Was Lee Oswald the passenger in Whaley's Taxi?

[Michael O'Brian]

There are lies, damned lies, and John Mytton's made up statistics.

Don't mind his made up lies and statistics he is just another one.
http://www.surnamedb.com/Surname/Mytton

[John Mytton]

Didn't you have a total of six witnesses?

Yep.

The first witness has a choice of 1 in 4 of choosing No.2.
The second witness has a choice of 1 in 4 of choosing No.2
And so on, for all 6 witness's.

The amount of options therefore is 4 raised to the 6th power.

4/4 to the 6th power = 1/4 to the 5th power = 1/1024



JohnM

[John Mytton]

4^6 equals 4096, so only one of the 4096 options is correct, right?

Not sure what you're expressing in your last line...

The first person can pick any random number then the next 5 have a 1 in 4 chance of picking the first persons random number hence we are left with a 1 in 4^5 chance of all the people picking the same number.



JohnM

[John Mytton]

They can not be dependent. Oswald's position in each lineup can be any one of four. Each witness has a 1/4 probability of a random correct pick.

Do you dispute the 6 eyewitnesses 1 in 1024 chance of all randomly picking Oswald?



JohnM

[John Iacoletti]

Nobody claimed that the lineup picks were random events.  Nor were they even independent events.  That's where your statistical analysis fails.

[John Mytton]

Nobody claimed that the lineup picks were random events. 

Well duh, of course there was nothing random about picking Oswald, he was the man they all saw.



JohnM

[John Iacoletti]

Well duh, of course there was nothing random about picking Oswald, he was the man they all saw.

I know that's what you believe, but your bogus argument here is that either Oswald was the man they all saw, or they all randomly picked the same guy.

[John Mytton]

As you state it yes, they are four times worse.

Like six persons rolling a four-sided dice looking for exactly one combination.

Probability of choosing the same number

Q. Assume n people choose a number between 1 and k uniformly at random, simultaneously. What is the probability that any two of the n people get the same number?
I tried: The probability that two people choose the same number is 1k. There are (n/2) different pairs. How to proceed from this?

Thanks.

A.
https://math.stackexchange.com/questions/509679/probability-of-choosing-the-same-number




Probability of choosing same number

Q. There are four friends ? Adam, Bella, Christopher and Drew. All of them are asked to choose any number in their mind. Now what is the probability that every one of them has the same number in mind? The chosen number must be in between 1 to 5 (including 1 and 5).

A1. Without loss of generality, let's say Adam picks first. The probability that the other 3 choose the exact same number as Adam is 1/5∗1/5∗1/5, which gives the result of 1/125.

Note: this answer presumes that the numbers are chosen at random.

A2. Another way to see the problem.
Total number of outcomes when 4 people are free to choose a number between 1 and 5 are equal to 5^4 (or 625)
Outcomes when everyone chooses the same number are 5.
Hence, probability = 5/625 = 1/125
https://math.stackexchange.com/questions/1384661/probability-of-choosing-same-number


JohnM