The key problem of the tight groupings

The flaw in Milam’s reasoning
      The criticism of Guinn’s heterogeneity is wrong because it is incomplete. It takes one step in logic when it needs to take two or three. After the critics of the heterogeneity saw that it apparently blends the groups together and destroys Guinn’s conclusions, they stopped there. It was as if they had done what they wanted to (destroy the dangerous NAA) or had gotten the answer that they wanted. In essence, they assumed (but never stated) that the two groups were just artifacts of chance from some undetermined number of bullets. The statistical analysis below shows this to be impossible, however. The critics did what critics of the official explanation have done for the past 37 years—they just criticized and stopped there. Had they taken that one extra step and looked for a real explanation, they would have seen that their view could not hold up. They would have realized that there was much more to the story, which they (and everybody else) had missed. What a pity that they defaulted and left the explanation for others to find.
      Here is the logical conundrum as it stood at the end of Milam’s criticism in 1994:

1. The two tight groups of fragments and the apparent clear separation of the groups is real—they were reproduced by five sets of analyses from two independent groups of analysts.
2. Guinn’s measurements of heterogeneity over quarters of three test bullets are also real—the differences among the quarters far exceed his analytical uncertainties, and ability to analyze properly (which should never have been questioned) is validated by his agreement with the FBI's earlier results.
3. The tight groupings and the physical meaning of the groups seem too solid to have arisen by chance.

Something has to give. These three points cannot all be true. I can see only two possible ways out of this dilemma: (1) heterogeneities of quarters of bullets do not apply to the five evidence fragments; or (2) the groupings did, improbable as it might seem, actually arise by chance.
    The rest of this section considers the second of these possibilities. The next section deals with heterogeneities on various scales. We begin with the groupings  because the statistical arguments used in dealing with it are understandable to more readers than are the arcane details of how jacketed bullets shatter.

Deriving the probabilities of the groupings in two different ways
    Anyone who works with probabilistic explanations faces the dilemma of presenting a simplified approach that everyone can understand but may not give the best possible answer versus presenting a rigorous approach that almost no one can understand even though it gives the best available answer. This section resolves this dilemma by presenting both approaches: a simplified version that I developed (in response to my habit of starting simple and only becoming complex as needed) and a much more rigorous approach that Larry Sturdivan has developed. We are gratified to report that the two approaches give essentially the same answers. We begin with my simplified version; Larry's approach is presented in his own words wherever possible.

A SIMPLIFIED APPROACH TO PROBABILITIES OF THE GROUPINGS (K. RAHN)

A simple formula for calculating probabilities of groupings
    The probability of the two groups of fragments arising by chance contains three components, one qualitative (the membership of each group) and two quantitative (having to do with the tightness of each group). Since these three properties are independent, the overall probability of the observed groups arising by chance is the product of each of its three properties arising by chance.

P_overall = P_grouping x P_tightness1 x P_tightness2

Calculating the membership probability P_overall
      We can define P_grouping as the probability that the two groups contained solely by chance the only combination of fragments that made physical sense: one group with the two fragments from the body shot and the other group with the three fragments from the head shot. To get the value of P_grouping, we have to find how many possible ways the five fragments can be grouped. This is the sum of the number of ways that the fragments can make one group, the number of ways that they can make two groups, three groups, four groups, and five groups. P_grouping will then be the ratio of the number of ways to make the observed group (1) and the total number of way to group the fragments:

P_grouping = 1/[total number of groups]

    Other than its value, the most important property of P_grouping is that its value is the same for all the scenarios considered below. No matter whether one considers all the fragments to be genuine or some of them to be false matches (from other bullets or from planting), the fact remains that five fragments were found, in only five places. It is these places that go into the calculation of P_grouping, not the validity of the fragments found in them.
    The following table shows all possible groups of five fragments taken in one to five groups. Assuming that each fragment has a distinct identity, there are 52 such groups.

Table 18. The 52 ways that five fragments can be grouped.

      Since the five fragments can be grouped in 52 different ways, the probability of getting the only group that makes physical sense, the right one of the 2,3 cases, is 1/52, or 0.0192 = 1.92%.

