Pawn tickets in Mitre Square

[David Orsam]

You are wrong, David.

I can't possibly be wrong Pierre. The tickets themselves no longer exist and we only know what was on them from the newspapers. Those newspapers tell us that only the names of Jane (or Anne) Kelly, Emily Birrell and Joseph Jones (the pawnbroker) are on those tickets. There are no other names on them.

Unless your suspect was called Jane Kelly, Emily Birrell or Joseph Jones, then your suspect's name is definitely not on either of the pawn tickets.

[Pierre]

[QUOTE=David Orsam;389235]Pierre, the points you list aren't even superficially convincing and are most certainly not "good historical reasons".

1. You start off with the assumption that Kelly and Eddowes were "maybe criminal" and that pawn tickets have a "specific market" but that's got nothing to do with anything because your hypothesis is that neither of them ever saw or knew anything about the pawn tickets isn't it? So why have you mentioned it?

Because it throws light on everything John Kelly said.

2. Our knowledge that the tickets were both pawned in the same pawn shop comes from the newspapers only – not from any 'official' source - but it's good, if amusing, that you are now happy to accept newspaper information as accurate.

I have told you it is a problem. This means a source problem. So I am not "accepting" it.

3. Given that Eddowes was already in possession of a pawn ticket in the name of Jones, there is quite possibly even nothing coincidental about her choice of pawn shop for the boots as she might have chosen it for that very reason. Or Jones might have been known for offering good prices for male items of clothing. Do you know anything about Jones' reputation in the local area? Of course not. So there's nothing in it this point.

What is "given" is your own interpretation based on John Kelly. And you know nothing about him and what he thought was the provenience of the two pawn tickets.

4. As for the pawn ticket bearing the date of 31st August you need to think statistically here.

Yes, letīs!

I suggest that most pawn tickets "in circulation" as at 30 September would have been produced within the previous month.

Four days where murders where committed. What is the chance that the pawn ticket would contain a date from one of those four days? 4/365 = 0,01.

We know that Friday was a popular day for people in the east end to pawn items so that they would have money for the weekend. Therefore there would be a high probability that a pawn ticket in someone's possession as at Saturday 29 September would be dated as one of the five Fridays in the 30 days preceding that date.

Since you do not have a representative sample for the population of all pawn tickets and their dates, there is no chance to test that hypothesis. You must have a sample and test it to see if there was an overrepresentation for some specific day and how large that would have been.

As it is now, you try, by pure arguments about a dayīs popularity, to deduce from your own argument to a murder site. That is not a statistical method but hypothesizing, i.e. it is where you start and then you have to test it.

Had one of the pawn tickets been dated 31 August and another dated 8 September there might be some kind of pattern here but there isn't so there is absolutely nothing remarkable about Eddowes having a ticket dated Friday 31 August in her possession.

And such a pattern would not have escaped the police. If you want to make the police look like fools and you want to feel superior to the police, you would not make it easy for them. Giving two of the murder dates would have made it easy.

5. It's been explained to you by others that people would often give a false name and address when pawning goods. It's notable that you were perfectly happy to accept that Polly Nichols pawned a flannel shirt in a false name and address but for some reason you don’t seem think that Eddowes would have done so even though the surname of Kelly was the surname of her partner.

Everything you write in this quote is wrong. I have told you that before but you do not understand this.

6. As for the name of "Jane Kelly", the next victim in the series was called Mary Jane Kelly and she was otherwise known as Mary Janet Kelly, Mary Jeanette Kelly and Marie Jeanette Kelly amongst possibly other variations. Jane and Kelly are common names. The next victim was not called Jane Kelly.

We do not know anything about that.

7. Further, when she was arrested, Eddowes gave the name of Mary Anne Kelly so there is absolutely nothing strange that she pawned boots in a similar name.

It is rather pointless for you to say that, isnīt it, since you have just been talking about "common names".

8. Dorset street was a very heavily populated street with a number of lodging houses, just like Fashion Street, so there is nothing remarkable about this address being on the pawn ticket. Had it said Jane Kelly of Miller's Court that might be considered a curious coincidence but there is no more confidence involved in the name on the pawn ticket than the fact that Eddowes called herself Mary Anne Kelly of 6 Fashion Street on the night of her death.

