08 Jun 08
Originally posted by DejectionWe have one of two situations.
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?
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?
08 Jun 08
1 edit
Originally posted by wolfgang59The 2/3 chance he will not take seat 1 is correct.
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.
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.
08 Jun 08
Originally posted by DejectionGeepamoogle said it in his post:
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?
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
08 Jun 08
1 edit
Originally posted by geepamoogleAssumption #2 is correct. Perhaps I could have been more clear in the problem.
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?
Also, the question doesn't ask for a proof, but many people seem to be skeptics so one might be useful to convince them.
09 Jun 08
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.
09 Jun 08
Originally posted by FabianFnasi think that this is more or less the correct solution.
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.
09 Jun 08
Originally posted by geepamooglerephrase/reword this, impossible to understand.
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.
09 Jun 08
Originally posted by TheMaster37thanks, geepamoogle likes to hide behind "evasive" terminology as you will soon witness yourself, and yes i did happen to read the entire thread.
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.