Calculating the tightness probabilities P_tightness1 and P_tightness2 
      The second and third numbers to be calculated are the probability that the observed degree of tightness of each group originated by chance. Here the heterogeneity of antimony plays no real role because the scatter that it imposes on each number will artificially bring two fragments closer together as often as it will more them apart. Thus, its effect will ultimately cancel out, and so may be neglected. To approach this problem in the simplest way, imagine that the concentration of antimony in fragments of WCC/MC bullets is distributed evenly over its range of about 0–1200 ppm (which it very nearly is—Fig 10). This means that all values between 0 and 1200 ppm will appear with about the same frequency. (Of course, real-world values would probably tend to cluster somewhat toward the middle part of the range, but the actual values measured by Guinn do not show much of that. Anyhow, it would not change the basic sense of the answer, which appears very clearly.)
      Consider first a group that contains one big fragment and one little fragment, and consider the big fragment to be the reference point. If the concentration of Sb in the big and little fragments are C_big and C_little, respectively, the probability that the little particle’s concentration by chance as close to the big one’s concentration as it did is just twice the difference between the two concentrations divided by the total range of possible concentrations (1200 ppm), because the little particle could fall above or below the big particle:

P_tightness1 = │2(C_little – C_big)/1200│

This situation is shown in Figure 19. (Since probabilities must always be positive, the sign of the difference term is reversed if it is negative.)

Figure 19. The scheme for calculating the probability that a little particle would fall randomly within its actual distance from a larger particle.

      If the group contains two little particles, the probability is the product of the two individual probabilities:

P_tightness2 = │[2(C_little1 – C_big)/1200] x [2(C_little2 – C_big)/1200]│

The actual case found by both Guinn and the FBI is one group with one little particle and the other group with two. The probability of both groups independently arising by chance is the product of the two probabilities given above. If we call the two groups 1 and 2 and the little particles in group 2 also 1 and 2, the product becomes:

P_tightness1 x P_tightness2 = │[2(C_little – C_big1)/1200] x [2(C_little1 – C_big2)/1200] x [2(C_little2 – C_big2)/1200]│

      This overall probability is extremely small. To get a sense of just how small, we can use the values from the FBI’s run 3 as an example:

P_tightness1 x P_tightness2 = │[2(C_little – C_big1)/1200] x [2(C_little1 – C_big2)/1200] x [2(C_little2 – C_big2)/1200]│

= │[2(773 – 813)/1200] x [(2(614 – 626)/1200] x [2(629 – 626)/1200]│

= 6.67 x 10^-6

Thus there is only about one chance in 10⁵, or 100,000, that the tightness of the two groups of fragments as measured in the FBI's run 3 arose solely by chance.
      When this very low probability is multiplied by the probability of getting the right fragments in the two groups by chance, the overall probability decreases by another two orders of magnitude:

P_overall = P_grouping x [P_tightness1 x P_tightness2]

= (1.92 x 10^-2) x (6.67 x 10^-6 ) = 1.28 x 10^-7

Thus the overall probability of getting these two groups at their observed tightness solely by chance is 1.28 x 10^-7, or about one in 10 million. This is equivalent to there being no practical possibility that these groupings and their tightness arose by chance. Even if we changed somewhat the way we calculated the probabilities, the sense of the result would not change: these two tight groups are not the work of chance, i.e., are not from some ill-defined number of bullets of highly heterogeneous composition that just happened to fall into physically meaningful tight groups. Some much more organized and meaningful process must have been at work.
    This first calculation is on the low side, however. The corresponding values for the FBI's runs 1, 2, and 4 and Guinn's data are, respectively, 5.2 x 10^-6, 1.15 x 10^-4, 1.95 x 10^-6, and 2.4 x 10^-6, equivalent to 5, 115, 2, and 2 chances in a million. The best value for the five results as a whole would be their geometric mean, because each value was determined multiplicatively and because the geometric mean handles the large and small extreme values equivalently. The geometric mean for the five sets of data is 3.2 x 10^-6, or about 3 chances in a million, of arising by chance. This number is still vanishing small. The groups are not a result of chance.

A more general suite of probabilities
    The above calculations are illustrative only, and for the specific case where all five fragments fell into place "randomly," whatever that means. Other less-restrictive scenarios are possible, such as that only one or two of the fragments were random, and the others were genuine. Let us now examine the full suite of possibilities in an orderly sequence.
    Six basic scenarios could explain the origin of the five fragments. In each scenario, the two large fragments Q1 and Q2 (stretcher and tip from front seat) are assumed to be genuine. That they represent different bullets is a given because Q1 is virtually a complete one, forcing Q2 to be a separate one. The three tiny fragments can be all genuine, all false positives, or some of each. Here are the scenarios, in order of decreasing probability:

    Scenario 5-0: All five fragments are genuine and just what they seem.
    Scenario 4-1: Four fragments are genuine; one of the three tiny ones (not known which) is a false positive from a third bullet or a plant (equivalent probabilistically).
    Scenario 4-1a: Four fragments are genuine; one specific one of the tiny ones is a false positive from a third bullet or a plant (equivalent probabilistically). The differences between scenarios 4-1 and 4-1a is that in 4-1 it is not known which of the tiny ones is the false positive, whereas in 4-1a it is known.
    Scenario 3-2: Three fragments are genuine; two of the three tiny ones are false positives from a other bullet or plants (again equivalent probabilistically).
    Scenario 3-2a: Three fragments are genuine; two specific ones of the three tiny ones are false positives. (Scenarios 3-2 and 3-2a are analogous to 4-1 and 4-1a.)
    Scenario 2-3: Two fragments are genuine; all three tiny ones are false positives.

