What are the Odds?

[Colin Roberts]

Assuming that our crystal ball tells us that fifteen cut-throat murders of female adults will occur in the coming year, throughout England, what are the odds against any two of them occurring on the same day?

They are much, much, much, … LOWER than I had realized.

In my attempts to formulate a meaningful set of odds against the occurrence of a multi-perpetrator 'Double Event', I have erroneously relied upon the concept of a simple Joint probability: i.e. the probability of outcomes A and B, where A and B are independent of each other.

Case in Point:

If our crystal ball were to tell us that fifteen cut-throat murders of female adults would occur in the coming year, throughout England, what would be the odds against any two of them occurring on the same day?

The concept of Joint probability dictates that the probability of outcomes A and B is equal to the probability of A, times the probability of B.

15/365 x 15/365

= 0.17%
= a chance of 1 in 592
= odds (against) of 1 to 591


This is all well and good, if in fact I am asking "what are the odds against any two of them occurring specifically on … let's say … September 30?"; which is precisely what I have been doing: Suggesting in turn that we should expect such an occurrence to take place once every 592 years.


I have recognized this shortcoming in my attempts to formulate a meaningful set of odds (against) that we should all be willing to perceive, with regard to the relative unlikelihood of a multi-perpetrator 'Double Event'; but I have mistakenly assumed it to be the nature of the beast. It's not! In fact, it is - so I have recently discovered – a simple process to formulate a set of odds against any two of the murders occurring on any given – i.e. unspecified – day:

- The probability that the first murder will occur on any given day, within the coming year: 365/365
- The probability that the second murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 364/365
- The probability that the third murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 363/365
- The probability that the fourth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 362/365
- The probability that the fifth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 361/365
- The probability that the sixth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 360/365
- The probability that the seventh murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 359/365
- The probability that the eighth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 358/365
- The probability that the ninth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 357/365
- The probability that the tenth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 356/365
- The probability that the eleventh murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 355/365
- The probability that the twelfth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 354/365
- The probability that the thirteenth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 353/365
- The probability that the fourteenth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 352/365
- The probability that the fifteenth murder will occur on any of the remaining (i.e. not yet selected) given days, within the coming year: 351/365

Multiplying each of these probabilities, we get: 74.71%, this being the probability that none of the fifteen murders will occur on the same day.

Subtracting this probability from 1, we get:

- a probability of 25.29%
- a chance of 1 in 4
- odds (against) of 1 to 3

This is tantamount to saying that if fifteen Casebook posters are online at any given moment, then there is a 25.29% probability, i.e. a chance of 1 in 4, that two of them will have the same birthday.

In formulating a set of odds against the occurrence of a multi-perpetrator 'Double Event', we must move beyond the realm of the number of days that exist within a year, and into the realm of the number of hours that exist within the same. We must also incorporate the demographic and/or geographic variables that are deemed to be applicable.

I will make a formal presentation of my calculations when I am able to set aside the necessary amount of time for doing so, but at this juncture I am seeing the following, with regard to the occurrence of a multi-perpetrator 'Double Event':

- a probability as great as 0.33%
- a chance as great as 1 in 301
- odds (against) as little as 1 to 300

These odds are hardly insurmountable, and make the prospect of a multi-perpetrator 'Double Event' substantially more believable.

