Plane Probability Problem

geep that reasoning is all good, but how do we know that there is a 50/50 chance between #1 and #100 being taken first? Well we do know it, but how to we prove it, not inductively?

Originally posted by Dejection
geep that reasoning is all good, but how do we know that there is a 50/50 chance between #1 and #100 being taken first? Well we do know it, but how to we prove it, not inductively?

We have one of two situations.

1) The problem isn't defined, because we are missing odds for the seats.

2) The first person, and any person whose seat is taken when they get on, will choose a seat at random with each available seat having equal likelihood of being chosen.

I was thinking the latter was the case, but if not, then it is the former and we still have no reason to believe one or the other is more likely to be chosen first, do we?

Originally posted by wolfgang59
I was calculating P(3) as a 1 in 3 chance he took seat 3 and a 2 in 3 chance he did not.

My solution may be wrong but that is not the flaw.

The 2/3 chance he will not take seat 1 is correct.

However, there is only a chance of 1/3 to be back in the two-seat problem, only when person 1 chooses seat 2. If person 1 chooses seat 3, you will NOT be back in the two seat problem,since person 3 cannot end up in seat 3 anymore.

(As I understand it we want to calculate the chance that in the n seat problem person n chooses seat n, so in the 3 seat problem that person 3 ends up in seat 3. If person 1 is in seat 1, person 3 cannot sit there anymore).

In a calculation

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 1/2
..........Person 3 does NOT end up in seat 3
Person 1 chooses seat 3 : chance 1/3
.....Person 3 does NOT end up 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.

Originally posted by Dejection
geep that reasoning is all good, but how do we know that there is a 50/50 chance between #1 and #100 being taken first? Well we do know it, but how to we prove it, not inductively?

Geepamoogle said it in his post:

The problem comes down to one simple question:

What is the chance that seat 1 will be taken before seat 100?

That clearly is a 50% chance

Originally posted by geepamoogle
We have one of two situations.

1) The problem isn't defined, because we are missing odds for the seats.

2) The first person, and any person whose seat is taken when they get on, will choose a seat at random with each available seat having equal likelihood of being chosen.

I was thinking the latter was the case, but if not, then it is the former ...[text shortened]... d we still have no reason to believe one or the other is more likely to be chosen first, do we?

Assumption #2 is correct. Perhaps I could have been more clear in the problem.

Also, the question doesn't ask for a proof, but many people seem to be skeptics so one might be useful to convince them.

Here's a follow-up question.

What are the odds that a person in any particular seat is displaced?

For example, the odds the 2nd person gets to keep their seat is 99%, because the only person who would have a chance to take his seat is the first guy, and assuming he picks randomly (a given), he'll only take it 1-in-100 times.

Originally posted by FabianFnas
Why not let person 100 enter the plane first, rather than last? He takes a seat in a random way. The probability that he take the seat #100 is 1 of 100. So the probability is 0.01.

What the others do does not have any bearing at all.

i think that this is more or less the correct solution.

Originally posted by geepamoogle
Here's a follow-up question.

What are the odds that a person in any particular seat is displaced?

For example, the odds the 2nd person gets to keep their seat is 99%, because the only person who would have a chance to take his seat is the first guy, and assuming he picks randomly (a given), he'll only take it 1-in-100 times.

rephrase/reword this, impossible to understand.

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

Wow, in two short posts you demonstrated a complete lack of any knowledge of probabilities AND an unwillingness to read all the other posts.

Way to go.

Originally posted by TheMaster37
Wow, in two short posts you demonstrated a complete lack of any knowledge of probabilities AND an unwillingness to read all the other posts.

Way to go.

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

Originally posted by TheMaster37
Wow, in two short posts you demonstrated a complete lack of any knowledge of probabilities AND an unwillingness to read all the other posts.

Way to go.

Yeah, this eldragonfly creature is a real brainiac.

Originally posted by LemonJello
Yeah, this eldragonfly creature is a real brainiac.

Hi lemmonjelly and top of the morning to you. 😉

geepamoogles explanation makes the most sense, but why isn't this more random or is this a terminology thing again.

Originally posted by eldragonfly
Hi lemmonjelly and top of the morning to you. 😉

Ad hominem and irrelevant!!

😀

Originally posted by PBE6
Ad hominem and irrelevant!!

😀

PBE6, your predictable interruptions are consistently getting tiresome and tedious, your general non-contributions are poorly worded. Get a grip my man.
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