# Brain Teaser: Crazy Airplane Guy

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- This topic has 46 replies, 31 voices, and was last updated 13 years, 12 months ago by aasif.

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- April 25, 2002 at 12:51 am #29322
I just came across this brain teaser and thought this forum could use a little statistics fun (aside from that of our daily lives):

“a line of 100 airline passengers is waiting to board a plane. they each hold a ticket to one of the 100 seats on that flight. (for convenience, let’s say that the nth passenger in line has a ticket for the seat number n.)

unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. all of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. if it is occupied, they will then find a free seat to sit in, at random.

what is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?”0April 25, 2002 at 7:11 pm #74846

Jack WelchParticipant@Jack-Welch**Include @Jack-Welch in your post and this person will**

be notified via email.You will give the answer (with solution) after a few days won’t you?

0April 25, 2002 at 7:17 pm #74847Did you get this off the GMAT? I remember questions like this….

0April 25, 2002 at 8:05 pm #74848

billybobParticipant@billybob**Include @billybob in your post and this person will**

be notified via email.Each of my possums (I have 53) have their own den. Sometimes one will enter another’s den first for the night which sets off a chain reactions of all the others having to find another den to sleep in. Actually this happens most nights because they are an ornary bunch. The question is; what is my chance of finding my favorite possum Ralphy in his own den on any given morning?

0April 25, 2002 at 8:11 pm #74849Depends, what color is the dirt?

0April 25, 2002 at 8:25 pm #74850

billybobParticipant@billybob**Include @billybob in your post and this person will**

be notified via email.The color of the dirt? Oh I get it….an X!

0April 25, 2002 at 9:53 pm #74851Is it correct?1/100 + 99/100(1/99)=

1/100 + 99/9900 = 2 %BR,0April 26, 2002 at 12:56 am #74855

Jack WelchParticipant@Jack-Welch**Include @Jack-Welch in your post and this person will**

be notified via email.Loks good to me….but i haven’t a clue if thats rght!

0April 26, 2002 at 2:06 am #74857

Jack StuartParticipant@Jack-Stuart**Include @Jack-Stuart in your post and this person will**

be notified via email.I put it at 50-50. The whole thing stops when someone else sits in Crazy Guy’s seat. The chances of that range from 1/99 (First person) to 1/1 for the last guy. Because this would require rate averaging, I want to do 1/99factorial divided by 99.

0April 26, 2002 at 4:04 am #74858I agree with the .5 probability. I worked out the formula:

1/99 + Summation from 98 to 2 of ( 1/ n(n+1) )

This returns the 0.5.

0April 26, 2002 at 8:39 am #74862Probability that the last (100th) person to board the plane will sit in proper seat = Probability that no one else sits in his seat =

= (99/100)*(98/99)*(97/98)*………………*(3/2)*(1/2) =

1/100 = .01 or 1 %

Is this correct

Nayan

0April 26, 2002 at 11:20 am #74867Ding Ding Ding Ding.

We have a winner. Congratulations Don, I believe that is the correct answer!

Very nice job to you, and kudos go out to Jack Stuart and Nayan for giving it a shot. I’m surprized that more people weren’t up for giving it a shot. Are we all too caught up in saving our companies money? :)

In any case, well done to everyone who tried it.

Best,

Todd0April 26, 2002 at 6:46 pm #74890I feel almost bad about keeping this message alive.

I do not think this is right…

The last fellow gets his seat in 2 scenarios:

(1) The first guy randomly selects his own seat OR

(2) The first guy chooses seats 2 through 99 AND one of the passengers in seats 2 through 99 sits in seat 1

NOTE: if the crazy guy randomly chooses seat 100, the game is over!

Looks like a classic P(A or (B and C)) = P(A) + P(B and C) since A and (B and C) are mutually exclusive.

P(A) is easy. It should be 1/100.

P(B and C) are the knucklebusters. You have to capture the right counting strategy, and it looks like it would be a tad bit wild.

