On Tuesday, October 4, 2022 at 11:48:59 PM UTC-4, henh...@gmail.com wrote:

On Tuesday, October 4, 2022 at 4:12:55 PM UTC-7, Gareth Taylor wrote:

In article <bd470200-70b0-46e7...@googlegroups.com>,
henh...@gmail.com <henh...@gmail.com> wrote:

( What's a probability value that (interestingly) Doesn't go to Zero
as N increases ? )

Suppose you have 26 cards labelled from A to Z. You shuffle them and
then deal them out in a row. What is the probability that none of the letters is in the correct position?

With just two cards, it's obviously 1/2.

How about three? The orders BCA and CAB have none in the correct
position, while the orders ACB, CBA, BAC, ACB have at least one in the correct position. So it's 1/3.

What happens for larger alphabets?

Gareth

(great problem!)

(possible Spoiler --- esp. if i'm thinking correctly)

5 passengers to board an airplane with numbered seats 1--5

passenger 1 will have 4 poss. seats
say, he sits in seat 3.

Next, p3 will have 4 poss. seats ...

Next, p? will have 3 poss. seats ...

the last 2 passengers will have no choice.

So the # of poss. P(5) = 4 x 4 x 3
________________________

with 100 passengers,

the # of poss. P(100) = 99 x 99 x 98 x 97 x ... x 4 x 3

________________________

i think... you must be suggesting that

P(n) / n!

Doesn't go to Zero as N increases ?

The probability approaches 1/e.
See https://math.stackexchange.com/questions/399500/why-is-the-derangement-probability-so-close-to-frac1e

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On Tuesday, October 4, 2022 at 4:12:55 PM UTC-7, Gareth Taylor wrote:

In article <bd470200-70b0-46e7...@googlegroups.com>,
henh...@gmail.com <henh...@gmail.com> wrote:

( What's a probability value that (interestingly) Doesn't go to Zero
as N increases ? )

Suppose you have 26 cards labelled from A to Z. You shuffle them and
then deal them out in a row. What is the probability that none of the
letters is in the correct position?

With just two cards, it's obviously 1/2.

How about three? The orders BCA and CAB have none in the correct
position, while the orders ACB, CBA, BAC, ACB have at least one in the correct position. So it's 1/3.

What happens for larger alphabets?

Gareth

(great problem!)




(possible Spoiler --- esp. if i'm thinking correctly)





5 passengers to board an airplane with numbered seats 1--5

passenger 1 will have 4 poss. seats
say, he sits in seat 3.

Next, p3 will have 4 poss. seats ...

Next, p? will have 3 poss. seats ...

the last 2 passengers will have no choice.

So the # of poss. P(5) = 4 x 4 x 3
________________________

with 100 passengers,

the # of poss. P(100) = 99 x 99 x 98 x 97 x ... x 4 x 3

________________________

i think... you must be suggesting that

P(n) / n!

Doesn't go to Zero as N increases ?

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On 05/10/2022 05:06, Ilan Mayer wrote:

The probability approaches 1/e.
See https://math.stackexchange.com/questions/399500/why-is-the-derangement-probability-so-close-to-frac1e

On RyanAir the probability quicky approaches 1

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On Tuesday, October 4, 2022 at 9:06:56 PM UTC-7, Ilan Mayer wrote:

On Tuesday, October 4, 2022 at 11:48:59 PM UTC-4, henh...@gmail.com wrote:

On Tuesday, October 4, 2022 at 4:12:55 PM UTC-7, Gareth Taylor wrote:

In article <bd470200-70b0-46e7...@googlegroups.com>,
henh...@gmail.com <henh...@gmail.com> wrote:

( What's a probability value that (interestingly) Doesn't go to Zero as N increases ? )

Suppose you have 26 cards labelled from A to Z. You shuffle them and then deal them out in a row. What is the probability that none of the letters is in the correct position?

With just two cards, it's obviously 1/2.

How about three? The orders BCA and CAB have none in the correct position, while the orders ACB, CBA, BAC, ACB have at least one in the correct position. So it's 1/3.

What happens for larger alphabets?

Gareth

(great problem!)

(possible Spoiler --- esp. if i'm thinking correctly)

5 passengers to board an airplane with numbered seats 1--5

passenger 1 will have 4 poss. seats
say, he sits in seat 3.

Next, p3 will have 4 poss. seats ...

Next, p? will have 3 poss. seats ...

the last 2 passengers will have no choice.

