09 Jun 08
Originally posted by twilight2007Seems like one of those "YES/No" trick questions which work out to a 50% probobility.
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.😀
10 Jun 08
Originally posted by eldragonflyObviously you didn't, as this problem has been solved AND proven. Yet you still claim a wrong answer to be right.
thanks, geepamoogle likes to hide behind "evasive" terminology as you will soon witness yourself, and yes i did happen to read the entire thread.
10 Jun 08
1 edit
Originally posted by wolfgang59Edit - Same as below.
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!
10 Jun 08
1 edit
Originally posted by TheMaster37Edited out because I misread the problem.
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.
11 Jun 08
2 edits
Originally posted by eldragonflyP(2) = 1/2
show me the proof, playing around with p(3) is not the same as proving p(100) = 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
11 Jun 08
4 edits
Originally posted by twilight2007Is 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.
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.😀
11 Jun 08
1 edit
Originally posted by eldragonflyThe 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.
rephrase/reword this, impossible to understand.
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.