The room belonged originally to a larger unit in Dorset Street. Even the journalists understood that.

9. There was absolutely no reason for the Emily Birrell ticket to be mentioned in "the original inquest sources" because it had no relevance to the death of Eddowes. Kelly had, in any case, already explained why Eddowes had it in her possession. There is nothing "historically questionable" about it. On the contrary, it has been historically explained.

You do not know anything about the "reasons" or "relevance". And again, "Kelly had already explained" - those sources might not be reliable.

"Historically explained" is a concept you should not use since you have no understanding of it.

10. As for there being "a name that should not have been in Mitre Square on the night of the murder of Catherine Eddowes", there was no name in Mitre Square that should not have been there on the night of Eddowes murder.

Actually, there was.

11. I assume you mean you have "found" a name on the pawn tickets, but you haven't even done that. All you have done is ignore the majority of words that would have been on the pawn tickets, which you have never even seen, and deliberately selected 42 characters in which some of those characters (discarding others) can apparently be rearranged to form the name of a person you suspect of being Jack the Ripper, just as they can be rearranged to form the names James Kelly, Joe Barnett and, as has been pointed out to me, Dr John Williams (another suspect), as well as other names such as Thomas Bond, John Trywhitt Drake, Walter Dew, Henry Moore, Rees Llewellyn, Myra Hindley, Rose West etc. etc.

That is the variance within the group of letters and that variance is not relevant. The relevant variance is between groups.

12. Clearly, given the fact that you have chosen the words to be rearranged from the tickets, in the knowledge that they contained the letters you were looking for, thus negating any random effect, statistics are never going to support anything connecting the pawn tickets to any particular individual and there is no "name that should not have been on the tin".

You, who have no knowledge about statistics, pretend to know the outcome of a statistical test before it has been done.

I, who have an education in statistics, say I do not know why the name was in the mustard tin, an therefore shall perform a test.

[John G]

[QUOTE=Pierre;389245]

Pierre, the points you list aren't even superficially convincing and are most certainly not "good historical reasons".



Because it throws light on everything John Kelly said.



I have told you it is a problem. This means a source problem. So I am not "accepting" it.



What is "given" is your own interpretation based on John Kelly. And you know nothing about him and what he thought was the provenience of the two pawn tickets.



Yes, letīs!



Four days where murders where committed. What is the chance that the pawn ticket would contain a date from one of those four days? 4/365 = 0,01.



Since you do not have a representative sample for the population of all pawn tickets and their dates, there is no chance to test that hypothesis. You must have a sample and test it to see if there was an overrepresentation for some specific day and how large that would have been.

As it is now, you try, by pure arguments about a dayīs popularity, to deduce from your own argument to a murder site. That is not a statistical method but hypothesizing, i.e. it is where you start and then you have to test it.



And such a pattern would not have escaped the police. If you want to make the police look like fools and you want to feel superior to the police, you would not make it easy for them. Giving two of the murder dates would have made it easy.



Everything you write in this quote is wrong. I have told you that before but you do not understand this.



We do not know anything about that.



It is rather pointless for you to say that, isnīt it, since you have just been talking about "common names".



The room belonged originally to a larger unit in Dorset Street. Even the journalists understood that.



You do not know anything about the "reasons" or "relevance". And again, "Kelly had already explained" - those sources might not be reliable.

"Historically explained" is a concept you should not use since you have no understanding of it.



Actually, there was.



That is the variance within the group of letters and that variance is not relevant. The relevant variance is between groups.



You, who have no knowledge about statistics, pretend to know the outcome of a statistical test before it has been done.

I, who have an education in statistics, say I do not know why the name was in the mustard tin, an therefore shall perform a test.

4/365=0,01? Or should that be !?

Good luck with your statistical test, though!

[Elamarna]

That is the variance within the group of letters and that variance is not relevant. The relevant variance is between groups.

Pierre,

When you say the "group" of letters it is obvious what you are talking about, that is the total letters used.

However you then go onto to discuss " between groups"

For those not as blessed as others in certain disciplines could you please tell us which groups these are?
How are they defined?

This would help people understand what you are talking about.

Regards

Steve

[David Orsam]

Four days where murders where committed. What is the chance that the pawn ticket would contain a date from one of those four days? 4/365 = 0,01.

