minimizes $\lambda_2$. For $c_2$, it depends on the details, but it's at least plausible that by adopting a more risky approach you can increase your own rate of scoring goals disproportionately more than your opponents', and thus maximize $c_1$. If this is so, then the team should indeed adopt a more risky approach when there's a lot of time left and a more conservative approach closer to the end of the match.
To derive a more quantitative result, let's assume for simplicity that one team's approach is fixed (say, they play conservatively throughout the game because a more risky approach would lower their proportion of goals scored); and given this, the other team has exactly two approaches, a conservative approach with goal rates $\lambda_1$ and $\lambda_2$ and a risky approach with goal rates $\lambda'_1$ and $\lambda'_2$ (again with $\lambda'=\lambda'_1+\lambda'_2$). Then there must be a crossover point $t_\mathrm c$ before the end of the match at which the team should switch from the risky to the conservative approach. To find $t_\mathrm c$, consider an infinitesimal time interval $\mathrm dt$ right before the switch. If the team takes the risky approach during this interval, it can lose either by conceding a goal during the interval, with probability $\lambda'_2\mathrm dt$, or by losing during the remainder of the game, with probability $(1-\lambda'\mathrm dt)c_2(1-\mathrm e^{-\lambda t_\mathrm c})$ (since it will take the conservative approach after the switch). If the team takes the conservative approach during the interval, the result will be the same with the primed quantities replaced by the unprimed quantities. The crossover point $t_\mathrm c$ is determined by the condition that these two probabilities are the same, i.e.
\begin{eqnarray}
\mathrm dt\lambda'_2+\left(1-\lambda'\mathrm dt\right)c_2\left(1-\mathrm e^{-\lambda t_\mathrm c}\right)&=&
\mathrm dt\lambda_2+\left(1-\lambda\mathrm dt\right)c_2\left(1-\mathrm e^{-\lambda t_\mathrm c}\right)\;,\\
\lambda'_2-\lambda'c_2\left(1-\mathrm e^{-\lambda t_\mathrm c}\right)&=&
\lambda_2-\lambda c_2\left(1-\mathrm e^{-\lambda t_\mathrm c}\right)\;,
\end{eqnarray}
\begin{eqnarray}
t_\mathrm c&=&-\frac1\lambda\log\left(1-\frac1{c_2}\frac{\lambda'_2-\lambda_2}{\lambda'-\lambda}\right)\\
&=&-\frac1\lambda\log\left(1-\frac\lambda{\lambda_2}\frac{\lambda'_2-\lambda_2}{\lambda'-\lambda}\right)\\
&=&-\frac1\lambda\log\left(1-\frac{1-\lambda'_2/\lambda_2}{1-\lambda'/\lambda}\right)\;.
\end{eqnarray}
To produce some numbers, let's assume that Germany takes a conservative approach throughout and the US has a choice of either doing the same, in which case they score one goal per $90$ minutes and Germany scores one goal per $60$ minutes, or taking risks, in which case they score one goal per $40$ minutes and Germany scores one goal per $30$ minutes (so the US can increase its share of goals by taking risks). Then we get $\mathrm t_c\approx86$ minutes, i.e. there would be about $4$ minutes of interesting play and then $86$ minutes of disappointment.
To improve that, let's assume that the US does a bit worse when playing conservatively, only scoring one goal per $120$ minutes. Then $\mathrm t_c\approx55$ minutes, so almost the entire first half would be interesting. To arrive at an even later crossover point, as predicted in the article, we'd have to assume an even greater advantage from taking risks, or assume overall higher goal rates (as $t_\mathrm c$ scales inversely proportionally if we scale all goal rates proportionally), or drop some of the simplifications. Thus, the continuation of the game after the first goal might favour risk taking, since a conceded goal can be compensated by a single goal whereas after taking the lead only two opposing goals can bring defeat; and certainly the incentive to win in order to get a weaker opponent in the next round favours risk taking.
