Plane Probability Problem

Originally posted by SwissGambit
PBE6, your predictable interruptions are consistently getting tiresome and tedious, your general non-contributions are poorly worded. Get a grip my man.
[/dragontits]

Lulz! 😀

Originally posted by twilight2007
Due to the stimulating questions here, this question might have been posted earlier. Sorry if it has. Also this question involves an airplane. I just used the word "plane" for the alliteration.

There are 100 people on a plane. Let's label them Persons 1-100. The order the people sit down is determined by his or her seat number. There are 100 seats, l ...[text shortened]... did, please feel free to say so.

If this isn't solved by tomorrow, I'll give a hint.😀

Seems like one of those "YES/No" trick questions which work out to a 50% probobility.

Originally posted by eldragonfly
thanks, geepamoogle likes to hide behind "evasive" terminology as you will soon witness yourself, and yes i did happen to read the entire thread.

Obviously you didn't, as this problem has been solved AND proven. Yet you still claim a wrong answer to be right.

Originally posted by wolfgang59
OK.
Lets call the probability of someone sitting in the nth seat on an n-seater plane P(n)

For n=2
P(2) = 1/2 (the first guy either takes seat 1 or seat 2)

For n=3
P(3) = the chance of first guy sitting in seat + chance of first guy NOT sitting in seat x 2seater scenario

Therefore
P(3) = 1/3 + 2/3 * P(2)
= 1/3 + 1/3
= 2/3

P ...[text shortened]... holder of ticket 100 getting his seat is 1/100.

Which is surprising .. so I could be wrong!

Edit - Same as below.

Originally posted by TheMaster37
The 2/3 chance he will not take seat 1 is correct.

Person 1 chooses seat 1 : chance 1/3
.....Person 2 chooses 2 and person 3 chooses 3. So in this
.....scenario, person 3 ends up in seat 3.
Person 1 chooses seat 2 : chance 1/3
.....Person 2 chooses seat 1 : chance 1/2
..........Person 3 ends up in seat 3.
.....Person 2 chooses seat 3 : chance 1 ...[text shortened]... in seat 3

Total chance of person 3 ending up in seat 3:

1/3 + 1/3 * 1/2 = 1/3 + 1/6 = 1/2.

Edited out because I misread the problem.

Originally posted by TheMaster37
Obviously you didn't, as this problem has been solved AND proven. Yet you still claim a wrong answer to be right.

show me the proof, playing around with p(3) is not the same as proving p(100) = 1/2

Originally posted by eldragonfly
show me the proof, playing around with p(3) is not the same as proving p(100) = 1/2

P(2) = 1/2
P(3) = 1/3+1/3*1/2 = 1/3(1+1/2) = 1/3*3/2 = 1/2
P(4) = 1/4+1/4*1/3+1/4*1/2+1/4*1/3*1/2 = 1/4*(1+1/3)*(1+1/2) = 1/4*4/3*3/2 = 1/2
P(5) = 1/5*(1+1/4)*(1+1/3)*(1+1/2) = 1/5*5/4*4/3*3/2*1/2 = 1/2
...

P(n) = 1/n*(1+1/(n-1))*(1+1/(n-2))*...*(1+1/2) = (1/n)*(n/(n-1))*((n-1)/n-2))*((n-2)/n-3))*...*4/3*2/3*1/2

The numerators and denominators cancel diagonally except for the first numerator and last denominator.

Hence P(n) = 1/2

but p(100) can end up in either seat 1 or 100 or 99 depending on which seat p(1) picks at random/displaces.

Originally posted by eldragonfly
but p(100) can end up in either seat 1 or 100 or 99 depending on which seat p(1) picks at random/displaces.

I just proved it for all n. If you don't understand the proof, then try again.

Originally posted by Palynka
I just proved it for all n. If you don't understand the proof, then try again.

yes, p-100 will either end up in seat 1 or 100.

Originally posted by twilight2007
Due to the stimulating questions here, this question might have been posted earlier. Sorry if it has. Also this question involves an airplane. I just used the word "plane" for the alliteration.

There are 100 people on a plane. Let's label them Persons 1-100. The order the people sit down is determined by his or her seat number. There are 100 seats, l did, please feel free to say so.

If this isn't solved by tomorrow, I'll give a hint.😀

Is this the correct stipulation? That is that only person-1 will choose a seat at random and all the rest of the passengers - except the displaced person will sit in their assigned seats.

nevermind

nevermind

Originally posted by eldragonfly
rephrase/reword this, impossible to understand.

The original question focused on the 100th and last man. My question was essentially if there was a formula for the odds for a specific other passenger to be displaced, if you knew his boarding number.

I gave you the for instance of Passenger #2, but I could have just as easily asked for the chance Passenger #99 or Passenger #50 got to sit in their proper seat.

So my challenge is to generalize a solution to this question.

What are the odds Passenger #X gets to sit in their own seat?

or if you prefer

What are the odds Passenger #X gets displaced by a previous passenger?

I will stand by my answer of 50% for the final passenger, and the explanation I gave.

And to the poser of the original question, I don't think your question was unclear on the matter of seating odds at all. Just covering all the bases.

BTW I forgot to mention only one person per seat. I say this now so that no one will say it later.

first approach: It seems that the intuitive solution is that person-100 can only be bumped by either person-1 or person-99, in either case he must take seat-1.