Why is the trajectory of the alpha particle in a cloud chamber almost straight? This question baffled me since high school. 
Presumably, the alpha particle has collided with macroscopically large number of molecules when it makes a macroscopically large displacement. One would expect that it has lost its memory of its initial velocity, specifically, its initial direction of motion. But the fact that the trajectory is a smooth, gently curved line indicates that it retains its initial direction of motion very well. 
So how to account for the apparently straight line? 
Both classically and quantum mechanically, the probability of deflecting the alpha particle by an angle bigger than some finite angle $\theta_0 > 0$ is finite. Even if it is as small as 0.0001, after 10000 collisions, it will lose its memory of initial direction. But 10000 is definitely microscopic compared with the real number of collision.  
Possibly there is some Zeno effect at work?

 A: Alpha particles are only significantly scattered by nuclei. Electrons are so much lighter than an alpha particle that it is hard for the alpha particle to transfer much momentum to them.
But nuclei are small. The radius of a nucleus is of the order of $10^{-5}$ times the radius of an atom, so the cross-sectional area of the nucleus is of order $10^{-10}$ times the cross-sectional area of an atom. The probability that an alpha particle will come close enough to a nucleus to scatter through any detectable angle is very small. That's why alpha particles can travel macroscopic distances without being scattered.
A: An important thing to recall about alpha particles is that at energies up for a few tens of MeV they range out in very short distances (less than a milimeter in many cases). Multiple scattering can be expected to generate non-trivial scattering angles only over larger ranges than the penetration depth of all alpha-decay alphas and a great many alphas that appear as a result of moderate energy beam reactions.
You haven't said what the source of the image is, but it looks like a fairly old bubble chamber image, in which cases the energy of the full event may well be in the few tens of MeV ranges. In that case there is no expectation of a visible kink in any given alpha track (or even any random selection of 100 such tracks).
The long and short of it is that you are overestimating both the length of the tracks and the mean expected deflection.

You can find rather a lot of information on alpha particle energy tloss (which is to say ranges) and multiple-scattering in the Review of Particle Physics chapter on The Passage of Radiation Through Matter (PDF link) prepared by the Particle Data Group
