Indeed you are right, the figure showing the "chance to reach, chance to knock-out" was accidentally an outdated one (from a test simulation with a lower number of trials). The real figure will be updated soon. They are quite close though, good job noticing that! -- it has been updated now

I figured as much, seeing as the chance to reach the group of 16 was below 1.0 for all teams. Looking forward to seeing the new charts!

Belgium probably has a better chance to beat the last two teams given that they make it to the quarterfinals. The US might have a better chance of getting to the quarterfinals but a high chance of being knocked out after that.

The "Winning Probabilities" chart is calculated "starting from the round of 16". But the "chance-to-reach/chance-of-knockout" graphs below seem to have been calculated at an earlier time - it shows the probability of reaching the round of 16 as less than 1 for each team.

I think the US lost their last group match, but Belgium won theirs, so maybe something like shifted the model probabilities between the time of the 2 graphs.

The US lost, but it was Germany that was knocked out.

No, we (Portugal) got knocked out, since the US had the same points and more goals scored.

No. The US lost, but both Germany and the US qualified for the next round, with Germany being group lead (Germany is 1G, US is 2G).

Not necessarily. You could say that Belgium has a lower chance of making it to the quarter finals, but, if they do, they have a great advantage, for example.

I'm not saying this is the case, but it explains how you could have a lower probability to reach the quarter finals but a higher probability to win overall.

The probability of winning all games is inherently conditional on the probability of reaching the quarter finals. To equate this to common examples of probability, winning at the knockout stage is like flipping a biased coin and getting 4 heads in a row (= four matches won).

As such, there is no way the probability of winning can exceed the probability of reaching the quarter finals.

This is a different situation.

We have

  P(Belgium reaches quarters) >= P(Belgium wins overall)

  P(USA reaches quarters) >= P(USA wins overall)

  P(USA reaches quarters) >= P(Belgium reaches quarter)

That does not imply

  P(USA wins overall) >= P(Belgium wins overall)

The coin won't get any advantage from playing in its home field, for example. I specifically said that I'm not arguing that this is the case here, but one can conceive of an example where it would.

I cannot conceive of an example. In fact I argued that it is impossible because of the fundamental rules of conditional probability.

Probability of losing is not transitive. If A is likely to lose to B, and B is likely to lose to C, it does not follow that A is likely to lose to C.

Unless we're using ELO to determine these probabilities.

Aw, you're right. I didn't read closely enough to understand their model.