So the # of poss. P(5) = 4 x 4 x 3
________________________

with 100 passengers,

the # of poss. P(100) = 99 x 99 x 98 x 97 x ... x 4 x 3

________________________

i think... you must be suggesting that

P(n) / n!

Doesn't go to Zero as N increases ?

The probability approaches 1/e.
See https://math.stackexchange.com/questions/399500/why-is-the-derangement-probability-so-close-to-frac1e

thanks! i thought i had at least a good start, ( and i still think so) according to what i had...

P(n) / n! === n / 2(n-1)
so this appraoches 1/2

( which implies that in Multiple-Loops cases, # of poss. is smaller -- which makes sense)



Richard Harnden wrote:

On RyanAir the probability quicky approaches 1

------------- thanks! .. . i didn't know there were lots of RyanAir jokes !


Paddy phones Ryanair to book a flight:
Operator asks, "How many people are flying with you?
Paddy replies, "How the f...k do I know?, It's your plane!




Two Ryanair pilots, Murphy and Seamus, are flying a Ryanair Ltd. jet on its final approach at Cork Airport.

Murphy says: “BuayJesus! Maury an' Josefff! Look how short this runway is.”

Seamus replies: “Yes, but look how fookin' wide it is!”

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On Thursday, October 6, 2022 at 9:31:49 AM UTC-7, henh...@gmail.com wrote:

On Tuesday, October 4, 2022 at 9:06:56 PM UTC-7, Ilan Mayer wrote:

On Tuesday, October 4, 2022 at 11:48:59 PM UTC-4, henh...@gmail.com wrote:

On Tuesday, October 4, 2022 at 4:12:55 PM UTC-7, Gareth Taylor wrote:

In article <bd470200-70b0-46e7...@googlegroups.com>,
henh...@gmail.com <henh...@gmail.com> wrote:

( What's a probability value that (interestingly) Doesn't go to Zero as N increases ? )

Suppose you have 26 cards labelled from A to Z. You shuffle them and then deal them out in a row. What is the probability that none of the letters is in the correct position?

With just two cards, it's obviously 1/2.

How about three? The orders BCA and CAB have none in the correct position, while the orders ACB, CBA, BAC, ACB have at least one in the correct position. So it's 1/3.

What happens for larger alphabets?

Gareth

(great problem!)

(possible Spoiler --- esp. if i'm thinking correctly)

5 passengers to board an airplane with numbered seats 1--5

passenger 1 will have 4 poss. seats
say, he sits in seat 3.

Next, p3 will have 4 poss. seats ...

Next, p? will have 3 poss. seats ...

the last 2 passengers will have no choice.

So the # of poss. P(5) = 4 x 4 x 3
________________________

with 100 passengers,

the # of poss. P(100) = 99 x 99 x 98 x 97 x ... x 4 x 3

________________________

i think... you must be suggesting that

P(n) / n!

Doesn't go to Zero as N increases ?

The probability approaches 1/e.
See https://math.stackexchange.com/questions/399500/why-is-the-derangement-probability-so-close-to-frac1e

thanks! i thought i had at least a good start, ( and i still think so) according to what i had...

P(n) / n! === n / 2(n-1)
so this appraoches 1/2

( which implies that in Multiple-Loops cases, # of poss. is smaller -- which makes sense)
Richard Harnden wrote:

On RyanAir the probability quicky approaches 1

------------- thanks! .. . i didn't know there were lots of RyanAir jokes !

Paddy phones Ryanair to book a flight:
Operator asks, "How many people are flying with you?
Paddy replies, "How the f...k do I know?, It's your plane!

Two Ryanair pilots, Murphy and Seamus, are flying a Ryanair Ltd. jet on its final approach at Cork Airport.

Murphy says: “BuayJesus! Maury an' Josefff! Look how short this runway is.”

Seamus replies: “Yes, but look how fookin' wide it is!”

https://en.wiktionary.org/wiki/derangement

i was noticing how even the most Deranged ppl would have lots of Correct Thoughts.... that it'd be very difficult to have all incorrect thoughts.


What else is an exmple of interesting discrepancy (or diff.) between a Word's ordinary and specialized meanings ?


_________________________________

re: RyanAir jokes

we no longer tell jokes about Yugo cars and Aeroflot ?

i get the feeling that ppl are telling fewer jokes now... that over the last 20 years... the Absolute freq. of Joke-telling is decreasing at the rate of about -10% every year.


i was also remembering the time Windows jokes were popular.... i can't remember the last time i saw a BLUE screen.... --- Not that Windows is now perfect...

