Congratulations to Mikey from the Archbishop of York Junior School who sent in the following solution based on theoretical probability:
We started off by thinking of 6 different objects and how many different ways there are of arranging them.
This gives 6 $\times$5 $\times$4 $\times$3 $\times$2 $\times$1 = 720 ways of arranging 6 different coloured balls.
But 3 of our balls are red - they can be arranged in 3 $\times$2 $\times$1 = 6 different ways that all look the same. (ABC, ACB, BAC, BCA, CBA, CAB)
Similarly there are 2 identical blue balls that can be arranged in 2 $\times$1 = 2 different ways that look the same. (AB, BA)
So although we have 720 different ways of arranging the balls only so many of them will look different in this question.
There are 720 / (6 $\times$2) different looking ways of arranging the balls in this question, giving 60 different looking triangles.
At first we thought there was only one way for the reds to all lose and that is for them to be in the middle of each side. But then we realised that the corner 3 balls could be arranged differently with the yellow ball in each of the 3 corners.
Hence there are 3 different ways to lose out of 60, or more simply 1 expected loss in every 20, ie 5% loss, 95% win.
The online scenario tester supports the 0.95, 95% chance of winning.