= Solution
Use the inclusive quota: a <coalition> wins at weight \b[at least $25$]. For each player, count losing predecessor <coalitions> that become winning when that player joins. Group these swing counts by predecessor <cardinality> $t=0,1,2,3$:
|| Player and weight
|| Swing counts $(n_{i,0},n_{i,1},n_{i,2},n_{i,3})$
|| <Shapley value>
|| <Banzhaf index>
| $1:\ 15$
| $(0,1,3,1)$
| $7/12$
| $5/8$
| $2:\ 10$
| $(0,1,2,0)$
| $1/4$
| $3/8$
| $3:\ 5$
| $(0,0,1,0)$
| $1/12$
| $1/8$
| $4:\ 5$
| $(0,0,1,0)$
| $1/12$
| $1/8$
For example, player one is pivotal for predecessor <coalitions> $\{2\}$, $\{2,3\}$, $\{2,4\}$, $\{3,4\}$ and $\{2,3,4\}$. Player three is pivotal only for $\{1,4\}$; player four has the symmetric counterpart.
Each swing contributes $1/8$ to the <probabilistic Banzhaf index>. For the <Shapley value>, a swing with $t$ predecessors contributes $t!(3-t)!/4!$, which is $1/4,1/12,1/12,1/4$ for the four possible sizes. Applying these weights to the table gives
$$
\boxed{\phi=(7/12,\,1/4,\,1/12,\,1/12),\qquad\beta=(5/8,\,3/8,\,1/8,\,1/8).}
$$
The <Shapley values> sum to one. The unnormalized <Banzhaf indices> sum to $5/4$; they must not be replaced by normalized shares in this part.
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