This is not a very typical MO question, but I hope you bear with me. It concerns a recent disagreement in the biology literature about how many different odors humans can discriminate. The authors of [a paper in Science](http://www.sciencemag.org/content/343/6177/1370) from March 2014 claimed that, based on their experiments, "humans can discriminate more than 1 trillion olfactory stimuli," a number which puts our other senses to shame. A recent critique of the Science paper, [posted on arXiv](http://arxiv.org/abs/1411.0165), takes issue with the way the Science authors interpreted their data, and claims that even a number as small as merely 10 discriminable stimuli is consistent with the data. From what I can tell, the disagreement boils down to a very clear difference in assumptions, which I will try to explain in a simplified way that probably misses a lot of the more minute details, but hopefully gives a good idea of the mathematical question which is at play.


The Science authors give a subject three vials to smell. Each vial contains a random, equal-parts mixture of 30 compounds out of a repertoire of 128 compounds. Two of the vials have identical mixtures, and the subject is asked to pick the odd vial out. Think of each mixture as a binary vector $\mathbf{x}\in X=\{0,1\}^{128}$ of Hamming weight 30. The authors estimate a critical Hamming distance $D$ at which 50% of mixture pairs at this distance are discriminable. Not many tests need to be run to obtain a very good estimate of $D$.


Think of each discriminable "odor" as a set $S\subset X$, and assume that the mixture space is completely partitioned into $N$ odors $S\_1,\ldots,S\_N$. Mixture $\mathbf{x}$ and $\mathbf{y}$ can be told apart by a subject if and only if they are not in the same odor $S$. The main task is to infer a likely value for $N$. If odors are all roughly Hamming balls of equal radius, we can directly obtain the radius from the critical distance $D$, from the radius we get the volume of the ball, and we obtain $N\approx|X|/|S|$. This analysis appears to give an estimate of $N>10^{12}$.


The critique points out that there is no reason to assume that the odors are roughly spherical in the absence of detailed mechanistic knowledge about the olfactory system, which apparently we do not have. One interesting section of the critique shows that a similar analysis applied to a hypothetical experiment using color stimuli yields nonsensical results. In the color vision system, we know that if $X$ is a binary vector space describing mixtures of optical spectra, then only the projection of $\mathbf{x}$ into a real three dimensional space can be sensed,
$(s,m,l)(\mathbf{x}) = \sum\_i x\_i (s\_i,m\_i,l\_i)$. Therefore, the colors are far from spheres in $X$; they are more like the preimages of balls under this projection. Clearly, for this kind of highly anisotropic shape, the same critical distance $D$ corresponds to a much larger volume. I don't know enough about the biology to tell whether it is reasonable to expect that something similar could happen in the olfactory system, but the critique points out that when allowing for nonspherical shapes, the same value of $D$ is consistent with many-order-of-magnitudes-larger odors. A particular construction even yields a number $N=10$ as being consistent with the data.


Now, ignoring the biology, measuring $D$ is clearly a bad way to constrain $N$, since one value of $D$ is apparently consistent with both $N>10^{12}$ and $N=10$. The question I have for the MO crowd is what extra quantity, preferably obtainable from the same kind of odd-vial-out tests (maybe even extractable from the existing data), can actually constrain $N$ to a reasonable range.



What measurable quantity can constrain the number of odors human can discriminate?
Suppose $\Omega = \{0,1\}^{128}$ is partitioned into $N$ different sets $S\_j$, with $|S\_j|/|X| = p\_j$. Suppose
you take $n$ random pairs of points of $\Omega$ and see how many of these pairs can't be distinguished (presumably because they are in the same partition).  

The probability that two given points are in the same partition is $S\_2 = \sum\_{j=1}^N p\_j^2$,
so the expected number of indistinguishable pairs is $n S\_2$; the fraction of pairs that are indistinguishable is an unbiased estimator of $S\_2$.
By Cauchy-Schwarz, $S\_2 \ge 1/N$ (which would be the value if all $p\_j$ are equal), so a good estimate on $S\_2$ gives you a lower bound on $N$.