Bearing that in mind, we should not expect such an occurrence to take place any more frequently than once every 300 years.

~~~

As arrogant as I am sure I must seem, I continually question and second-guess my own conclusions. I reevaluate in perpetuity. That comes with being obsessive compulsive.

As I have already stated above; I was aware of the shortcomings of my earlier efforts in this arena, but believed that nothing more was attainable. I also believed that my conclusions were nonetheless quite meaningful. They really weren't.

Again:

I will make a formal presentation of my calculations when I am able to set aside the necessary amount of time for doing so.

[Barnaby]

Hi Colin,

Your second analysis still assumes all of the murders are independent of one another. I'm wondering what would happen if we consider one serial killer (a Jack who kills - let's say - 4), and 11 others that are independent. I realize that some of the others might not be independent (e.g., the Torso murders) but let's keep this relatively simple. So the question becomes what is the probability of an independent murder occuring on the same day as one of Jack's (to simulate a double event that really wasn't)?

In a year, the probability of a specific one of these independent murders occuring on the same day as any of Jack's would be 4/365. That's about a 1% chance. But if there are 11 other murders and if you ask what the probability is that any one of them (not a specific one) could cooccur on the same day as any one of Jack's then you got to add this 11 times, and you get a probability close to 12%. Unlikely, but certainly within the realm of possibility. This probability decreases to about 10% if we attribute only three murders to Jack with 12 other independent murders.

Now let's consider just the 137 days from Tabram to Mylett. For the sake of argument, let's say that Jack killed three people (Nicholls, Chapman, Eddowes). There are six unrelated murders (Tabram, Stride, Whitehall, Kelly, Farmer, Mylett). What is the probability that a specific one of these unrelated murders would occur on the same day as a Ripper murder is 3/137, or about 2%. The probability that any one of these would fall on the same day as a Ripper killing then becomes 3/137 x 6 or about 13%. Again, this isn't altogether unlikely. This increases to about 15% if we grant Jack five murders (adding Tabram and Kelly) and ask what is the probability that any of the remaining four could occur on any of Jack's murder dates.

To add further complexity, due to a copycat effect, I'm sure that a Ripper killing would momentarily increase the probability of other killings. And so the independent killings might not really be independent. This would also increase the probability of cooccurences.

Having said all of this I believe that Stride and Eddowes were killed by the same hand. I'm just making the point that it is not wildly improbable that they were not.

[Colin Roberts]

= 0.17%
= a chance of 1 in 592
= odds (against) of 1 to 591

...

- a probability of 25.29%
- a chance of 1 in 4
- odds (against) of 1 to 3

...

- a probability as great as 0.33%
- a chance as great as 1 in 301
- odds (against) as little as 1 to 300

Corrections:

= 0.17%
= a chance of 1 in 592
= odds (against) of 591 to 1

...

- a probability of 25.29%
- a chance of 1 in 4
- odds (against) of 3 to 1

...

- a probability as great as 0.33%
- a chance as great as 1 in 301
- odds (against) as little as 300 to 1

I had erroneously expressed odds (for), rather than odds (against), in each of the quoted instances.

Your second analysis ...

Thanks, Barnaby.

I'll have to read your post more carefully later this evening.

Having said all of this I believe that Stride and Eddowes were killed by the same hand. I'm just making the point that it is not wildly improbable that they were not.

All things considered, ...

I have been - and still am - inclined to perceive a single-perpetrator 'Double Event' as being perhaps twice as likely as a multi-perpetrator one.

- a probability of 66.67%
- a chance of 2 in 3
- odds (against) of 1 to 2

Again:

I will make a formal presentation of my calculations as soon as I am able to set aside the necessary amount of time for doing so.

… which could be several weeks from now.

[Ginger]

I don't think the date of September 30th is meaningful to the calculation, in that the date doesn't argue for or against a Ripper killing. He might as easily have killed two victims on Sept 29th. What I think might ought to be taken into account though is the day of the week, since the Ripper (so far as we know) stuck to weekends and holidays, and the 'double event' did take place on a Saturday.

While I do believe that Jack was responsible for the double event, I'm also mindful of the example of the Brighton Trunk Murders. Sometimes you just have weird coincidences.

[Errata]

Just because the odds are astronomical, doesn't mean it doesn't happen. My grandfather's heart was on the right behind the lungs. The odds of that were pretty slim, but if it hadn't been in the wrong place he would have died at 16 when his father shot him point blank with a hunting rifle. Not by accident, but because the guy was a psychopath. He also survived two separate lightening strikes, but since he was a test pilot I'm not sure how much more likely that made a lightening strike.

