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4 We's per case
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GrOwInG_FeAr
GrOwInG_FeAr

Dec-7-2006 09:33

alright , the fact of having 3 WE's in 1 case caused my other detective to retire :(:(:( ... i was on a ridiculously hard case ... and i had 4 alibis i dnt know ... 1 of which i ddnt even get from the suspect .. the other 3 i do have , but i had to bribe for them ... anyways , i found a WE for one suspect , i bribed and found that he had a real alibi .... then i found another alibi belonging to the person that i ddnt even get an alibi frm , i accused and it turned out to be wrong ... so RAnstetts theory is correct ... there can be 4 WE's per case

Replies

AraLives
AraLives
Battered Shoe

Dec-7-2006 15:18

Okay, where does the contention that 4 WE are possible come from? Is it the recent retirement? Because in the case I am thinking of, the person submitted the case to Ben and was told that their assumption (of 3 WE? I don't actually know what the assumption was) was incorrect. So has Ben actually confirmed the whole 4WE thing?

R Anstett
R Anstett

Dec-7-2006 16:56

No Ben has not confirmed anything about what the numbers are at what levels or anything except what is in the FAQ.

"There is a quite simple formula for knowing when to accuse a suspect. First, you need to have no or a fake alibi for the suspect. Then, you'll need one of two things. You will need either one piece of physical evidence (bloody footprint, hair, thread, threatening note) pointing to the suspect or one piece of witness evidence (when another suspect believes your suspect is guilty). "

So we have always abbreviated this to:
false/liar + PW = guilty
false/liar + WE = guilty

According to the FAQ there is no basis for the WE + WE = guilty part of the formula.

We know that it works based on our observations. Which is why I remain skeptical on the possiblity of 4 WE in a case. No one has observed this until that retirement. If it were part of the way cases are structured all along someone would have observed it long ago.

henryc
henryc
Old Shoe

Dec-7-2006 18:55

As I said yesterday, though I haven't found 4 WE in a case, I have twice found multiple pieces of bogus WE in a case. I verified this by checking the case file and seeing WE against three different people.

This means either there were 4 WE, or there was only one WE pointing to the guilty person. If the latter is true, then Serges' final point is incorrect. It does require every other person to deny knowledge of someone before you can eliminate them as a suspect. If there are 3 WE in these cases, the formula for solving a case stated by Ran is still correct. There just is no instance of WE + WE in these cases.

I think this situation hasn't been observed more often because of the way players solve their cases. To positively detect it requires asking for WE against two people who end up having valid alibis.

henryc
henryc
Old Shoe

Dec-7-2006 19:49

I was able to find an example of multiple bogus WE in the first case I solved after writing the above note. I found 5 !! WE on a RH case with 7 people. There were 3 people with valid alibis; all had one piece of WE against them. The remaining two WE were against the murderer.

I created a .bmp of the case file. How do I copy it here?

Tz_BG
Tz_BG
Well-Connected

Dec-7-2006 20:15

In regards to Serges post and henryc's reply, I just finished a case where the same situation occurred. I asked around town and learned that 3 people thought they knew. Two of the three did not give evidence against the actual killer, who I already knew from PE evidence. Under the assumption that there are always 2 WE against killer, there must be at least 4 WE for this case.

henryc - I would recommend putting the file on a web site somewhere and posting a link here.

Tz_BG
Tz_BG
Well-Connected

Dec-7-2006 20:19

As I thought about what I wrote and has been written by others, I realized there are two basic assumptions here which if untrue could throw things off.

1) Killer always has 2 WE against them.
2) When the townies say someone thinks they know, they always have WE against someone.

Have either of these been stated somewhere as true by Ben?

AraLives
AraLives
Battered Shoe

Dec-7-2006 20:21

No, Tz. I and many others have been accusing on 2 WE for a long time with (at least for me) no adverse effects.

Tz_BG
Tz_BG
Well-Connected

Dec-7-2006 20:52

That isn't quite what I meant Ara. I'm not talking about the assumption that 2 WE = guilty. It seems pretty established that 2 WE = guilty. I'm referring to the assumption that there are even 2 WE against the killer on every case. Is this always the case? I don't personally know of a situation where it doesn't happen but there is no reason that I've seen that indicates it must happen.

Serges
Serges
Vigilante

Dec-8-2006 05:02

Tz_BG, I wholeheartedly believe the killer always has multiple pieces of WE against them because of Ben's assertion that there are always at least 2 ways to solve a case. Thousands of cases have been solved with 2 WE as the deciding factor; there's no reason to doubt this aspect of the formula as far as I can see.

As far as your 2nd assumption, that seems mostly common-sense. Why would a townie confirm that someone has WE when in fact they don't suspect anyone at all? It seems as though the "Doesn't know anything" part of the townies' information covers that possibility.

With those being "truths" about Sleuth, the examples discussed previously confirm the suspicion that cases can have more than 3 pieces of WE, and multiple bogus WEs. Which means again that the only way to eliminate a suspect entirely through WE is to ask ALL other suspects (minus one) if they suspect the person. Since a killer has 2 pieces of WE and can't suspect himself, getting "X"-1 suspects to not suspect him/her must mean acquittal. (X is the total number of suspects in the given case, excluding your suspect in question).

R Anstett
R Anstett

Dec-8-2006 05:59

Henry, if you email that to me I would be glad to post it up on an agency website for everyone to be able to see.

Perhaps we can add this to the mythbusters thread now?

There will be at a minimum 3 WE per case with as many as 6 possible?


We still do not know a definative number but will help settle any other debates on the matter.

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