 To see why, let \(\lambda=0.11\), \(Q_1\) (the “consequence” common to \(P_1\) and \(P_2\)) be a lottery yielding $\(1\)M for sure, \(Q_2\) be a lottery yielding $\(5\)M with probability \(10/11\) and \(\$0\) otherwise, and finally \(Q_3\) (the “consequence” common to \(P_3\) and \(P_4\)) a lottery yielding \(\$0\) for sure. \(P_1\) turns out to be a \(\lambda\)-mixture of \(Q_1\) and \(Q_1\), \(P_2\) one of \(Q_2\) and \(Q_1\), \(P_3\) one of \(Q_1\) and \(Q_3\) and \(P_4\) one of \(Q_2\) and \(Q_3\). This is probably best seen by considering the decision trees representing the corresponding compound lotteries: