Let's play a game. You have two biased coins: coin A has a $0.4$ and $0.6$ for H and T probability and coin B has the opposite ($0.6$ and $0.4$ for H and T). These coins must be flipped one at a time, regardless of previous result.
Now, start off with $1$ written on a piece of paper, and start flipping with A. Every time you flip one coin and get H, you add 1 to the previous number, w/e the previous number was (in this case, the first number is one. So if you get H on the first try on coin A, you add one to one and if you get T you subtract one and get zero, and so on). However, every time you get T, you not only subtract one, but you also divide by the nth Fibonacci corresponding to the $n$-th flip.
So, flipping one at a time, if this is your $15$-th throw and you get T, you subtract 1 from the number that you got after the $14$-th throw, and you divide this new number by the $15$-th Fibonacci. If you get H, you add one to the $14$-th throw number, and then continue to the $16$-th throw.
What is the probability that the 100th throw ends up with a number greater than 1?