Is this Seat Taken?
100 people are waiting to board a plane. The first person’s ticket says Seat 1; the second person in line has a ticket that says Seat 2, and so on until the 100th person, whose ticket says Seat 100. The first person ignores the fact that his ticket says Seat 1, and randomly chooses one of the hundred seats (note: he might randomly choose to sit in Seat 1). From this point on, the next 98 people will always sit in their assigned seats if possible; if their seat is taken, they will randomly choose one of the remaining seats (after the first person, the second person takes a seat; after the second person, the third person takes a seat, and so on). What is the probability the 100th person sits in Seat 100?
NO — email me at [email protected] for a hint.
It would be 1/10,000
Please email and do not post solns!
email me at [email protected] for a hint (email to the address you listed bounced)
This is tough
higher.
1/100 ?
any1 kno the anser?
correct
RR: correct – email me at sjm1 AT williams.edu to discuss
sirigiri.psfc.mit.edu: correct (email me at sjm1 AT williams.edu to chat about riddles)
Andy: correct (sjm1 AT williams.edu)
pawan:yes (sjm1 AT williams.edu)
no, it’s not 1/24 (sjm1 AT williams.edu)
If there is 4 is the answer 1/24
Ravi: yes (sjm1 AT williams.edu)
Sergio: correct!
no, it’s a nice, simple number. try doing just 2 people, then just 3, to sniff out a pattern
Is the answer (1/100)(1/1!+1/2!+1/3!+…+1/99!)?
double-zero: correct; email me at sjm1 AT williams.edu to chat about the problem ..s
no, the answer is a nice simple fraction — email me at sjm1 AT williams.edu for a hint. /s
im just wondering if this answer is a series, more specifically the series 1/100+1/99+1/98+…+1/(100-n) where n is obviously the number of seat remaining after the person has taken their seat?
try doing a small case first
imagine there are only 2 people
first person takes seat one 50% of the time, seat 2 50% of the time
imagine there are 3 people
first person takes seat 1 33% of the time, seat 2 33% of the time, seat 3 33% of the time
if took seat 1, then person 2 takes seat 2 and then 3 takes 3
if took seat 2 then 50% of the time person 2 takes seat 1 and
50% of the time person 2 takes seat 3
if took seat 3 then doesn’t matter as can’t win
I don’t understand this puzzle.
No, it’s much higher: I’ll email you a hint.
actually, 1/(99(100!/100))
it’s not 2/3rds — I’ll email you a hint ..s
setup: passenger 1 = P1; passenger 100 = P100; seat 1 = S; and seat 100 = S100
if P1 = S1 then P100 = S100 (this is given)
if P1 does not equal S1 then there are two possible outcomes:
P100 = S100 and P100 not = S100
with these 3 possible outcomes, passenger 100 has 2/3 chance of getting seat 100 or 66.66666666…% possibility.
Close, but not quite. Email me at sjm1 AT williams.edu to chat more.
the answer is a little more than 50% (about maybe 51%)
no, it’s much higher than that
try to do just 2 people, then just 3, then just 4. you’ll see a pattern….
I think the answer would be 98/9801
The first guy has 99 options to choose from, and the rest of the people have 99! Ways to arrange themselves. The last guy chooses his seat 98 of the times (When the first guy doesn’t choose the 100th seat). There are then 98 times 98! Factorial ways the others can arrange themselves. 98×98!/99×99! Gives us the answer
email me at sjm1 AT williams.edu if you want to discuss. //s
i see the constistent answer with 1,2,3,4 (with the “brute force” approach), but 5 throws me out – close but not exactly…
is your second answer 1/5050? If yes, that’s not right. Try the hint, if that doesn’t help drop me an email (sjm1 at williams.edu). //s
That’s not the answer — also, the sum of the reciprocals is not the reciprocal of the sum. Try the following hint first: try to solve the problem when there are just 2 people, then just 3, then just 4. You should see something remarkable about these three answers. Do these cases out by brute force, seeing where people sit….
scratch the 1st answer thats completly wrong i am going to stay with the scond answer
is the answer like: 1/100 +1/99+1/98+………+1/2+1 ?
or is it 1/5050 chance. Can you send me the answer and explanation and where I can brush up on probability?
Stan: your answer is correct but not sure about your logic. email me at [email protected] for my soln. //s
Would it be a 50% chance since technically the problem is just between the first and last person? Because it’s all based on if the first person sits in a different seat or his own?
Would it be a 50% chance since technically it’s only based on where the first person sits? So it’s between the two of those people?
it’s not that high — try doing smaller cases. imagine there are just 2 seats, then just 3 seats, then just 4 seats. look for a pattern. //s
Well the probality is 99/100.