    The probability for each of these scenarios contains the three terms described above, although each is somewhat different in form. Recall that the membership probability P_grouping  is the same 1/52 for all scenarios. To calculate the tightness probabilities, the fragments considered genuine are assigned probabilities of unity, and only the false positives are calculated out. Here are the probabilities for each scenario, using Guinn's NAA data as the best set. Recall that these data (for concentrations of Sb) are Q1: 833 ppm, Q2: 602 ppm, Q4,5: 621 ppm, Q9: 797 ppm, Q14: 642 ppm. These values lead to the three probabilities of tightness for the tiny fragments as:

P_tightness4,5 = │2(C_4,5 – C₂)/1200│=  │2(621 – 602)/1200│= 38/1200 = 0.032 (3.2%)

P_tightness9 = │2(C₉ – C₁)/1200│=  │2(797 – 602)/1200│= 72/1200 = 0.060 (6.0%)

P_tightness14 = │2(C₁₄ – C₂)/1200│=  │2(642 – 602)/1200│= 80/1200 = 0.067 (6.7%)

The full calculations for each of the scenarios are:

    Scenario 5-0
        P_grouping50 = 1/52.
        P_tightness50 = 1.
        P_overall50 = 1/52 = 0.0192 = 1.9%. This is well below the 5% level that is often used to state scientifically that something is "true."
    Scenario 4-1
        P_grouping41 = 1/52.
        P_tightness41 = P_4,5 + P₉ + P₁₄ = 0.032 + 0.060 + 0.067 = 0.159 (15.9%)
        P_overall41 =  (0.0192)(0.159) = 0.0031 (0.3%)
    Scenario 4-1a
        P_grouping41a = 1/52.
        P_tightness41a = P_4,5 or P₉ or P₁₄ = 0.032 or 0.060 or 0.067 (3.2% or 6.0% or 6.7%)
        P_overall41a =  (0.0192)(0.032 or 0.060 or 0.067) = 0.0006 or 0.0012 or 0.0013 (0.06% or 0.12% or 0.13%). Average = 0.1%.
    Scenario 3-2
        P_grouping32 = 1/52.
        P_tightness32 = P_4,5P₉ + P_4,5P₁₄ + P₉P₁₄ = 0.00192 + 0.00214 + 0.00402 = 0.0081 (0.8%).
        P_overall32 = (0.0192)(0.0081) = 1.56x10^-4 (0.016%)
    Scenario 3-2a
        P_grouping32a = 1/52.
        P_tightness32a = P_4,5P₉ or P_4,5P₁₄ or P₉P₁₄ = 0.00192 or 0.00214 or 0.00402 (0.19% or 0.21% or 0.40%)
        P_overall32a = (0.0192)(0.00192 or 0.00214 or 0.00402) = 3.69x10^-5 or 4.11x10^-5 or 7.72x10^-5 (0.0037% or 0.0041% or 0.0077%) Average = 0.0052%.
    Scenario 2-3
        P_grouping23 = 1/52.
        P_tightness23 = P_4,5P₉P₁₄ = 1.29x10^-4 (0.013%).
        P_overall23 = (0.0192)(1.29x10^-4) = 2.47x10^-6 (0.0002%)

    These results can be summarized in a short table:

    Now if these six scenarios are the only possible ones, their overall probabilities (P_overall) must sum to 1. The situation can be expressed as the Venn diagram for the union of P_overall4,5, P_overall9, and P_overall14, as shown in Figure 19a:

    The mathematical expression of this diagram is:

P_overall4,5 + P_overall9 + P_overall14 -P_overall4,5P_overall9 - P_overall4,5P_overall14 - P_overall9P_overall14 + P_overall4,5P_overall9P_overall14 + P_overall50 = 1

    This is equivalent to: 

Scenario 4-1 - Scenario 3-2 + Scenario 2-3 + Scenario 5-0 = 1,

or

P_overall41 - P_overall32 + P_overall23 + P_overall50 = 1

    Solving for P_overall50 gives:

