Let E, L, K, and C be the chances that Eddie, Layne, Kurt, and Chris win the game, respectively. There is a 100% chance that one of them will win, so E + L + K + C = 100%. We know that K + C, the probability that Kurt or Chris will win, is 40%. This tells us that E + L = 100% - 40% = 60%. Now, we also know that the probability that Eddie will win is twice the probability that Layne will win. In numbers, this means that E = 2*L, i.e. E – 2*L = 0%. Knowing these two facts lets us find out what Eddie’s chance of winning is: add the equation E + L = 60% twice to the equation E – 2*L = 0%. This tells us that 3*E = 120%. Therefore E = 40%. We are almost there. We know that the conditional probability that Eddie wins, given that Chris loses, is 40%/70%. This probability is equal to the probability that Eddie wins and Chris loses, divided by the probability that Chris loses. But notice: CyberChance has only one winner, so the probability that Eddie wins and Chris loses is the same as the probability that Eddie wins. Also notice that the probability that Chris loses equals the probability that anybody else wins. In numbers, the probability that Chris loses is 100% - C. So we have the following equation:
40%/70% = E / (100% – C) = 40% / (100% – C)
Solving for C tells us that C = 30%. Recall that K + C = 40%. Therefore K = 10%. So Kurt has a measly 10% chance of winning CyberChance. No wonder the Kurt fan was sulking in the corner!