I do not think the sum from 2 to 98 of (1/n(n+1)) does the trick. Remember, if one of passengers 2 through 99 finds that their seat is occupied, they will then randomly select another seat. If they do not select the 1st seat, then it will start the whole seat shuffle again.0April 26, 2002 at 7:01 pm #74893

Jack StuartParticipant@Jack-Stuart**Include @Jack-Stuart in your post and this person will**

be notified via email.It becomes easier if you don’t get bogged down in the pardigm of trying to solve every decision that could possibly be made. The crux of the problem is that someone will sit in either Mr. 100s seat or in Crazy guy’s seat. Who or when is irrelevent, except for The Crazy guy. He by definition has not sat in his own seat. But I think we can pretty much ignore the change in the odds. It is therefore just about 50-50.

0April 26, 2002 at 7:26 pm #74895

Jack WelchParticipant@Jack-Welch**Include @Jack-Welch in your post and this person will**

be notified via email.Forget it! I’m taking the train; derailments or not!

Jack0April 27, 2002 at 12:00 am #74908

GabrielParticipant@Gabriel**Include @Gabriel in your post and this person will**

be notified via email.Jack:

The crazy guy can seat on seat 1. He just ignores the assigned seat and pick one randomly among all seats, including the 1st seat.

I have to admit that I could’t derive a formula, so I made a spreadsheet for a Monte Carlo experiment that simulates the boarding a given number of times (if any one wants the Excel file post your e-mail here, I will send it). I tried three times with 1000 trials and got arround 48 – 49%, so it is close to 50% as you said. Even when you seem to be right, I still do not fully understand your logic. We agree that if someone before Mr 100 takes seat number #1 (Mr 1 taking the seat #1 is one of these cases), Mr 100 won. If someone before Mr 100 takes the seat #100, Mr 100 loses. Mr 100 will seat either in seat #1 or #100, no further choice. Now, this appears to be a 50 -50 case, but I was cought before with tricky statistics where a result seems to be obvious (for example, the three curtain problem, do you know it?). I will appreciate that you or someone else could post a demonstration (either mathematical or logical). I saw that someone derived a formula for this 0.5 probability, but I have to admit that I have no idea about how to get it.0April 27, 2002 at 2:43 pm #74913

Jack StuartParticipant@Jack-Stuart**Include @Jack-Stuart in your post and this person will**

be notified via email.Gabriel, Here is how I think it goes. Crazy guy sits in a seat not his own. 1/99 chance it is seat 100. Guy number 2 will do one of three things: Take his own (96/98 but we don’t care), take crazy guy’s seat (1/98) or take 100’s (1/98). Guy three has the same three choices but the odds change slightly 95/97;1/97;1/97. The only thing we are worrying about are the last two alternatives and they always have a 50:50 chance of happening. The total odds of 100 not getting 100’s seat are figured by a weighted average 1 event of 1/99 and 98 events of 50%. That will be about 49.01%. That is a practical answer. 50 vs 1 or 2 %. A true statistical answer will require each event to be estimated, with one string having replacement and the other two not. The middle string always leads to the side strings.

0April 28, 2002 at 6:41 pm #74919The answer is 1/2. Here’s how:

Let p(x) denote the probability that the xth passenger finds his/her seat occupied when coming on board the plane.

When the first person gets on board, his seat is definitely NOT occupied. so…

p(1) = 0

The second person will find his seat occupied if the first guy took it, and the probability of that is 1/100. so …

p(2) = 1/100

The third person will find his seat occupied if EITHER the first guy occupied it OR the second guy occupied it. so …

p(3) = 1/100 + p(2).1/99

Arguing similarly, let’s p(4) and p(5):

p(4) = 1/100 + p(2).1/99 + p(3).1/98p(5) = 1/100 + p(2).1/99 + p(3).1/98 + p(4).1/97

You should now see a pattern emerging:

p(2) = 1/100p(3) = p(2) + p(2).1/99p(4) = p(3) + p(3).1/98p(5) = p(4) + p(4).1/97……..p(x) = p(x-1) + p(x-1).1/(100-x+2)……..p(100) = p(99) + p(99).1/2

Infact, it’s easy to see why we get this pattern. Think about what happens when person ‘x’ boards the plane. That depends on what happened when the (x-1)th person boarded the plane. EITHER the xth seat was already occupied OR the (x-1)th person occupied the xth seat because the (x-1)th seat was taken.

