If there is an alliance formation (there can be several such cases), then in the worst case the rest of the pirates form an alliance and go against $P_1$. In this case, $P_1$ can ensure:
$\left\lceil\frac{1986}{40}\right\rceil = 50$ Gold Coins (by dividing initial pile into $1986$ and $1$ piles)
Thus, after $40$ divisions, we would have:
Twenty-six piles of $50$ coins, fourteen piles of $49$ coins, one pile of only $1$ coin.
However, if we assume NO any alliance formation (which also ensures no fruitful communication among pirates), and that every pirate always tries to maximize individual wealth whatever the current situation is, then $P_1$ can ensure:
$62$ Gold Coins
This is because:
There is an invariant hidden in the individual strategies.
Interesting claim (Invariant):
Each pirate will try to maximize the second shortest pile that remains after their division.
This is because:
After $i$ divisions, the current second shortest pile is the current $i^{th}$ tallest pile. The $i^{th}$ maximum can increase later on only when some following pirate is biased towards $P_{i}$, but in the worst case that pirate can always keep the $\delta$ aligned with some other pile and it won't affect their own share. So $P_i$ must greedily maximize the size of the $i^{th}$-tallest pile to ensure that the current $i^{th}$ maximum is as high as possible.
Notice that if we think backwards, this is very trivial for the last pirate $P_{40}$ as the second minimum overall will be $P_{40}$'s share. Similarly, $P_{39}$ will try to maximize the $39^{th}$ tallest pile...and so on.
Finding the answer to above strategy is simple. This is because:
$P1$ divides the initial pile into $1986$ and $1$ as it maximizes the second minimum. For the rest of the pirates, dividing the tallest pile into two halves always maximizes (or does not affect) the second minimum. Sometimes it would also be possible to divide in a different manner, but dividing the maximum into halves is the worst case for the rest of the pirates.
So now, the divisions occur as follows:
$P_1$ creates piles of sizes $\{1986, 1\}$
$P_2$ creates piles of sizes $\{983, 983, 1\}$... and so on.
The maximum after $2^{i}$ divisions is:
$\left\lceil\frac{1986}{2^{i}}\right\rceil$
So after $32$ divisions, we would have:
Two piles each of $63$ coins, thirty piles of $62$ coins, one pile of only $1$ coin.
Thus, after $40$ divisions, we would have:
Twenty-four piles of $62$ coins, two piles of $32$ coins, fourteen piles of $31$ coins, one pile of only $1$ coin.
Note that this gives the maximum possible value $P_{1}$ can ensure, given that all are trying to maximize their individual score. That is $P_{i}$ won't bother about how the earlier pirates have performed their division, but rather focus on maximizing her guaranteed wealth from the current coin-set she has. Everyone is aware that there is no alliance formation anywhere, and that all other pirates are intelligent enough.
