Sure, but what's the probability that you call timeout in OT for no reason and miss the playoffs?

In the Riddler Football League, you are coaching the Arizona Ordinals against your opponent, the Detroit Lines, and your team is down by 14 points. You can assume that you have exactly two remaining possessions (i.e., opportunities to score), and that Detroit will score no more points.

For those unfamiliar with American football, a touchdown is worth $6$ points. After each touchdown, you can decide whether to go for $1$ extra point or $2$ extra points. You happen to have a great kicker on your team, and your chances of scoring $1$ extra point (should you go for it) are $100$ percent. Meanwhile, scoring $2$ extra points is no sure thing — suppose that your team’s probability of success is some value $p$.

If the teams are tied at the end of regulation, the game proceeds to overtime, which you have a $50$ percent chance of winning. (Assuming ties are not allowed.)

What is the minimum value of $p$ such that you’d go for $2$ extra points after your team’s first touchdown (i.e., when you’re down 8 points)?

Let's set up the problem. Say $S_i$ is the score after your $i$th touchdown and extra point(s), with $S_0 = -14.$ You are faced with two choices, $u_1$ and $u_2,$ each of which can be K (kick) or T (try for two-point conversion). So the transitional probabilities are $$\mathbb{P} \{ S_i = s \mid u_i \} = \begin{cases} 1, &\text{if $u_i = K$ and $s = S_{i-1} + 7$;} \\ p, &\text{if $u_i = T$ and $s = S_{i-1} + 8$;}\\ 1-p, &\text{if $u_i = T$ and $s= S_{i-1} + 6$;}\\ 0, &\text{otherwise.}\end{cases}$$

Here we assume that losing by any number of points, or winning by any number, while potentially valuable to degenerate gamblers who are following the RFL, is of no use to you the coach (so I guess there are no weird tie breakers or other such secondary incentives to score the most points possible). So the utility function that we are trying to maximize is $$V(S_2) = \begin{cases} 1, &\text{if $S_2 \gt 0$;}\\ 0.5, &\text{if $S_2 = 0$} \\ 0, &\text{if $S_2 \lt 0$}\end{cases}$$ and we have the following optimization problem $$\max \{ \mathbb{E}[V(S_2) \mid u_1, u_2 ] \mid u_1, u_2 \in \{K, T\}, S_0 = -14\}.$$

The boring strategy would be $u_1 = u_2 = K,$ essentially playing for a tie, which give $S_2 = 0$ w.p. $1$ and so $\mathbb{E}[V(S_2) \mid u_1 = u_2 = K] = V(0) = 0.5.$ We needn't really check due to the prompt, but the strategy $u_1 = K, u_2 = T$ would give $S_1 = -7$ w.p. $1$ and so $$\mathbb{E}[V(S_2) \mid u_2 = T, u_1 = K] = p V(1) + (1-p) V(-1) = p,$$ which would only be better than $u_1 = u_2 = K$ if $p \gt 0.5.$

If instead, we go for $2$ after the first touchdown, then there are two possible outcomes. If we succeed and $S_1 \mid u_1 = -6,$ then we can safely pick $u_2 = K,$ since this will give $S_2 = 1$ w.p. $1$, so $$\mathbb{E} [V(S_2) \mid S_1 = -6, u_1 = T, u_2 = K ]= 1$$ which is a better outcome for any value of $p \lt 1$ than $$\mathbb{E}[ V(S_2) \mid S_1 = -6, u_1 = u_2 = T ] = p V(2) + (1-p) V(0) = 0.5 (1 + p).$$ If on the other hand $S_1 \mid u_1 = -8$ because the intial $2$ point conversion try was unsuccessful, then we should also do $u_2 = T,$ since $S_2 \mid S_1 = -6, u_2 = K \lt 0$ w.p. $1.$ When we go for two a second time, we have $$\mathbb{E} [V(S_2) \mid S_1 = -8, u_1=u_2=T] = pV(0) + (1-p) V(-2) = 0.5 p.$$ Therefore, if $u_1 = T,$ then we have $$\begin{align*}\mathbb{E}[ V(S_2) \mid u_1 = T ] &= p \mathbb{E}[V(S_2) \mid S_1 = -6,u_2 = K] + (1-p) \mathbb{E} [V(S_2) \mid S_1 = -8, u_2=T]\\ &= p + 0.5 p(1-p) = \frac{3p - p^2}{2}.\end{align*}$$ The choice to go for $2$ after your first touchdown will be optimal whenever $$\frac{3p - p^2}{2} > \frac{1}{2},$$ or $p^2 - 3p + 1 \lt 0,$ that is, whenever $$p > \frac{3 - \sqrt{9 - 4}}{2} = \frac{3 - \sqrt{5}}{2} = 0.381966\dots.$$