The Oswald bullets, the Tippit case, and a probability consideration

[Greg Doudna]

A narrative of the Tippit case which may offer better explanatory power of the facts than the existing narrative might run something like this:

• Oswald went to the Texas Theatre to meet someone (Oswald did not go to the crime scene where there is no known reason why he would have gone there)
• the person Oswald would have met was Carl Mather of Collins Radio company
• Tippit was looking for Oswald not to harm him but likely to pass on a message from Carl Mather that would have saved his life
• rather than Tippit being killed by Oswald, Tippit was killed by the same interests who killed Oswald
• rather than Oswald being the gunman who killed Tippit, Oswald was the gunman's next intended victim
• the Tippit killing was a professional execution, not a random encounter even if it looked that way (the killing did not need to involve an appointment or luring of Tippit to the location but rather a lying in wait in keeping with habitual appearances of Tippit at that location, an ambush when he did appear)
• the crime scene eyewitness identifications, and that of Brewer to police at the theatre, all are no-previous-acquaintance/brief-sightings-at-a-distance in genre known to have high rates of error which have resulted in many wrongful convictions of innocent persons. Those identifications of the gunman as Oswald are not secure. (This is an extension to witnesses in the Tippit case of the phenomenon illustrated by Tippit crime scene witness Tatum's mistaken identification of witness William Smith as Oswald; deputy sheriff Courson's mistaken identification of a man he encountered coming down from the balcony in the Texas Theatre as Oswald; deputy sheriff Craig's mistaken identification of a running man in front of the TSBD as Oswald, and so on.)
• the Tippit murder weapon was the paper-bag revolver found the morning of Nov 23, 1963 in downtown Dallas which disappeared in Dallas Police custody (and was never forensically examined for relationship to a murder done by the same kind of handgun less than 24 hours earlier, the killing of officer Tippit) 
• the man who went into the Texas Theatre balcony without buying a ticket after passing Brewer's store on Jefferson was not Oswald but the killer of Tippit entering the theatre to kill Oswald
• the hulls of the shells fired from Oswald's revolver, that the FBI lab found were fired from Oswald's revolver, were not the hulls found at the crime scene (not one officer of the five who marked the hulls found at the crime scene gave a positive identification of their own mark on any one of those evidence hulls, in their testimony to the Warren Commission under oath, or in any other form under oath, raising non-trivial chain of custody issues in that police evidence against Oswald)
• supported by a probability analysis to follow it is suggested the bullets originally found in Oswald's revolver and spare bullets on his person at the time of his arrest were of a single manufacturer's make, all Winchester-Western (W's), prior to police substitutions of three Remington-Peters (R) in the revolver bullets

The five Winchester-Western live cartridges found in Oswald's pants pocket: a probability consideration

Under the theory and assumptions that Oswald was the gunman who killed Tippit; that he started with a fully loaded revolver; the number of shots fired was 5; before the killing he had 10 spare bullets in his pants pocket of mixed W's and R's; immediately after killing Tippit he reloaded 5 of those 10; after this reloading the 5 that were left in his pocket were all W's, no R's . . . As I thought this through in my mind I thought, that is a little odd: why wouldn't there be mixed W's and R's remaining in his pocket, if he was grabbing bullets at random from mixed W's and R's, if the scenario was correct. But that wasn't what was found in Oswald's pocket. The bullets officer Boyd found in Oswald's pocket were all five W's.

I wondered if there was a way to quantify the chances of that happening naturally. Calculating the math on this proved challenging, so I enlisted a friend of mine with requisite expertise, Russell Gmirkin to solve this presented in the form of a story problem I gave him (with several variants differing only in numbers in the starting conditions). He returned to me a report which appears below.

It turns out that the chances that a remaining 5 out of 5 bullets in Oswald's pocket would be all W's instead of any mixture of W's and R's, if he had taken 5 out of a starting 10 from his pocket at random, would be about 22% likely to happen through ordinary chance without tampering of evidence, not counting any other factors; and about 78% likely not to have happened unless there was tampering of evidence, not counting any other factors.

The alternative is that Oswald, obtaining his revolver and remaining supply of bullets from his room in Oak Cliff to carry for self-defense, never had other than only Winchester-Western's (W's) in both revolver and pocket.

Of course there are uncertainties in initial conditions in the story problem; also, 78% odds is not probative of anything. Like so much in the Tippit case--the fingerprints; which revolver was the murder weapon; the eyewitnesses--this item goes into the mix of a cumulative-weight judgement. Following is the report of Russell Gmirkin.

[Edit note: unfortunately some of the math below is not posting. Nothing I can do about it. If anyone is seriously interested contact me and I will forward by email.]

Math Problem

 Notation: The number of ways n items can be chosen k at a time is given by 

[missing formula]

Where n! = 1•2•…•n 

0! = 1

1! = 1

2! = 2

3! = 6

4! = 24

For instance 2 of 4 items ABCD can be chosen 6 ways 

[missing formula]

AB, AC, AD, BC, BD, CD

(A) Person L has a revolver which holds six cartridges (bullets). He has 11 cartridges in his pocket, 3 Remingtons and 8 Winchesters. He loads 6 out of his pocket into the revolver, one after the other, selected at random from the 11. After loading, 5 remain in his pocket. Of the 5 remaining in his pocket, what are the probabilities/chances that:

-- 0 in pocket are Remingtons (5 Winchesters)

-- 1 in pocket are Remingtons (4 Winchesters)

-- 2 in pocket are Remingtons (3 Winchesters)

-- 3 in pocket are Remingtons (2 Winchesters)

First, we count the number of ways these combinations of cartridges can occur.

[four lines of equations missing here]

Total = 56+210+168+28 = 462

For instance, 1 of 3 Remingtons can be chosen any of 3 ways, 4 of 8 Winchesters can be chosen any of 70 ways, so a 1+4 combination can be chosen 3•70=210 ways.

Either 0, 1, 2 or 3 Remingtons can be chosen, so the total number of possibilities are 56+210+168+28 = 462. Then the individual odds are

0+5: 56/462

1+4: 210/462

2+3: 168/462

3+2: 28/462

And the sum of these odds are 

56+210+168+28/462=462/462=1=100% 

In chart form:

Combinations	Odds
[missing equation]	56/462
[missing equation]	210/462
[missing equation]	168/462
[missing equation]	28/462
Total = 462	462/462

(B) Do the same analysis as "A", this time starting with 10 cartridges in his pocket, 2 Remingtons and 8 Winchesters, and loading only 5 into the revolver (5 remaining). What are probabilities/chances that of the 5 remaining:

-- 0 in pocket are Remingtons (5 Winchesters)

-- 1 in pocket are Remingtons (4 Winchesters)

-- 2 in pocket are Remingtons (3 Winchesters)

Combinations	Odds
[missing equation]	56/252
[missing equation]	140/252
[missing equation]	56/252
Total = 252	252/252

(C) Do the same analysis as "A", this time with 10 cartridges in his pocket, 3 Remingtons and 7 Winchesters, and loading only 5 into the revolver (5 remaining). What are probabilities/chances that of the 5 remaining:

-- 0 in pocket are Remingtons (5 Winchesters)

-- 1 in pocket are Remingtons (4 Winchesters)

-- 2 in pocket are Remingtons (3 Winchesters)

-- 3 in pocket are Remingtons (2 Winchesters)

Combinations	Odds
[missing equation]	21/252
[missing equation]	105/252
[missing equation]	105/252
[missing equation]	105/252
Total = 252	252/252

Edited August 1, 2022 by Greg Doudna