Making a Predicate/fact that depends on an inequality

[rickshilling]

% Consider
.85::color(b1,purple); .15::color(b1,yellow).
.90::color(b2,purple); .10::color(b2,yellow).
.05::color(b3,purple); .95::color(b3,yellow).

% This computes that all balls are all the same color
same_color([],_).
same_color([Ball|Tail],Color) :- color(Ball,Color), same_color(Tail,Color).

query(same_color([b1,b2,b3],purple)). % Probability == 0.03825
query(same_color([b1,b2,b3],yellow)). % Probability == 0.01425

% How can I create a predicate majority_color(Group,Color) so that it
% compares the same-color probabailities so that
% query(majority_color([b1,b2,b3],purple)) with prob 1.0
% query(majority_color([b1,b2,b3],yellow)) with prob 0.0?

% I was trying
majority_color(G,purple) :- same_color(G,purple) > same_color(G,yellow).
majority_color(G,yellow) :- same_color(G,yellow) > same_color(G,purple).
query(majority_color([b1,b2,b3],purple)).
% but an error is reported:
% ArithmeticError: Error while evaluating '>'(same_color([b1, b2, b3],purple),same_color
% ([b1, b2, b3],yellow)): Unknown function 'same_color'/2 at 19:50.

[rmanhaeve]

The probables of 2 atoms cannot be compared like that. This requires a meta-query, which is offered in ProbLog through the subquery(Goal,Probability) predicate. I've included an example below, is this what you wanted to do?

% Consider
.85::color(b1,purple); .15::color(b1,yellow).
.90::color(b2,purple); .10::color(b2,yellow).
.05::color(b3,purple); .95::color(b3,yellow).

% This computes that all balls are all the same color
same_color([],_).
same_color([Ball|Tail],Color) :- color(Ball,Color), same_color(Tail,Color).

query(same_color([b1,b2,b3],purple)). % Probability == 0.03825
query(same_color([b1,b2,b3],yellow)). % Probability == 0.01425

majority_color(G,X) :- 
    subquery(same_color(G,X),P1),
    subquery(same_color(G,Y),P2),
    X \== Y,
    P1 > P2.
query(majority_color([b1,b2,b3],purple)).
query(majority_color([b1,b2,b3],yellow)).

[rickshilling]

Thanks.