= Solution
Use a simultaneous two-player <normal-form game>, with X choosing the row and Y the column. Each player's <pure strategies> are $b$ and $nb$; a <payoff> pair lists X first. Measure every dataset size in millions and use the stated normalization $M=1$. The source's reference to a shared half as “points” is interpreted in the same million-point units as the corpus, so the shared addition is $n/2$ to each initial dataset.
The <bimatrix game> has <payoffs>
$$
\begin{array}{c|cc}
&Y:b&Y:nb\\ \hline
X:b&
\left(\dfrac{n+2}{2(n+3)}-\dfrac P2,\
\dfrac{n+4}{2(n+3)}-\dfrac P2\right)&
\left(\dfrac{n+1}{n+3}-P,\ \dfrac2{n+3}\right)\\[6pt]
X:nb&
\left(\dfrac1{n+3},\ \dfrac{n+2}{n+3}-P\right)&
\left(\dfrac13,\ \dfrac23\right).
\end{array}
$$
For example, joint purchase gives dataset sizes $1+n/2$ and $2+n/2$, total $3+n$, and each buyer pays $P/2$. A sole buyer gets the full corpus and pays $P$. The total <payoff> is $1-P$ whenever a purchase occurs and $1$ when neither buys: the <game> is not a <zero-sum game>, even though gross market shares always sum to one.
For the following calculations, set
$$
t=\frac n{n+3},\qquad 0<t<1.
$$
The <payoff> differences “buy minus do not buy,” with the other player's action held fixed, are
$$
\begin{array}{c|cc}
&\text{opponent does not buy}&\text{opponent buys}\\ \hline
\Delta_X&\dfrac{2t}{3}-P&\dfrac{t-P}{2}\\[4pt]
\Delta_Y&\dfrac t3-P&\dfrac{t-P}{2}.
\end{array}
$$
These four differences give all the <best response> comparisons without mixing up row and column <payoffs>.
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