[Guest]

There are 100 seats on an airplane. To be seated is a high-profile CEO named Bob, 98 old ladies and a young man named Jimmy. Bob sits down first but he has misplaced his boarding pass and arrogantly takes a seat at random. After Bob come the 98 old ladies who are polite and if their seat is available will sit in it but if it isn't they will take another seat at random. Finally it is Jimmy's turn to sit down after the other 99 people have been seated. What is the probability that Jimmy's assigned seat will be available for him to sit in?

[Guest]

100%, since the ladies will move and Bob will be forced to if he's in Jimmy's seat

[Guest]

After someone sits down, they don't move.

[unreality]

question: do the ladies sit down at the same time (ie, a lady displaced by Bob must pick between Bob's original or Jimmy's) or do they sit down 1 by 1?

[Guest]

they sit down one by one and if their seat is taken they choose randomly out of all the remaining seats. (i.e. if Bob sits in the first lady's seat she will pick randomly between the remaining 99 seats)

Edited July 16, 2008 by griffin

[kingofpain]

My first instinct says this... Not based on any real prob. theory

3 scenarios:

a) Bob sits in his own seat (P = 0.01). All the ladies sit correctly and Jimmy gets his seat

b) Bob sits in Jimmy's seat (P=0.01). Ladies sit in their correct seats and Jimmy does NOT get his seat.

c) Bob sits on a lady's seat (P=0.98). Of the 99 available seats, probability that Jimmy's seat is taken when he gets to his seat = 98/99

Therefore, prob that jimmy gets his seat = 1-(98/99)-0.01

[Guest]

unfortunately that is not correct

the first two make sense but the third doesn't

Edited July 16, 2008 by griffin

[unreality]

great riddle I had to think a bit before getting the answer

50%

the answer is 1/2

[Guest]

great riddle I had to think a bit before getting the answer

50%

the answer is 1/2

How did you get that answer?

Oh wait, I think I see your logic. Clever.

Edited July 16, 2008 by Frost

[Guest]

Nice job unreality

[Guest]

Well, there are 100 seats, so...

I say 50%.

The CEO only has a 1% chance of choosing Jimmy's seat, and a 1% chance of choosing his own seat, so those two cancel each other out.

Assuming that the CEO chooses a seat that belongs to one of the ladies (98% chance), then the ladies will all sit in their seats , unless their seat is taken. If their seat is taken, then they will randomly choose another seat.

So the first lady who chooses a seat that wasn't assigned to another lady has a 50% chance of choosing either Bob's seat or Jimmy's. It all boils down to that one lady who chooses a seat that wasn't assigned to her group.

[Guest]

UGH...

Unreality beat me to it while I was typing up my explaination...

Good job

Edited July 16, 2008 by Big Red

[unreality]

what I did was broke the 98% down (the 1%s canceled, like you said) and I figured it out to be 49%,49% (after some probability attempts with the complicated sequence I realized that there was no bias of Bob over Jimmy or vice versa) which coupled with the 1%s makes 50-50

[Guest]

Forgive me as I'm new to this but aren't you assuming that in the scenario (Bob chooses one of the ladies seats) that the 1 lady who's seat is filled by Bob only chooses from either Bob's seat or the Students seat (50/50)? I think there is a thrid choice which is that the lady chooses another ladies seat at randomn and a cascade effect occurs.

[Guest]

Forgive me as I'm new to this but aren't you assuming that in the scenario (Bob chooses one of the ladies seats) that the 1 lady who's seat is filled by Bob only chooses from either Bob's seat or the Students seat (50/50)? I think there is a thrid choice which is that the lady chooses another ladies seat at randomn and a cascade effect occurs.

Yes, but the cascade takes effect until one lady takes either Jimmy or Bob's seat, at which point all the remaining women will get their scheduled seat (and Jimmy will get either his seat or Bob's). Even in the off-chance it cascades down the chain to the final woman, there would only be two seats available: Jimmy or Bob's.

Try it with 6 people instead of 100 and play it out by hand; you'll see Jimmy has a 50% chance of either seat each time.

[bushindo]

This gives the same results as unreality, for those who like to see equations and stuffs

Let T(n) represent the probability that the n-th person's seat is already taken.

T(1) = 0

( Since Bob, n=1, has access to an empty plane. His assigned seat is empty. )

T(2) = 1/100

( The second person has a 1/100 chance that Bob took her seat )

T(3) = 1/100 + T(2)*1/99

( The chance that the 3rd person's seat is taken is equal to the chance that Bob took it or that the second lady took it )

T(4) = 1/100 + T(2)*1/99 + T(3)*1/98

.

.

.

In general

T(n) = T(n -1) + T(n-1)/(102-n)

T(n) = T(n-1) * (103-n)/(102-n)

The solution amounts to finding T(100). This can be found as a solution of the above recursive function T(n) where T(2) = 1/100. The answer is .5, agreeing with what unreality and Big Red found earlier.