J-crew: yes, but not posting your answer as it’s correct.
It’s not 1 — that would mean the person ALWAYS gets their seat. Try special cases. Imagine that there are just 2 people and find the answer. Then do just 3 people. Then 4. You should hopefully see an interesting pattern.
well the answer is 1.
Alejandro: correct. Not posting as it’s the soln, well done
Jim: correct! not posting as it’s the soln
Correct — glad you’re enjoying the riddles (not posting your response as it’s the soln). Email me at [email protected] if you’d like more involvement in the site.
Well, I think it’s fine as is. Remember there are 100 people and 100 seats. After the first person, after the next 98 then only one person remains, and they MUST sit in whatever seat is free. If, however, you want to view it as the next 99 that’s fine.
“The first person ignores the fact that his ticket says Seat 1, and randomly chooses one of the hundred seats (note: he might randomly choose to sit in Seat 1). From this point on, the next 98 people will always sit in their assigned seats if possible”
Do you mean the next 99 people?
close, but too high
email me at [email protected] if you want a hint. /s
42/50 chance?
To: SirMordred: Correct! Not posting as it’s the soln. Email me at [email protected] if you want to discuss it further.
100!
Try doing small cases. Imagine there are just 2 people, or 3 people, or 4. Look for the pattern. If that doesn’t help, email me at [email protected].
You need more details, but yes….
please could I have the answer in like a spoiler box or something?
it is 1/2 because whenever someone comes on the plane and his or her seat is open they have to take it, so i guess 50% is your answer.
Aldo: Correct soln, well done (not posting as it’s correct)
Now that you know the right answer, try going back and carefully looking at what you wrote to see what’s wrong. Go through your argument with say 4 people.
So I’ve done 3, 4 and 5 and I now see the answer, but I am curious if you can explain what is wrong with my previous approach.
Nice try, and very detailed attack, but not quite. Try doing just 3 or 4 people, and seeing what happens in those cases.
You have to calculate the odds of each person having the possibility of choice and then multiply this probability by the odds that that person randomly selects seat 100. Once you have calculated these sums for all 99 people, then you can sum them, and this is the probability the final seat is taken.
For the first person the odds of having a choice are 100% and the odds of selecting seat 100 is 1/100. This gives person 1 a 1% chance of filling seat 100.
Person 2 will only have the option of choice if person 1 selects seat 2. This will happen 1 in 100 times and person 2 will select seat seat 100 1 in 99 times. The odds of person 2 selecting seat 100 in 1 in 99 or 1/9900
Person 3 will have the option of choice if person 1 or person 2 take seat 3. The odds of person 1 doing this is 1 in 100. The odds of person 2 doing this is 1 in 100 times 1 in 99. The justification behind this is that person 2 can only take seat 3 if person 1 takes seat 2 and then person 2 randomly selects seat 3. We can then sum the different probabilities of person 3 having a choice and multiply this sum by the odds person 3 select seat 100, or 1 in 98.
This process becomes very long and tedious. I used excel and mapped the whole thing out in a 100×100 grid and found the answer to be approximately
5.2199% chance seat 100 is already taken,
ie. person 100 will get his/seat 94.7801% of the time
Much higher than 1/50
See previous comment: it’s not 100! as not all arrangements are possible — if someone can sit in their seat, they do. try special cases — look at 3 people, 4 people….
it’s not 100! as not all arrangements are possible — if someone can sit
in their seat, they do.
try special cases — look at 3 people, 4 people….
1/50
actually 100! different arrangments
so p of each arrangment is 1/100!.
The p of 100 being in his place is
1/100. and it happens 99! times
so p of being in his place 99!/100! = 1/100
The answer is
100 seats that can have a 100! different arrangments. 100 factorial = 9.33262154 × 101^57 that includes the 100th being in his 100th seat.
Since each is an independent event, then they are all equally likely to happen.
P( of the 100th seat to being empty for the 100th passenger ) = 99*99/ 9.33262154 × 101^57
much lower than 90%, much higher than .9% (couldn’t tell what you wrote). will email you.
oh yeah my bad…
I caculated again whole thing and get ansrew… but there is one problem.
My Ansrew is 0,901 %
Problem is that i am not sure how many 0 there should be between 9 and 1
If this is irrcorrect, i have give up cause this is too damn hard to me 🙂
You’ve read the problem incorrectly. All 3! = 6 permutations are NOT possible. Remember every person after person 1 MUST sit in their seat if available. Thus if person 1 sits in seat 1, then person 2 MUST sit in seat 2. Your case of 1 3 2 is thus not possible; other cases may be impossible as well.
u have to give me a hint cos i cant understand what i caculate wrong… :/
in my caculates with 3 seats and 3 person:
Here is 6 different scenarios, what it could be. Anything else combination is not possible.
