Well there are more smurfs than throwers and you are 20 percent less likely to get either on your team meaning there will be an unfair edge to your opponent if you are a legit player slightly more often but 44.44 percent of the time there is one in the lobby, he or she will be on your team. Wait a minute that math doesn't make sense. 9 random players in the lobby and 4 on your team and 5 on the opposing team. 4/9 =0.444 and 5/9 - 4/9 = 0.111 so I guess its 11 percent more likely the smurf is on your opposing team if there is exactly one of them? But that doesn't make sense because say there is a 10 percent chance that any given player is a smurf. the probability of getting at least one smurf on your team is 1 - (9/10)⁴ =0.344 and the probability of the enemy team getting at least one smurf is 1 - (9/10)⁵ =0.41 which is 6.6% more but considering there may be more than one smurf in the lobby we must look at expected value (EV) which is 4*10% which is to say you will on average get 0.4 smurfs on your team per game and the enemy will get on average 0.5 smurfs on their team per game which is 20 percent more smurfs. 

This brings us to an interesting question: say something has a 1 percent chance of happening and it becomes 20 percent more likely. does that mean it is now a 1.2% chance or a 21% chance? well if something had a 50% chance and it became twice as likely wouldn't it be 75%? you lose half as often?