That is certainly not correct. It's rubbish maths. You've got two pawn tickets in the tin. You should go back and watch some tutorials on You Tube.

And that's just for starters.

[David Orsam]

Let's say it was my birthday yesterday (it wasn't, but let's imagine).

If I walk out into the street now and see a shop receipt lying on the floor and note that the date on the receipt is 27 July 2016, is the probability of the receipt bearing that date 1/365 = 0.002, so that it is an amazing coincidence that the receipt is the same date of my birthday?

Of course not! While the date on the receipt in theory could have been any date of the last year (or even further back) the probability that it is going to have been dated today or yesterday is much higher than the probability of it being dated, say, 3 January 2016.

Just another reason why Pierre's maths fails.

[David Orsam]

Although Pierre's calculation divides four days, being the days of the murders (one of which was a Sunday), by the number of days in the year (and ignores the number of tickets) he clearly assumes that the pawn tickets could have been dated any day of the year.

But the pawn shops were closed on Sundays so that clearly affects the probability.

Then you need to factor in which days were the busiest days for pawnbrokers. I suggest these were Fridays, Saturdays and Mondays, so there needs to be a weighting to reflect that a pawn ticket is more likely to be dated one of these days than any other.

Then you need to factor in seasonal variations. According to Melanie Tebbutt, Making Ends Meet: Pawnbroking and Working Class Credit (1983):

"The heaviest pledging usually occurred in summer and autumn - between June and November" (p. 31).

So a weighting needs to be given to these months.

Then you have to consider what number of pledges were redeemed within a month. Tebbutt suggests the majority were, so the further back you go from the date of the murder the less likely the pawn ticket will be from that date.

This is why I suggest that a pawn ticket found in someone's possession on 30 September 1888 has a much higher probability of being dated Friday 31 August 1888 than, say, Wednesday 8 February 1888 and certainly than Sunday 2 September 1888!

[David Orsam]

You, who have no knowledge about statistics, pretend to know the outcome of a statistical test before it has been done.

I, who have an education in statistics, say I do not know why the name was in the mustard tin, an therefore shall perform a test.

Firstly, I do have a knowledge of statistics, and education in that subject, so please stop telling lies about me (as you would phrase it).

Secondly, do you seriously believe that you can perform a test which will tell you "why the name was in the mustard tin"? You are delusional if so.

[Joshua Rogan]

Since the hopping season traditionally started in the first week of September, it makes perfect sense that someone about to go hopping would pop into the pawn shop on 31st August to obtain a bit of ready cash for the journey.

[Elamarna]

Four days where murders where committed. What is the chance that the pawn ticket would contain a date from one of those four days? 4/365 = 0,01.

Pierre

I have an issue with the formula used here.

While I understand your premise is looking for the odds of one of the four dates of the murder showing up.There are some issues here.

The use of 365 days, while I understand the reasoning, the use of a year does not seem the best figure to use here for two possible reasons:

1. The murders took place over a 2 month period, which you seem not to have factored in. Mitre Square of course occurred only a month after Buck's Row.

2. To use 365 days when we are talking about pawn tickets does again not seem right.
Given that items pawn a year before a given date would have been sold, tickets dating 365 days before a given date are therefore highly improbable to still exist.

The probability that tickets older than 3 months would still exist has not been factored in.

For those reasons 365 does not seem the appropriate figure to use.


I now see a problem with the formula with regards to the murder dates:

You are using 4, however the fourth date did not exist at the time of the Mitre square killing, and it would seem sensible to use 3 dates.


While I do not wish to challenge your expertise in this area; I would question
if the figures used in formula, particularly the 365, are the most appropriate to use in this case?
Are they not very simplistic?


Steve


David,
while i was posted you did and I see you said this and much much of this in your post, and included a source reference.

s

[John G]

Dr John Williams, one of numerous suspects, performed an abortion on one Mary Anne Nichols in 1885, the same name as a JtR victim. Wow! Surely that can't just be a coincidence. I mean, considering how few women of the period would have shared the same forename and Surname -and an almost identical middle name-the odds against must be immense.

But there's more. There's a letter written by him, dated the 8 September 1888, stating that he won't be able to attend a prearranged meeting because he will be attending a clinic in Whitechapel. Of course, 8 September was the date of the Annie Chapman murder! Just what are the odds that he would be coincidentally visiting Whitechapel on that date?