P_overall50 = 1 - (P_overall4,5 + P_overall9 + P_overall14 -P_overall4,5P_overall9 - P_overall4,5P_overall14 - P_overall9P_overall14 + P_overall4,5P_overall9P_overall14),

which is equivalent to

P_overall50 = 1- P_overall41 + P_overall32 - P_overall23 =

1 - 0.0031 + 0.00016 - 2.47x10^-6 =

1 - 0.0033 = 0.9967 (99.7%)

    This result means that the total probability of all ways to incorporate one to three false positives (from MC bullets) into the basic grouping of the five fragments, whether by stray fragments from additional bullets or by deliberate plants (falsification of evidence by a conspirator who didn't know about the distinctive properties of antimony in WCC/MC ammunition), amounts to no more than 0.3%. Thus there is a 99.7% chance that all the fragments are genuine and the groupings are real.
    But we should not take these probabilities seriously, for they represent (all except the base case 5-0) hypothetical situations with absolutely no supporting evidence. They are answering the question "Suppose I have solid evidence that one or more of the tiny fragments come from additional WCC/MC bullets or are tampered evidence. What is the chance that they could then randomly fall into the "proper" groups?" Since this question presupposes evidence that does not exist, we may not legitimately ask the question. Thus all the above probabilities are moot.

A RIGOROUS APPROACH TO PROBABILITIES OF THE GROUPINGS (LARRY STURDIVAN)

The data were taken from Vincent Guinn’s Neutron Activation Analysis (NAA) published in the House Select Committee on Assassinations (HSCA) Hearings Volume I, Appendix B, Table I, p. 538 and Appendix D, Table II-A, p. 547. Following Guinn’s and Rahn’s lead, I concentrated on the antimony level as the main discriminator among individual rounds from the four lots of ammunition manufactured by Western Cartridge Company for use in the 6.5mm Mannlicher-Carcano Rifle, referred to by Rahn as WCC/MC. Other metals are ignored (though relevant to Rahn’s more comprehensive argument).
    Table II-A from the referenced hearings contains Guinn’s NAA analysis of samples from the open base of two bullets from Lot 6000 and four bullets each from Lots 6001 - 6003. From the wide variety of levels of antimony content within the bullet lots, it appears that there are no lot-to-lot differences. Assuming, for the moment, that this is true, we can look at the means, standard deviations, etc., of the antimony levels of all 14 samples in the first entry of the Appendix. Notice the large gap between the mean of 406.9 ppm and the median level of 250.5 ppm. This is one indication of skewness in the data. A normal plot of these data is shown in Figure 20. The tight curvature at the left end and the very low Anderson-Darling P-statistic of 0.03 indicates that the data are far from Gaussian in distribution. For the natural logs of the same antimony concentrations, the mean and median are 5.567 and 5.523 (which are equivalent to a geometric mean of 261.6 and median of 250.4). A normal plot of the log values is given in Figure 20a. Here the increased linearity is evident and the Anderson-Darling P-value increases to a respectable, but not great, 0.77.  The reason it isn’t better is the point above the line on the right, indicating that the upper tail is truncated, and the point above the line on the left, indicating that the lower tail is heavier than “normal.” Both of these are understandable when one considers the way that different concentrations of antimony in the original constituents move toward the mean with increased mixing. The larger quantity of low-antimony lead in the mix can leave relatively larger pockets of low-concentration alloy, while the smaller pockets of high-antimony alloy form a secondary peak that shifts downward and flattens with mixing, but retains something of its shape as it merges into a unimodal, skewed distribution. This leaves the upper tail truncated.  (A pencil sketch would help here, but is hard to do in a word processor.)
    The bottom line is that we want to stay away from the tails of this distribution when using a Gaussian approximation to do statistics. Though the recovered bullets and fragments have cores that are above the geometric mean of antimony concentration, they are on the flat part of this curve, well below the outlying point on the right in Figure 20a. To check the assumption mentioned above that there was no lot-to-lot difference in the antimony content, look at the General Linear Model analysis of the data from Guinn’s Table II-A in the Appendix. Not only is the lot number variable (LotNo) not significant, the Weight covariate is far from being significant also. Thus, the systematic error observed in the FBI data is not present - or at least not found - in Guinn’s NAA results.
    The data from Guinn’s Table I in the above reference are his measurements of the metal content of metals other than lead in the same recovered bullet and bullet fragments previously analyzed by the FBI. We see that these are a bit different than the FBI data in run 4 (analyzed in the previous paper) and even further from the FBI’s other three runs. This resulted from the systematic error in the FBI data found by Guinn and further explained by Rahn.  Thus, let us work with Guinn’s measurements, as they are more accurate and more compatible with his similar measurements on the population as a whole. Guinn’s data are reproduced in Table 20 below. Data on the concentration of metals other than antimony (Sb) are left out. We use the natural log of the Sb concentration, LnSb, to calculate the standard normal variate, x.  This x is calculated simply by subtracting the estimated population mean of 5.567 and dividing the difference by the estimated standard deviation of the population, 1.074 (both also measured on the log scale - see Appendix). These standard normal variates are used to calculate the cumulative probability distribution, P, up to the point x.

