Bicycle counter-intuitive: in which direction it will move? I saw this puzzle in a local newspaper:

Consider a normal bicycle set to stand in its upright position, and
  its pedal is set to the position as shown in this figure. One man
  slightly hold the seat to keep the bicycle from falling. (Actually
  this is not important, there can be many ways to keep the bicycle from
  falling without affecting the experiment result). A non-elastic string
  is tied to the end of the pedal crank arm as in here. Another man then pulls the
  string backward.
The question is: In which direction will the bicycle move? Forward,
  backward, or standstill? Discuss possible cases. 

Hint: The answer is
  non trivial. The readers are encouraged to try this experiment before coming to a
  hasty conclusion, since it's quite simple to carry out.

Despite the "Hint" part, and the "counter-intuitive" in the title, which is a quite clear suggestion that the bicycle will move in a counter-intuitive direction (backward instead of forward), many people still submitted answer such as "Forward, why not!!", or "Standstill" or "Forward, then as the crank move to 9 o'clock position, it will move backward". Of course they are not the correct answer.
Some people answered that it will move backward, but couldn't give a decent explanation.
My best guess is that this must have something to do with the size (radius) of the wheel, the radius of the crank arm, and the ratio of the crank gear and the wheel gear as well. If the bicycle moves forward, the crank bolt/shoulder will move as well, then the displacement of the crank does not simply equal to the displacement cause by the wheel rotating...
So what can be a simple, short explanation? And what can be a detail explanation with math and equations involved?
 A: I think the way to look at the problem is this: The tricky thing about this problem is that the force $F$ is doing two things. First, it is pulling the entire bicycle backwards and, secondly, it is exerting a torque on the pedals. So let's try to separate those two effects. Consider, then, a person just sitting normally on the bicycle with his feet on the pedals and exerting the same force $F$ on the pedals (or, equivalently, a torque $rF$ on the pedal gear where r is the effective radius at which the pedals are located with respect to the center of the pedal gear). Also, assume that there is another man pulling backwards on the bicycle with a rope attached not to a pedal but to the frame of the bicycle. Now we have a situation which is equivalent to the problem originally posed. We have (1) the same torque as originally stated acting on the pedal gear and (2) we have the same force as originally stated acting to pull the bicycle backwards. Any problems with this picture thus far?
OK, so with this new but equivalent situation, what is the answer? Imagine that you are on the bicycling trying to pedal forward while exerting the torque stated above on the pedals while at the same time someone is trying to pull you backwards by pulling with a force $F$ on a rope attached to the frame of the bicycle. Remember that he is required to pull with exactly the same force $F$ that you are exerting on the pedals. No more and no less. Do you go forward or does he pull you backwards? 
I think that the answer is clearly that it all depends on what bicycle gear you are in. If you are in low gear, you will be able to move forward. If you are in high gear, he will pull you backwards. In physics terms, it depends on whether the (clockwise) torque that you are exerting on the rear drive wheel by means of your feet on the pedals is greater than or less than the (counter-clockwise) torque that the other man is able to effectively exert on the rear drive wheel by pulling the bicycle in the backwards direction with a rope.
P.S.: With the single-speed bicycle shown in the picture, the gearing is such that the bicycle would most likely move backwards.
EDIT - SOLUTION FOUND
Found a discussion and solution for this puzzle on the Scientific American web site. It turns out that the bicycle can go either backwards or forwards, depending on the gearing ratio (which is the same result that I arrived at above). For normal gearing the bicycle will most probably go backwards, but for very low gearing the bicycle can also go forward. The argument they use is based on noting whether the point where the rope is attached to the pedal goes forward or backwards with respect to the ground. Here's the web link with an explanation video: Scientific American: Bicycle Puzzle
A: In the reference frame fixed to the bike, a backward movement of the pedal by $dx$ moves the ground backward by $\alpha dx$. So the pedal moved backward $(1-\alpha)dx$ with respect to the ground. 
If $\alpha>1$ then this means it either 


*

*moved forward, which would deliver energy to the string puller and thus makes no sense

*didn't move at all ($dx=0$)
If $\alpha<1$ then the pedal moves backwards so the string puller is doing work on it, which makes sense.
Therefore the bike either stays still or goes forward depending on $\alpha$, which is a function of pedal length $L$, gear ratio $g$ and wheel radius $R$: $\alpha=g\frac RL$. The condition on $\alpha$ is the same as looking at the trajectory of the pedal: if it ever has backwards ground speed when you're riding, then pulling the string moves the bike forward.
A: Here is another way to look at this problem: we should consider the total torque relative to the two points where wheels touch the ground. Clearly in your picture you rotate the whole bicycle backwards regarding the back as well as front wheel (points where wheels touch the ground). Imagine that the wheels are nailed. If you would pull the rope forward the bicycle would go forward.
Me and my brother went to our bicycle to test this hypothesis. We put the bottom pedal slightly towards the front wheel and pulled it  a lot down and slightly back so that the total projection of the force is backwards. But in this case the torque on the back wheel is such that it tends to rotate forward. And this torque is larger than the torque on the front wheel. So overall, even though the force projection was backwards the bicycle went forward.
Also this video confirms the idea: if the rope touches the pedal under the ground level, the torque on both wheels will make the bicycle move forward.
This explanation is good because it does not make you think about gears and wheels. What you need to know is the shoulder and the force for both wheels.
A: It will definitely move backwards.
The angular velocity of the wheel, multiplied by the radius of the wheel is much bigger then the angular velocity of the pedals, multiplied by their length (the ratio of angular velocities is constrained) so the backward motion of the bicycle is much more significant then the rotation of the pedals. When the horizontal pedal position is exceeded (string above pedal axis), the bicycle will unambiguously move backward (effects add instead of subtracting).
I assume the bicycle does not have any ratchet mechanism and no extremely low gears.
[sarcasm]Also, Physics SE is for questions and answers, not discussion or puzzles. By posting this question you are doing evil to the community! This post must be deleted immediately and the servers must be incinerated to stop the contamination![\sarcasm]