My question is simply where did you get your statistics for throat cuttings? Because the modern number is very different from the 1888 number. Ways of killing swing in and out of fashion. Not things like beating someone to death, that's a constant. But not that many people get hanged by another person anymore, and that was very popular in the 20s and 30s. Drowning people has also fallen off in the past century, not nearly as many garrottings, not as many dangerous chemical murders either. I mean you see acid attacks with domestic violence and stalking to disfigure, but not murders. Bludgeoning is still popular, shooting people is obviously far more popular... throat cutting, not so much.

[caz]

Just because the odds are astronomical, doesn't mean it doesn't happen.

Of course, but Colin has graciously admitted that the odds are not, as he previously thought, quite so hugely stacked against two independent cut-throats active in Whitechapel within an hour of each other and 15 minutes' walk.

However, I still wonder why anyone would reject an uncommon, but well documented phenomenon - a repeat offender's double event - in favour of a happenstance the like of which nobody can say they have ever heard.

Only a week ago a serial sex attacker, Clive Sharp, aged 46, was sentenced to at least 37 years in prison for the weekend murder of his second victim in one night. He had been jailed three times before for sex offences going back to the age of 17 when he raped a girl of 15 after holding a piece of glass to her neck. He was later jailed for eight years for tying up a woman and threatening her with a knife.

On the weekend of his double event, he went to see a woman for sex and tied her up. When she refused to take part in certain sexual acts he didn't force her - unusually - but he did leave her tied up and drove off to the second woman's home, where he subjected her to four hours of rape and sexual abuse before killing and dismembering her.

Love,

Caz
X

[Errata]

Of course, but Colin has graciously admitted that the odds are not, as he previously thought, quite so hugely stacked against two independent cut-throats active in Whitechapel within an hour of each other and 15 minutes' walk.

However, I still wonder why anyone would reject an uncommon, but well documented phenomenon - a repeat offender's double event - in favour of a happenstance the like of which nobody can say they have ever heard.

Only a week ago a serial sex attacker, Clive Sharp, aged 46, was sentenced to at least 37 years in prison for the weekend murder of his second victim in one night. He had been jailed three times before for sex offences going back to the age of 17 when he raped a girl of 15 after holding a piece of glass to her neck. He was later jailed for eight years for tying up a woman and threatening her with a knife.

On the weekend of his double event, he went to see a woman for sex and tied her up. When she refused to take part in certain sexual acts he didn't force her - unusually - but he did leave her tied up and drove off to the second woman's home, where he subjected her to four hours of rape and sexual abuse before killing and dismembering her.

Love,

Caz
X

My reminder that incredibly rare things do happen was in no way meant to reprimand Colin's analysis. Which I should mention is something I couldn't do with a gun to my head I'm so bad at organizing numbers. It was essentially saying that 20 to 1 odds may mean 20 against, but it guarantees 1 for. Not in a literal sense, but if I have a 1 in 350ish chance of dying in a car wreck then 1 of 350ish people is going to die in a car wreck. So I could be in the 350, or I could be in the 1. But some poor bastard is going to be that one.

Ted Bundy had a double event the night of the Chi Omega attacks. He bludgeoned 4 women at Chi Omega, and then a woman 8 blocks away who survived. Ed Kemper sort of had one. He killed his mother, and then called her friend to come over, which can be interpreted as a separate act. I think Richard Ramirez killed twice in one night two or three times. And H.H. Holmes had an assembly line going, so that could be two or three kinds of death each night. In the case of Bundy and Kemper, their double events were some of their last murders. For Kemper it was the last murder, and for Bundy the second to last but clearly he was devolving at that point.

[caz]

Hi Errata,

Ted Bundy had an earlier double event too, when an abduction attempt went wrong and the woman escaped. He immediately went looking for another victim and this one was never seen again.

I think many people underestimate the levels of frustration and determination that can grip repeat offenders who go out on the prowl and don't get their way first time, every time.

I have now seen at least three cases in recent years in the UK involving a spectacular amount of violence when the offender finds a second victim very soon after an earlier botched or less than satisfactory attack.

Stride and Eddowes bear all the hallmarks, with 'Lucky Liz' being cut down where she stood, in an impossible location for anyone to set about mutilating her, then Eddowes being meekly led to the slaughter and ripped up 'like a pig in the market' within the hour.

Love,

Caz
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