[BruceZ]

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Now they calculate risk of ruin r as:

r = q + (1-q)*r^2.

I'm really confused by what r would be on the right side of this formula. Is it possible to give p,q, and r a value at this point on the right side of the formula in order to come up with a value for r on the left side?

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r is the risk of ruin for playing forever. It is the same on both sides of the equation.

You have found the correct derviation.

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ok...

I'm still stuck at the "first neat trick", where r=q+(1-q)*r^2...

This shows that we can bust either by winning once then busting, or by busting outright on the first flip, if we have a bankroll of 1 unit.

However, there's some chance that we're going to win 3 flips and then bust also, right?

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That is included in the second term where we win the first flip and then go bust after that. There are only 2 cases. The second term includes all cases where we win the first bet and then go broke at any time after that, no matter how many wins and losses that takes. The probability of this is r^2 because however we go bust, we must lose a $1 bankroll twice, and r is the probability of losing $1 once.