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On 10/8/2022 11:28 AM, henh...@gmail.com wrote:

i get the feeling that ppl are telling fewer jokes now... that over the last 20 years... the Absolute freq. of Joke-telling is decreasing at the rate of about -10% every year.

Have you /seen/ Twitter?

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On 05/10/2022 04:48, henh...@gmail.com wrote:
[..snip..]

5 passengers to board an airplane with numbered seats 1--5

passenger 1 will have 4 poss. seats
say, he sits in seat 3.

Next, p3 will have 4 poss. seats ...

Next, p? will have 3 poss. seats ...

the last 2 passengers will have no choice.

So the # of poss. P(5) = 4 x 4 x 3
________________________

with 100 passengers,

the # of poss. P(100) = 99 x 99 x 98 x 97 x ... x 4 x 3

________________________

i think... you must be suggesting that

P(n) / n!

Doesn't go to Zero as N increases ?

This reminds me of an old puzzle posted here, um, many years ago...

A plane with 100 passenger seats has 100 passengers, but while queueing to get on, the first
passenger drops his ticket and it is whisked away by the wind! The plane crew decide to let him on
anyway, so he just boards and chooses one of the 100 seats at random. Subsequent boarders go to
their assigned seat, but if it's already occupied they just choose another unoccupied seat at random
and take that one.

What is the probability that the last passenger to board gets to sit in his originally assigned seat?


Mike.

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On 10/28/2022 1:18 PM, Mike Terry wrote:

A plane with 100 passenger seats has 100 passengers, but while queueing
to get on, the first passenger drops his ticket and it is whisked away
by the wind!  The plane crew decide to let him on anyway, so he just
boards and chooses one of the 100 seats at random.  Subsequent boarders
go to their assigned seat, but if it's already occupied they just choose another unoccupied seat at random and take that one.

What is the probability that the last passenger to board gets to sit in
his originally assigned seat?

[spoiler space]































All that really matters is whether the first passenger's assigned seat
or the last passenger's assigned seat (hereafter S1 and S100) ends up
occupied first. If it's S1, then the last passenger gets to sit in his
assigned seat; if it's S100, then he doesn't.

By the time the last passenger boards, exactly one of (S1, S100) will
be occupied.

* At least one because of the pigeonhole principle; you can't fit 99
people into 98 other seats (S2 through S99).

* At most one because S1 and S100 start out empty, and once one is
occupied (after passenger N boards), no one will be bounced into
the other one (because seats N+1 through 99 are all unoccupied and
their assigned passengers just sit there).

Regardless of which random choice causes one of those seats to be
occupied, the odds of which one gets occupied are 50/50. (Each random
choice has some probability p that S1 is chosen, the same probability
p that S100 is chosen, and probability 1 - 2p that some other seat is
chosen, in which case some later random choice will instead cause
either S1 or S100 to become occupied.)

Another way to view the situation is to imagine that, whenever one of passengers 2 through 99 finds their seat occupied, they take the seat
and the previous occupant (always passenger 1 in this scenario)
randomly chooses another unoccupied seat. This leads to the same set
of seats being occupied at each step (just by different people); and,
again, each random choice (always by passenger 1 in this scenario)
is equally likely to choose either S1 or S100 (and if it chooses
neither, then it just kicks the can down the road a bit).

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Mike Terry:

A plane with 100 passenger seats has 100 passengers, but
while queueing to get on, the first passenger drops his
ticket and it is whisked away by the wind! The plane crew
decide to let him on anyway, so he just boards and chooses
one of the 100 seats at random. Subsequent boarders go to
their assigned seat, but if it's already occupied they
just choose another unoccupied seat at random and take
that one.

What is the probability that the last passenger to board
gets to sit in his originally assigned seat?

I think it was not myself, yet I used this problem in an
exam, with the mad old lady at the head of the que, as a
demonstration for the total probability formula.

The mad old lady gets her seet with a probability of 1/n,
and then the last passenger is sure to get his. If she take
the last passenger's seat, then he has no chance. If she
take one of the other seats, the problem repeats with the
corresponding passenger in the role of the mad old lady.
The total probabity formula comprising these alternatives
is:

F(n) = [ 1 + F(2) + ... + F(n-1) ] / n =>
F(n) = 1 / 2

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