@mrmac,

Based on the chi square tests I ran, the probability that your data is random is less than 0.000032.

Having your dog make random choices on a group of 3 wines is equivalent to rolling a fair 6-sided die with the 6 permutations of (E)xpensive, (M)id, and (C)heap on each side:

EMC, ECM, MEC, MCE, CEM, CME

Without the data, I don't know the distribution of the 56 (14*4) "rolls." However, since the dog must get 0, 1, or 3 correct, this can also be modeled with an unfair 3-sided die with the following probabilities for each side:

0: 1/3, 1: 1/2, 3: 1/6.

The expected outcome of 56 (14*4 participant/varietal parings) "rolls" of the 3-sided die is:

0: 18 2/3, 1: 28, 3: 9 1/3

Given that 90 (53.5% of 168) of the tastings were correct, we can construct this table of possible distributions of outcomes of the rolls of the 3-sided die:

Code:

0 1 3 X^2

------------------------------------

0 39 17 29.2857142857

2 36 18 25.2142857143

4 33 19 22.4285714286

6 30 20 20.9285714286

8 27 21 20.7142857143

10 24 22 21.7857142857

12 21 23 24.1428571429

14 18 24 27.7857142857

16 15 25 32.7142857143

18 12 26 38.9285714286

20 9 27 46.4285714286

22 6 28 55.2142857143

24 3 29 65.2857142857

26 0 30 76.6428571429

You said, "So, with 14 of us hard at work, at least 9 of us should have got 1 wine correct in each flight. Sadly for even the proudest wine tasters, we did statistically no better than that in any of the four flights." That would seem to indicate that at least 5 zeros were "rolled" in each of the 4 flights for a total of at least 20 rolls of 0, which would put you somewhere in the bottom 4 rows of the table.

The fourth column is chi squared. If you google chi square test, you can read about how to calculate this (it's trivial) and what the value means. Since there are 3 possible outcomes, there are 2 degrees of freedom. The closer X^2 is to zero, the more likely the data is random.

The smallest value in the table (20.7142857143) corresponds to about a 0.000032 chance of randomness. The values for the bottom four rows correspond to chances that are orders of magnitude smaller than 0.000032.

So, your score of 90 correct far exceeded the expected value of 56, and your results were not random.