what do you think the odds are of abduction in this case?

[slartibartfast]

Good point IMO, which raises the question, is there an innocent explanation for the abducted child's father's failure to look for Smithman and the possibly abducted child he was carrying?

Items to evaluate

Dog Alerts
Wiped Cell Phone records
Inconsistent Statements
Deep Sleeping Twins
Refusing to Answer Questions
Refusing to do reconstruction
Leaving PDL quickly after saying would never leave
Excessive use of Litigation
Body Language
Unrestricted Fund
Smith Sighting
Crying in the Night
Leaving Twins for extended periods (plus to raise alarm)

[Mr Gray]

The best way to use probability and statistics in this case IMO, are to judge innocent reasons for the anomalies and assign a probability. The probability of an innocent reason for all anomalies is the product of the individual probabilities.

statistics are intelligence not evidence..the statistics tell the police where they should look.....at family and friends first. The problem in this case is that the statements were taken in such a way thtat their accuracy cannot be guaranteed. Further statenets were given to control risks and apssed to the Pj.

SY will have had the chance to see these and re interview the mccanns which Im sure they will have done. They will then have the chance to decide on the significance of the anomolies...and it appears they dont find them significant

[Mr Gray]

Items to evaluate

Dog Alerts
Wiped Cell Phone records
Inconsistent Statements
Deep Sleeping Twins
Refusing to Answer Questions
Refusing to do reconstruction
Leaving PDL quickly after saying would never leave
Excessive use of Litigation
Body Language
Unrestricted Fund
Smith Sighting
Crying in the Night
Leaving Twins for extended periods (plus to raise alarm)

things to consider...and Im sure they have been. Nothing there that cannot be easily explained...and nothing to build a case on

[Mr Gray]

Is there an innocent explanation for the abducted child's father's failure to look for Smithman and the possibly abducted child he was carrying?

there obviously is

[Robittybob1]

im well aware of zero and double zero...the point I was making is taht if red comes up 19 times in a row...on a perfectly true wheel ...although the odds of 20 reds in arow are around 250,000 to one ...the chances of red coming in a twentieth time on this occasion are 50/50 approx....some find that hard to beleive...and lose a lot of money

The thing I remind myself when playing roulette is that the past has no influence on the future.  Each throw has exactly the same odds.

[Robittybob1]

This has got me thinking about the evidence again and especially the Smith's sighting down beside the medical centre. IMO that was Maddie they saw being carried by an as yet unidentified man that night and most probably was on his way to that centre but unfortunately it was closed. He must have had a vehicle nearby which he most likely returned to. What happened next is anyone's guess.

That is what I've been saying for years.  It was a bonus to see you are finally thinking along the same lines.  If she needed medical help who would be the next best to help?

[Vertigo Swirl]

Is there an innocent explanation for the abducted child's father's failure to look for Smithman and the possibly abducted child he was carrying?

When were you expecting Gerry to be looking for Smithman exactly?

[Robittybob1]

Presumably in NZ you select 6 numbers on your ticket?

If so you clearly have 6 chances of getting the first number correct.  So that's 49/6, not 49/1.

You then have 48 balls in the pot, and 5 remaining numbers on your slip.  So that's 48/5, not 48/1.

Then continue the (boring) iteration a further 4 cycles.

Do we care?  The only thing I'm learning here is which posters don't understand probabilities and statistics.

You might be right there.  That gave an answer of 13,983,816 or approx 14 million:1 of picking the right numbers.  I have often wondered how to calculate that.  Thank you.

[Mr Gray]

You might be right there.  That gave an answer of 13,983,816 or approx 14 million:1 of picking the right numbers.  I have often wondered how to calculate that.  Thank you.

I explained that to you yesterday.. 49 factorial to six places

[Mr Gray]

Well? Go for it?

I've better things to do as you obviously wouldnt accept my explanation... I'm sure Grange have already discounted all these points... As it seems have the pj

[ShiningInLuz]

The thing I remind myself when playing roulette is that the past has no influence on the future.  Each throw has exactly the same odds.

These are called independent events in statistics.

The lottery events were not independent, because they depended on the outcome of the previous round.

[Robittybob1]

What's the odds that this one father would fail to look for this one man while he was conducting his search for his abducted daughter? See what I mean.

The odds simply do not favour the abducted child in this case, but there is always the ONE.

We are all on our own - each of us are the ONE, for you think differently to me and to others.  We are all individuals. 
Had Mr Smith reported the sighting the very next day, once he heard people were looking for a 3 year old, I would give him better odds, but to me, IMO, it is the Smiths who are the suspects.

[Mr Gray]

Presumably in NZ you select 6 numbers on your ticket?

If so you clearly have 6 chances of getting the first number correct.  So that's 49/6, not 49/1.

You then have 48 balls in the pot, and 5 remaining numbers on your slip.  So that's 48/5, not 48/1.

Then continue the (boring) iteration a further 4 cycles.

Do we care?  The only thing I'm learning here is which posters don't understand probabilities and statistics.

If you're allowed  to state other posters don't understand statistics then it's OK for me to say it's clear some posters don't understand the evidence

[ShiningInLuz]

I explained that to you yesterday.. 49 factorial to six places

Wrong.   

[Mr Gray]

Wrong.   

No right...