Martingale, physical probability and arbitrage-free probability

So I was working through a problem in the book Heard on The Street. It goes like this:

"Suppose that the riskless rate is zero...a stock is at $100...one year from now will be at either $130, or $70, with probabilities 0.80 and 0.20...What is the value of a one-year European call with strike $110?"

My first response is: A-ha! Since r is zero, there is no drift and hence the process is a martingale. For a martingale, physical probability = arbitrage-free probability, and so

c = 0.8*(130-110)=16

Before seeing why the red part (and hence the result) is wrong, let me point out how I got that impression. Consider a game in which you bet $1 and get $2 or $0 if a head or tail turns up when a coin is flipped, respectively. This is obviously a martingale, since there is no drift; and clearly, r is zero. And unfortunately, in this particular case, the statement physical probability = arbitrage-free probability is correct.

What about in general? It turns out that it is not true. In fact if we think about the relationship between the two probability measures on a binomial tree, the arbitrage-free probability p is

p = (exp(r)-D)/(U-D)

where U and D are the up- and down-scaling factors. Hence even when r is zero, we still need to compute the non-trivial arbitrage-free probability. It is by accident only that in the coin-flipping game physical probability = arbitrage-free probability since

(1-0)/(2-0) = 0.5

What an ugly coincidence! The moral of the story:
As one of my professor has told me, never plug in 1's and 0's when checking your calculations - and never think of the coin flipping game when checking your martingale pricing. Use other cases as examples.