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 ?
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
The probability approaches 1/e.
See https://math.stackexchange.com/questions/399500/why-is-the-derangement-probability-so-close-to-frac1e
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
On RyanAir the probability quicky approaches 1
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.thanks! i thought i had at least a good start, ( and i still think so) according to what i had...
See https://math.stackexchange.com/questions/399500/why-is-the-derangement-probability-so-close-to-frac1e
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!”
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.
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 ?
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?
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?
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