Expected loss in sequential prediction with nonoblivious opponent

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In Cesa-Bianchi's book $Prediction, learning, and\ games$, page 69 introduces expected loss for non-oblivious opponent (i.e, in the $t^{th}$ round of the game, the opponent will give a feedback, according to the player's past action: $Y_t=g_t(I_1\ldots I_{t-1})$, where the $I$'s are the player's actions). He defines the expected loss as: $\bar{l}(p_t, y_t)=E[l(I_t, Y-t)|I_1\ldots I_{t-1}]=\sum_{i=1}^N l(i, Y_t)p_{i,t}$, where $p_t$ is the distribution the player draws his action $I_t$ from, and claims: $\bar{l}(p_t, y_t)$ is solely determined by $I_1\ldots I_{t-1}$.

In the following paragraph he also says "$\ldots$the probability distribution $p_t$ is fully determined by the past outcomes (of the opponent) $Y_1\ldots Y_{t-1}$; that is, it doesn't explicitly depend on the past plays (of the player) $I_1\ldots I_{t-1}$. This seems to contradict the previous paragraph which says "$\bar{l}(p_t, y_t)$ is solely determined by $I_1\ldots I_{t-1}$".

Am I missing something here? I know this problem is very specific (wrt to that book), and I may omit some details here. Hopefully, someone has read this part on stack exchange could help me, much appreciated.