Figure 20. N-Plot of Guinn’s 14 Measurements of Antimony Levels in WCC Bullets Made for the M-C Rifle

Figure 20a. N-Plot of Guinn’s 14 Antimony Measurements on the Natural Log Scale

    Differences in the P-values in Table 20 may be used to define how “close” two samples are to each other, as the difference is a measure of what proportion of the population lie between the two levels. For instance, the difference between Q1 and Q9 is 0.0084, indicating that less than 1% of the population lies between these two concentrations. Two different determinations were made of Commission Exhibit (CE) 840, so the difference between determinations 1 and 2, 0.0036 gives us a measure of the lowest detectable difference in antimony concentration, about 1/3 of 1%. The mean of the two determinations is indicated by (m).
    The point of these calculations is that not all conspiracy scenarios justify assuming that the samples are drawn from five independent WCC/MC bullets. Some conspiracy scenarios frankly admit that Oswald was a probable shooter, but that only one of the five samples being derived from a different source would prove a conspiracy. The other four could be the very non-random samples the Warren Commission (the HSCA , and many others, including us) claimed they were.  How likely is it, then, that the one random sample would match one group or the other.
    Consider the scenario that someone “planted” CE 399 in Parkland to frame Oswald for the crime. This would only work, of course, if they somehow gained possession of a bullet that had been fired from Oswald’s rifle (and ignores the fact that CE 567 also has engraving from the Oswald rifle on it). Then we need to estimate the probability that a randomly selected bullet would have an antimony concentration at its base that was as close as the antimony concentration of CE 339 is to that of CE 842. But the difference in the P-value of the two, in Table 20, only gives us about half the desired probability estimate. The concentration could actually be lower than that of CE 842 and still be as close to CE 842 as CE 399 is. So we need to consider the standard normal variate in the interval as far below as CE 399 is above. Subtracting the difference from that of CE 842 gives us a standard normal variate, x = 0.9960 with a corresponding cumulative probability of P of 0.8403. Thus, the total interval width is the difference 0.8595 - 0.8403 = 0.0192, or about 1.9%. A sample from the base of a random bullet from any box of WCC/MC ammunition would have less than a 2% probability of being as close to the recovered fragments from John Connally’s wrist as CE 399 is.
    Random matches to members of Group 2 are a bit more involved as well as more difficult to use in constructing a conspiracy scenario. The unconvincing argument that somebody planted fragments in the upholstery that would match the head fragments has little probability (about 1.8%), and even less credibility. Table 2 has the relevant calculations. As in the last paragraph, we need to calculate a standard normal variate as far above CE 843 as CE 840 is below. The row labeled “Other side” of CE 840 has the relevant x-value and the corresponding P. The difference is: 0.8073 - 0.7894 = 0.0179. Planting a bullet fragment with engraving from the Oswald rifle (i.e., CE 567) that would match the other two samples makes more sense, in spite of the difficulty of obtaining such a fragment to plant. If we calculate a standard normal variate that is as far above the mean of CE 840 and CE 843 as CE 567 (the engraved fragment) is below, we get the x and P values listed on the fourth line of Table 21. Subtracting the P-value for CE 567 listed in Table 20 (0.8066 - 0.7758 = 0.0308) leaves about 3% probability of this close a match by chance alone.

    We have noted that all the fragments and bullets have an antimony concentration above the geometric mean for the population. So how probable is the random selection of two bullets that the Warren Commission concluded caused all the injuries to the two men? For this we need to select one group as the “reference” and calculate how probable it is that a random sample would be as close as the other is to it. The two different calculations, with two different reference groups, give slightly different results. Let us use the fragments from JFK’s head and car as the reference (call it Group 2), as almost everybody would agree that this shot is the fatal shot that was shown so dramatically in the Zapruder film. It is also very likely that some of the remnants of this bullet would be found in the car.
    The x-value of the last entry in Table 21 is the same distance from the x-value of Group 2 as the x-value of Group 1 is, but on the lower side. The difference spanned in the P-variable is (0.8548 - 0.7101 = 0.1447) about 14.5% probability of selecting two bullets that would fall this near to one a