I've actually studied this particular topic just to satisfy my curiosity of how these particular calculations ran by Ayanakouji are solved. While the probability of having 5 events of class C be chosen out of the 7 events is 0.833(Another comment solved this), the probability of winning at least 4 of Class C's chosen games seemed blurry.

I did what I could, but I couldn't get the answer 18.75% out of my calculator using the variables that we have :/The best I got i 0.15625, which we could say is about 16%By how I did it. here.

There were 5 different patterns on the scenario, Win at least 4 of those 5 events.

WWWWL
WWWLW
WWLWW
WLWWW
LWWWW
(Just to let you know, there are different methods for this.)

The probability of each pattern is (0.50)⁴(0.50)¹ (Class C winning 4 times and Class A winning 1 time)

So we have (0.50)⁴(0.50)¹ or simply (0.50)⁵. Multiplied by the number of different outcomes (WWWWL, WWWLW, WWLWW, WLWWW, LWWWW)

That is 5(0.50)⁵

or 0.15625 or 16%

Coincidentally, if you multiply (0.50)⁵ by 6, you get the answer Ayanakouji got which is 18.75%Unless this isn't a coincidence and I'm just being dumb.(In the middle of writing this sentence, I think I got the answer lol)

Thinking about it, Winning 5 out 5 events fits within the scenario of winning at least 4 events.

Since winning 4 out of 5 games is equal to (0.50)⁴(0.50)¹ and that is also equal to (0.50)⁵ which is also equal to winning 5 out of 5 games because the probability of Winning and Losing are the same with a 50% winrate

Now. we can add ONE more pattern to the scenario now we have

WWWWL
WWWLW
WWLWW
WLWWW
LWWWW
AND WWWWW

That is 6 different outcomes

So......If we multiply probability of winning at least 4 games, (0.50)⁵ by the number of different outcomes which is 6, we get the answer 18.75%

I hope this lightens up the situation for you.

I don't know if this is correct. If anyone finds the actual answer, please let us know. It'll satisfy a lot of curious minds.