Iowa went undefeated and won the conference in 1921 and 1922. They've never gone undefeated and won the conference again. They have played football for 130 years. 2 / 130 = .015 = P(A).
The CFP has been played five times, so whoops on the .25, it's actually .20. In those 5 times, 3 undefeated P5/ND teams has occurred once, so 1 / 5 = .20 = P(B).
1 - loss Big ten champs have made it in 2 / 5 times, so that's actually 2 / 5 = .4 = P(C).
The probability of any two independent events is equal to the probability of event A and the probability of event B occurring. You get this by multiplying P(A) by P(B). P(A) AND P(B) = P(A) x P(B).
Multiplication is associative. This means that X x Y x Z = (X x Y) x Z = X x (Y x Z). This means that the probability of 3 independent events occurring simultaneously is P(A) AND P(B) AND P(C) is P(A) x P(B) x P(C). Thus, to get the chance that op asked fo we multiply .15 x .2 x .4 = .012.
The percentage formula is X x 100. Thus a 1.2% chance of Iowa being undefeated and being a Big Ten Champ; three other P5/ND being undefeated; and a Big Ten Team making the playoffs.
We can get even more detailed, since we can independently calculate the odds of Iowa going undefeated this year with partial knowledge of being 4 - 0 and how the seasons ended before. Iowa went undefeated twice, and didn't a bunch of other times (6 I think, it would take some digging). Let's assume 8 times, so 2 / 8 = .25. That gets us to a 2% chance of going undefeated and winning the playoffs.
We could look at all the teams in our conference and figure out the dependent probability of going undefeated, since there can be only one, and get a really accurate estimate, but that's just too much work. I'll leave it at 1.5%.
@Fryowa.
I do data analytics for a living.