= Linear probability score
{title2=$S(p,x)=xp+(1-x)(1-p)$}
The <linear probability score> pays the reported <probability> of whichever binary outcome occurs. With report $p$ and observed $y\in\{0,1\}$, its reward is
$$
S_{\rm lin}(p,y)=yp+(1-y)(1-p).
$$
It looks sensible: assign more <probability> to what happens and receive more reward. But the <expected value> under a <Bernoulli distribution> with true success <probability> $q$ is
$$
\mathbb E_q[S_{\rm lin}(p,Y)]=qp+(1-q)(1-p)=1-q+(2q-1)p.
$$
This is an <affine function> of the report. When $q>1/2$, the maximum over $[0,1]$ is at $p=1$; when $q<1/2$, it is at $p=0$. At $q=1/2$, every report ties. Thus the <linear probability score> encourages reporting certainty about the more likely outcome and is not a <proper scoring rule>.
For the stipulated $q=0.7$, the <expected value> of the reward is $0.3+0.4p$. Honest reporting $p=0.7$ earns $0.58$ in <expectation>, whereas $p=1$ earns $0.7$. The forecaster improves expected reward by replacing a 70% forecast with a 100% forecast. The <Brier score> and <logarithmic scoring rule> avoid this incentive, as the expected-score diagram under <proper scoring rule> shows.
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