Soccer game outcome, are chances equal between win/loss/tie? In soccer (or any discipline with similar scoring rules), is there statistically equal chance for win, loss and tie?
So, there are 3 possible outcomes, but are all 3 outcomes equally possible?
This does not take into account better/worse teams, so teams are equal for the chances... So ranking, player setup or physique, etc, should not be taken into account. Also, neither should any external condition (quality of playing field, weather, ...).
The reason I thinks so, is because a win is everything where scoreA > scoreB, which seems (intuitively) more likely to happen than scoreA = scoreB. 
If scoreB is 1, then a tie is only possible when scoreA is also 1. However when scoreB is 1, a win for A is any possible combination where scoreA > scoreB. Which is much more than the 1 single option for a tie...
 A: First of all, it only makes sense to consider a situation in which both teams are equally good. Then, losses and victories are equally probable, so consider the comparison between winning and drawing for illustrative purposes. Then outcomes for which the two teams draw are for instance 0-0, 1-1, 2-2, ..., whereas outcomes for which team 1 wins are for instance 1-0, 2-0, 3-0, 4-0,..., 2-1, 3-1, 4-1,... As you can see, the number of outcomes for draws and the number of winning outcomes is not equal (and hence the probabilities are also unequal). You could formalize this idea by assuming that team $X$ and team $Y$ both score a number of goals in a time period of for instance 90 minutes that is Poisson distributed with a certain parameter $p$. Then you could compute the probabilities that $X-Y>1$ (so that team one wins) and that $X=Y$, so that the teams draw.
A: This is not a question that can resolved using statistics or stochastics. First off, it obviously depends on the player of each team, of the conditions on the field, etc.
Let's make an easy scenario: We have two duplicates of exactly the same team playing against each other, with sides of the field determined by random choice. In this case, it would follow by symmetry that each team is equally likely to win. 
These assumptions do not suffice to determine the likelihood of a draw - there are psychological factors at play. For example, depending on the team they might become especially motivated or demotivated if they are behind.  
Sure, as @rbm could you might model the number of goals as Poisson-Distributed, but this would assume that the performance of the teams would be unaffected by the current score and the time remaining. That wouldn't be soccer anymore. If you would go that route, it would depend on the lambda parameter (which roughly described the rate of goals per team). Depending on lambda a tie become more or less likely. My intuition is that the higher lambda is the less likely a win becomes - if both team tends to score many goals it becomes unlikely that they end up at the exact same number. If both tend to not score any goal at all it is likely the game will end 0:0.
I then tested my intuition with some simulation code in R:
lambda <- seq(0, 30, by = 0.1)
draws <- rep(0, length(lambda))


for (i in 1:length(lambda)) {
lamb <- lambda[i]
team1 <- rpois(1e6, lamb)
team2 <- rpois(1e6, lamb)
draws <- team1 == team2
draw.likelihood[i] <- sum(draws)/length(draws)
}

plot(draw.likelihood ~ lambda, type = "l", lwd = 2)
grid()

You can interpret lambda as average goals per side per game. The results look like this and matches my intuition:

Since even good teams seldom score more than three goals per game you get a substantial draw rate. At least it matches real observations so far. For reference for lambda = 0, 1, 2, 3 the draw rates are 100%, 31%, 21% and 17% respectively.
