FIDE norm chance expectation calculation question What is the expected probability that someone rated 2150 will have a performance rating of 2450 over a 9 round tournament and earn an IM norm?
Given the new requirements, I tried to do a calculation. I couldn't find one on the internet suitable for this task, but this one suggests that for one game between a 2150 and a 2450 player, the following probabilities hold (from the standpoint of the weaker player):

win   4.5%
loss 75.1%
draw 20.4%


To get a score of 4.5 or more, which is needed to score a TPR of at least 2450, the following results work:

5 or more wins              (0.0023%)
4 wins, and 1 or more draws (0.0434%)
3 wins, and 3 or more draws (0.0842%)
2 wins, and 5 or more draws (0.0490%)
1 win,  and 7 or more draws (0.0045%)
            9 draws         (0.0001%)
-------------------------------------
                             0.1835%


(you need some combinatorics to get the percentages)

So he will succeed once every 545 tournaments, on average.

It will depend on the strength of the opposition. To take two extremes: if your opponents all have a rating of 1000, you'll always win but never be able to get a TPR of 2450 (9/9 gives a TPR of +700). Conversely, if you're playing nine Stockfishes/Houdinis, rated 3300, you'll score 0/9 but that would give you a TPR of 2600. So the probability you're looking for can be 0%, 100% and everything in between.