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03-17-2007, 03:13 PM | #1 |
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There is a fairly tough brainteaser inside this thread.
Suppose 100 people are about to board Air Force One (which we will assume has 100 seats) and you are the last person to board. Dubya is the first person to board. All 100 people have tickets and a seat number assigned to them. Since Dubya cannot read, he picks a random seat. Everyone else boards and sits in their assigned seat unless Dubya is sitting there. In that case the person is too terrified to tell Dubya to move, so he just sits in a random seat as well. This process continues of people sitting in their assigned seat or sitting in a random seat if they have been displaced by Dubya or by a person who took a random seat because there's was taken. You are the last person (number 100) to board. What is the probability that you sit in your assigned seat?
If you need clarification just ask. A proper explanation of your solution is required if you want me to tell you if you're correct . This was a question asked by Deutsche bank to a job applicant looking for a position as a junior quant trader. The interviewee who I got the question from solved it within 3 minutes...but he also has his PhD in algebraic geometry. Update: The answer is not 1/100. If it was I wouldn't have put "fairly tough" in the thread title. Update 2: Solution in post 28.
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Last edited by Physics Junkie; 03-17-2007 at 10:31 PM. |
03-17-2007, 03:15 PM | #2 |
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you are the pilot.
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03-17-2007, 03:24 PM | #3 |
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1/100
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03-17-2007, 03:26 PM | #4 |
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id just grab a stewardess and sex her up in the bathroom...if i get tired i can use the seat in the bathroom
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03-17-2007, 03:29 PM | #5 |
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i couldn't stand to hear George babbling.
i'd wait for the next flight. |
03-17-2007, 03:29 PM | #6 |
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Ugh, breaking out excel for this bad boy...
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03-17-2007, 03:34 PM | #7 |
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something something nPr(99,1)
syntax error ....... |
03-17-2007, 03:34 PM | #8 |
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Why would the President pick a random seat when he has his own suite and office to stay in?
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03-17-2007, 03:41 PM | #9 |
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BTW, W would be the last to enter
1/10,000? |
03-17-2007, 03:41 PM | #10 |
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03-17-2007, 03:44 PM | #11 |
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1/10000000000000000000000000000000
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03-17-2007, 03:48 PM | #12 |
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So far none of you are close. I won't tell you guys if the answer is correct unless there is an explanation to accompany it.
Keep working on it! Once you figure out how the situation works it's not terribly difficult to model. Last edited by Physics Junkie; 03-17-2007 at 03:55 PM. |
03-17-2007, 03:49 PM | #13 |
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the same chance of you getting laid tonight.
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03-17-2007, 03:55 PM | #14 |
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uh.....wait....... what?
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03-17-2007, 03:56 PM | #15 |
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03-17-2007, 03:57 PM | #16 |
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Thinking about this too hard, aren't we?
1***37; chance since your seat was one of 100 and 99 people are already in almost random seats that may or may not include the one you have a ticket for. Your seat is static while the people are the variable. oh, I see someone else said that. There better not be any trick "Yore actually this, so it means this" answer either. |
03-17-2007, 03:58 PM | #17 |
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03-17-2007, 03:59 PM | #18 |
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OK here how you figure this out. Compute the number of permuations for N people. for 100 people this is 100 factorial (a very large number), of that set see how many have you in the correct seat (also a very large number). This works out to 1/100 chance that you will be in the correct seat.
Edit: actually its tranpositions not permutations.... but you know what I'm talking about. Last edited by dmpi; 03-17-2007 at 04:09 PM. |
03-17-2007, 04:39 PM | #19 |
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50***37; chance.
Either your seat will be taken, or it won't. I really didn't give this answer any thought. -A |
03-17-2007, 04:59 PM | #20 | |
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Quote:
The calculation stops when someone else sits in W's seat. The chances of that are from 1/99 (the first person to get on the plane) to 1/1 for the last guy. So: 1/99 + Summation from 98 to 2 of ( 1/ n(n+1) ) This returns 0.5 or 50%. |
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03-17-2007, 05:05 PM | #21 |
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Now I want to check the math... but I've got more pressing work to do
-A |
03-17-2007, 05:30 PM | #22 |
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For the dummer than a sack of wet hammers ,namely me, whats the answer?
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03-17-2007, 05:34 PM | #23 |
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You don't have a seat, W is a robot, so there are 100 people and one robot, your seat is gone, sorry.
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03-17-2007, 06:00 PM | #24 |
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Now "Dubya" could by dumb luck wind up in his assigned seat.
In which case each subsequent passenger including you, if your the last person, will get the right seat. But what's much more likely is that he takes someone else's seat. Now, here's the counterintuitive part. There are only two seats that count, your seat and his seat. All the other seats don't make a difference. Whoever gets on the plane subsequent to him, those people are going to take some seat or another. Nothing matters until either your seat gets taken or his seat gets taken. If your seat gets taken by some displaced passenger, then you have zero chance of getting your seat. If his seat gets taken by some displaced passenger then every other passenger who walks onto the plane including you, will have his assigned seat. And how often does that happen? One time out of two. Half the time your seat's going to get taken by some displaced passenger, and half the time his seat is going to get taken by a displaced passenger. So your chances are one out of two. |
03-17-2007, 06:08 PM | #25 | |
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Quote:
Last edited by richde; 03-17-2007 at 06:39 PM. |
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