I don't think P_1 and P_2 are calculated correctly.
Looking at P1. Realistically you only have one way to succeed making the first attempt and fail the others (assuming that the 2nd and 3rd attempts both use w_2). You succeed with w_1 with a probability of p_1 and you fail 2x at w_2 with a probability (1-p_2)^2.
P1 = Prob(w_1 && !w_2 && !w_2) = p_1 ( 1-p_2 )^2
The way p_1 being calculated in the paper would be equivalent to
succeeding the first time the selecting the same weight 2x more and failing OR
succeeding the first time the selecting the same weight failing and then selecting the second weight and failing OR
succeeding the first time then selecting the 2nd weight and failing 2x.
Only the last one is needed because no one succeeds an attempt and then takes the next attempt at the same weight because it would be pointless.
So
P0 = (1-p_1)^3
P1 = p_1 ( 1-p_2 )^2
P2 = p_1 p_2 (1-p_3)
p2 = p_1 p_2 p_3
Also a note on common definitions
w_1 P1 + w_2 P2 + w_3 P3 is the expected value of the weight lifted not the average value of the weight lifted
(w_1 + w_2 + w_3 )/3 is the average value of weight lifted.
Possible outcomes. Let S = success, F = Failure
FFF : P = (1-p1)^2 : W = 0
FFS : P = (1-p1)^2 : W = w_1
FSF : P = (1-P1)p2(1-p2) : W = w_1
SFF : P = p1(1-p2)^2 : W = w_1
FSS : p = (1-p1)p1p2 : W = w_1 + w_2
SFS : p = p1(1-p2)p2 : W = w_1 + w_2
SSF : p = p1p2(1-p3) : W = w_1 + w_2
SSS : p = p1p2p3 : W = w_1 + w_2 + w_3
This should be the correct expected value
Expected Value = (1-p1)^2 w_1 + (1-P1)p2(1-p2) w_1 + p1(1-p2)^2 w_1 + (1-p1)p1p2 ( w_1 + w_2 ) + p1(1-p2)p2 ( w_1 + w_2 ) + p1p2(1-p3) ( w_1 + w_2 ) + p1p2p3 (w_1 + w_2 + w_3 )