XGBoost Classification Probabilities higher than RF or SVM?
Rather than answering why XGBoost give very confident predictions, I will answer why random forest and SVM give not-so-confident predictions.

Random forest probability estimates are given by the percentage of the forest that predicted a particular class. For example, if you have $100$ trees in your forest and $81$ of them predict some class for some example, the probability estimate for that example belonging to that class is calculated to be $\frac{81}{100} = 0.81$. Because of the random nature of the ensemble members, it's very unlikely that each individual tree will end up with the correct prediction, even if the majority do. This makes probability estimates from random forests shy away from the extreme ends of the scale.

SVM is a slightly different case, because they are unable to produce probability estimates directly. Typically, Platt scaling (essentially logistic regression) is used to scale the SVM output to a probability estimate. This has the added benefit of calibrating the probability estimates, meaning the predicted probability is quite accurate - in other words, if a probability of $0.8$ is given for a prediction, it actually has approximately an $80\%$ chance of being correct. For a problem like this where there's a lot of noise (underdog teams do win sometimes, and there's a lot of evenly matched games that are hard to predict), these predictions will tend not to be overconfident.

I don't have a good reason as to why XGBoost is possibly overconfident, but it has been observed in the past that additive boosting models tend to provide distorted probability estimates without applying post-training calibration e.g. here, here &amp; here.