We know that the probability that the (x-1)th seat was occupied when person ‘x-1’ got on board is p(x-1). THE KEY IS to realize that the probability that the xth seat was occupied at that time is also, infact, p(x-1). Why? because every occupied seat had the same likelihood of getting picked (because people chose their seats at random if their seats were taken). If we add the two probability in the EITHER/OR case in the preceding paragraph, we get the formula:

p(x) = p(x-1) + p(x-1).1/(100-x+2)

Anyway, the rest is easy.

Simplifying our final pattern, we get the following:

p(2) = (1/100)p(3) = (100/99).p(2)p(4) = (99/98).p(3)p(5) = (98/97).p(4)…p(100) = (3/2).p(99)

Since each probability is > 0, we can safely multiply all the terms on either side and we get:

p(2).p(3)…p(100) = (1/100)(100/99).(99/98).(98/97)…(3/2).p(2).p(3)…p(99)

Canceling numerators & denominators on the right side and p(2).p(3)…p(99) on either side, we get:

p(100) = 1/2.

So, the probability that the 100th guy sits in his/her own seat = 1 – p(100) = 1/2.

Thanks again for everyone’s thoughts. Anyone have another?0April 29, 2002 at 2:38 am #74921

Jack StuartParticipant@Jack-Stuart**Include @Jack-Stuart in your post and this person will**

be notified via email.Todd, I am sorry to do this to you, because I know you want to put this to rest, but 50% is not quite right. Probability-wise everything looks good, but logically we have a problem. What if the plane only had two seats? The last guy has 0% chance of getting his seat. What if 3 seats? Now it is 75% (1/2 the time crazy guy takes your seat, 1/2 of the rest, the second guy might take your seat.). The problem is two sets. The 50% is good for everyone’s decision but Crazy guy(Because he will never take his own seat). The second set is the crazy guy’s set which is 1/(N-1). For two seats it is 1, for 3 seats it is 50%. For an infinite set the answer drives to 50% because Crazy guys set drives to nothing. I would like a new puzzzle also. This was a very good one.

0April 29, 2002 at 4:34 pm #74951

GabrielParticipant@Gabriel**Include @Gabriel in your post and this person will**

be notified via email.Jack:

First, thanks for explining me your view before.

Second: I do not see from the puzzle that the crazy guy wil not take his own seat by cahnce. He just ignores that he has a seat assigned, ad choses any seat, his own seat may be the lucky one. In this case 50% would be ok for any number of seats.

Third: I subscribe your request for more puzzles.

Gabriel0April 29, 2002 at 4:59 pm #74954

jimggggParticipant@jimgggg**Include @jimgggg in your post and this person will**

be notified via email.You have come upon an elegant answer.

2 – If there were two seats there is clearly a 50% chance #1 will pick the right one.

3- If 3 seats, 1/3 chance he is right, 1/3 he is #3s seat. If in #2’s seat, 50-50 that #2 will sit in #1vs #3. So net to 50% chance also.

4 etc. always a 50% chance regardless of the number of seats on the plane.0May 2, 2002 at 7:26 am #75051The chances that the first (abnormal) person taking a correct seat or an incorrect seat is 50%.

If he (Mr abnormal) takes the correct seat then the 100th person also gets the correct seat.

If Mr Abnormal takes a wrong seat then the 100th person also gets a wrong seat.

Hence the probability of the 100th person getting the 100th seat is also equal to 50%.

Ans : So the probabilty is 0.5

QED0May 2, 2002 at 8:12 am #75054The probability that the crazy passenger occupies seat #100 is 1/100. Therefore the probability of the 100th passenger finding his seat empty is 1-1/100.

0May 3, 2002 at 12:59 pm #75106

GatorfishParticipant@Gatorfish**Include @Gatorfish in your post and this person will**

be notified via email.94.8%?

0May 3, 2002 at 1:30 pm #75108

Jack StuartParticipant@Jack-Stuart**Include @Jack-Stuart in your post and this person will**

be notified via email.It is not demonstrated. The problem states that the first person did not sit in his own seat. Therefore, his odds are different than everyone else’s. Please see my analysis in the thread a few steps up. Otherwise you are correct about the 50-50 thing.