Surely must be a case of case closed! Or then again, maybe resorting to selective statistics is not the ideal way to solve this mystery.

[Pierre]

Pierre,

When you say the "group" of letters it is obvious what you are talking about, that is the total letters used.

However you then go onto to discuss " between groups"

For those not as blessed as others in certain disciplines could you please tell us which groups these are?
How are they defined?

This would help people understand what you are talking about.

Regards

Steve

Hi Steve,

OK. It was a comparison. In statistics you can compare mean values between individuals in one group or you can compare mean values between groups. You test those mean values with hypothesis testing to see if differences in the samples are significant.

But in this case, what I was trying to tell David, was that I will not do a test which resembles the first type, i.e. I will not compare names generated from the mustard tin source (consisting of two names and two streets) to each other, but I will discuss the probability for the particular name to be found in the mustard tin source in relation to the probability for the same name to be found in a larger sample of, say, 100 units (consisting of cases from Old Baileys where there is two names and two streets per case).

So this is the simplest form of test.

Thanks for asking, Steve. I appreciate it.

Regards, Pierre

[Pierre]

Although Pierre's calculation divides four days, being the days of the murders (one of which was a Sunday), by the number of days in the year (and ignores the number of tickets) he clearly assumes that the pawn tickets could have been dated any day of the year.

But the pawn shops were closed on Sundays so that clearly affects the probability.

Then you need to factor in which days were the busiest days for pawnbrokers. I suggest these were Fridays, Saturdays and Mondays, so there needs to be a weighting to reflect that a pawn ticket is more likely to be dated one of these days than any other.

Then you need to factor in seasonal variations. According to Melanie Tebbutt, Making Ends Meet: Pawnbroking and Working Class Credit (1983):

"The heaviest pledging usually occurred in summer and autumn - between June and November" (p. 31).

So a weighting needs to be given to these months.

Then you have to consider what number of pledges were redeemed within a month. Tebbutt suggests the majority were, so the further back you go from the date of the murder the less likely the pawn ticket will be from that date.

This is why I suggest that a pawn ticket found in someone's possession on 30 September 1888 has a much higher probability of being dated Friday 31 August 1888 than, say, Wednesday 8 February 1888 and certainly than Sunday 2 September 1888!

The problem is we do not have data to construct any control variables. We would prefer regression analysis but canīt do it (but your suggestions are not correct, we should have to construct other models). The reason is the way data collection did (not) work in the 19th Century. Durkheim was one of those researchers who had a really big problem with that.

[David Orsam]

But in this case, what I was trying to tell David, was that I will not do a test which resembles the first type, i.e. I will not compare names generated from the mustard tin source (consisting of two names and two streets) to each other, but I will discuss the probability for the particular name to be found in the mustard tin source in relation to the probability for the same name to be found in a larger sample of, say, 100 units (consisting of cases from Old Baileys where there is two names and two streets per case).

It's not a question of a name being found in the mustard tin. It's a question of a name being found on a pawn ticket.

Your test will never work because of the bias you have shown by ignoring anything other than the names and addresses on the pawn tickets to play your game of word rearranging. By which I mean: what if the solution to the puzzle is to be found in the item descriptions?

Also, a sample of 100 groups of names at the Old Bailey will not necessarily reflect a sample of 100 groups of names on pawn tickets. The reason? Well take note of this press article from the London Daily News of 1 September 1908:

'The Assistant Commissioner of the Metropolitan Police has called the attention of pawnbrokers to the improper use on pawn tickets of the names of Jane, John and Ann for brevity. Pawnbrokers should write on the ticket the correct first name and surname of the person pledging, even if it is Melchizedek Featherstonehaughton.'

The consequence of what appears to be a practice of writing shortened, Anglicised names on pawn tickets, is that a sample of 100 names at the Old Bailey will not necessarily reflect the types of names on 100 pawn tickets. Certain letters such as "z" may be overrepresented at the Old Bailey against what is written on pawn tickets.

You also haven't said what result will satisfy you that the results you have found on the mustard tin tickets can be produced by chance.

[David Orsam]

The problem is we do not have data to construct any control variables.

If it's a problem, it's a problem for you, not me.