(number means persons ticket number, and u have to read them up to down)
seat 1 – 1 1 2 2 3 3
seat 2 – 2 3 1 3 1 2
seat 3 – 3 2 3 1 2 1
And there are in line “seat 3” two time 3 so it 2 out of 6 combination that person 3 gets to seat nro. 3
you miscalculated what happens with 3 people (as well as the higher ones). you missed some cases, as with three it is higher than 2/6.
I used ur way to count that. I stared from 2 seat, then goed 3 seat an so on.
I noticed one thing.
With one seat it is 1 out of 1 that 1st man gets seat nro. 1
With two seats probality is 1 out of 2 that second man gets seat nro. 2
With three seats it is 2 out of 6 that 3th man gets seat nro. 3
With four seats it is 6 out of 24 to 4th man gets seat nro. 4
so here is little more easily version.
with one seat it is 1/1
1 out of 2 is same as 1/2
2 out of 6 is same as 1/3
6 out of 24 is same as 1/4
so…
in hundred it will be 1/100
so it is 1% change with this caculation
if that is not correct please tell me why this caculation does not work. (cause my caculator says that this is correct)
4945 is a strange number — how did you get that? the answer is significantly larger. Try doing special cases: just 2 people, just 3 people, …. look for a pattern. //s
My first quess would be that it is 1 of 4945. I am pretty sure that it is wrong. But it is just a first quess.
not quite — it’s actually much higher. //s
1/ 10000
For those stating 1/100 it is very easy to see that the probability is much higher. Consider that if the 1st man chooses his own seat (a 1/100 occurrence), you already have reached 1% chance of the 100th man sitting in his own. However consider the possibility that the 1st man chooses to sit in seat m, and the mth man chooses to sit in seat 1. In this case the 100th man still gets his seat which establishes the lower bound of this problem well above 1%. I’m not strong in probability so I’ll have to play around longer to find an exact solution myself.
Whomever just posted (using author name of pi): correct, well done! Not showing the post as it gives away the soln.
Michael W: incorrect, not 99%.
the chance is 99%
when n is the number of seats the formula at the end is (n-1)/n
Nope, it’s higher.
there is 1 out of 99 chances
Actually, it’s much higher than 1 out of 98. HINT: Try to do some special cases. See what happens if there are 2 seats, or 3 seats, or 4 seats. These cases are small enough to analyze by brute force, and you might see a pattern.
There is a 1 out of 98 chance.
I get a different answer with a significantly simpler formula and a higher probability. //s
197/9900. the nth case after the trivial case n=1 is given by (2n-3)/(n(n-1)).
There is an elegant answer that doesn’t take too long to state. Try doing small cases first to get a feel for the problem. Imagine there are only 2 seats and apply your logic. Imagine only 3 seats. Only 4 seats. Do you see a pattern?
I started calculating this, and it seems extremely complex. I have no lengthy background in math, so I’m doing it more manually, but am I missing something that should make it simpler?
Here’s what I started out with, their is a pattern but the complexity seems to get multiplied each level
1/100 he sat in 100
+
(1/100 chance he sat in 2 * 1/99 2 sits in 100)
+
(1/100 he sits in 2 * 1/99 2 sat in 3 * 1/98 3 sits in 100) + (1/100 he sits in 3 * 1/98 3 sits in 100)
+
(1/100 chance he sits in four * 1/97 picks 100) + (1/100 he sits in 2 * 1/99 2 sat in 4 * 1/97 4 sits in 100) + (1/100 he sits in 2 * 1/199 2 sits in 3 * 1/98 3 sat in 4 * 1/97 4 sits in 100) + (1/100 he sits in 3 * 1/98 3 sits in 4 * 1/97 4 sits in 100)
this is to calculate chance seat 100 is taken
so it goes
(1/100)
(1/100 * 1/99)
((1/100 * 1/99 * 1/98) + (1/100 * 1/98))
((1/100 * 1/99 * 1/98 * 1/97) + (1/100 * 1/99 * 1/97) + (1/100 * 1/98 * 1/97) + (1/100 * 1/97)
so I predict the next line will be
((1/100 * 1/99 * 1/98 * 1/97) + (1/100 * 96) +
No, it’s not 1%. There are definitely only 100 seats on the plane. Your assumption that if the first person sits someplace other than seat 1 then the 100th person doesn’t get their seat is wrong.
answer 1%
If the first person to sit down does not sit in seat 1, then in the end seat 100 will be taken before the 100th person boards.
However, if the first person to board sits in seat 1, then all the following passengers will go to their assigned seats.
exception: if there are more than 100 seats on the plane then the solution cannot be found