0May 3, 2002 at 1:36 pm #75109

GabrielParticipant@Gabriel**Include @Gabriel in your post and this person will**

be notified via email.Jack, this is from the original post:

“the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat”

In which way does this excludes the possibility that the crazy guy picks his own seat at random?0May 3, 2002 at 2:06 pm #75114

GabrielParticipant@Gabriel**Include @Gabriel in your post and this person will**

be notified via email.I agree with the number (50-50), but I don’t understand your explanation, because it seems from that that the luck of Mr 100 depends from the choice of Mr 1 only. If Mr 1 picks seat #1 (1/100 cahnces) Mr 100 will get seat #100. If Mr 1 picks seat #100 (also 1/100 cahnces), Mr 100 will not get his seat. If Mr 1 picks another seat (98/100 chances) then the luck of Mr 100 depends on what seat will pick the person that was supposed to seat on the seat picked by Mr 1.

0May 3, 2002 at 6:48 pm #75130

Jack StuartParticipant@Jack-Stuart**Include @Jack-Stuart in your post and this person will**

be notified via email.I stand corrected. I over-interpreted the description to mean he intended not to sit in his own seat. That makes everyone’s significant (To Mr.100) decision to be 50-50. (Mr. 100’s seat or Crazy guy’s seat). It does not matter how many decisions are made. Whichever of those two seats gets picked ends the problem. And the chance of that happening at random is 50%.

0May 3, 2002 at 10:33 pm #75142

GabrielParticipant@Gabriel**Include @Gabriel in your post and this person will**

be notified via email.Thanks, Jack.

Despite the liittle missunderstandig about the possibility of Mr 1 picking seat #1, your explanations along the thread showed me the “logic” behind this 50-50, which was apparent for me but I could not explain with conviction.

Shortly. Any passanger before Mr 100 picking seat #1 leads to Mr 100 seating in his seat. Any passanger before Mr 100 picking seat #100 leads to Mr 100 seating in seat #1. And any time the seat #1 has a chance to be occupied, also seat #100 has the same chance. So it is 50-50.

By the way, and just in case, I run the Monte Carlo experiment spreadsheet with 10.000 trials, and the result was 4995 times Mr 100 seated in seat #100. Now that I understand that the answer is 50-50, this experiment tells me that the Excel spreadsheet and the macro are well programmed!0September 12, 2002 at 3:36 am #78829I don’t think 50-50 is right.

If person #1 picks the right seat, then person #100 will also get the right seat, but that probabililty is 1/100. I #1 picks the wrong seat, then somebody else will also have to pick the wrong seat so that the preson after him (her) will also have to pick the wrong seat, etc. The error will propagate to person #100. Thus, if #1 picks the wrong seat, then person #100 will find his seat occupied. Consequently, overall P=1/100.0January 30, 2003 at 10:58 pm #82575

H. FloresParticipant@H.-Flores**Include @H.-Flores in your post and this person will**

be notified via email.The hard part of the problem is the “big” number 100… there exist a general strategy, namely. Simplify the problem. In this case choose small numbers to get insight.

The case for 2 passengers is too easy.

In the case 3 we start getting some idea

Call passengers 1,2,3 . And seats 1,2,3.

Probability of pass 1 to take seat 1 = 1/3 (then pass 3 takes seat 3)

Probability of pass 1 to take seat 2 = 1/3. In this case: Probability of pass 2 to choose seat 1 =1 / 2 total probability is 1/3 x 1/2 = 1/6 .. (then pass 3 takes seat 3)

We can see that the total probability is 1/3+1/6=1/2… as many of you claim.

In this kind of problems, i.e. when the small cases suggest you an answer. Mathematical Induction is a good option to continue see this references

http://www.cut-the-knot.org/induction.shtml

http://www.cc.gatech.edu/people/home/idris/AlgorithmsProject/ProofMethods/Induction/UnderstandingInduction.html

If you are not familiar with this tool, I suggest you to read some of the previous links (specially the second which explains the induction with falling dominos)… If not, the explanation will be perhaps more confusing.

General Induction consist of two steps:

#1:Prove the statement for some small number (in this case 2 or 3).

#2:Assume that the statement is true for every number less than k… and using this assumption derive a proof for the next case… (k+1)

For the airplane problem the step 2 is tricky… assume that if the (any possible) plane has less than k seats, then the probability of the last passenger to get his seat is 1/2… now we should prove that, if the plane (another plane if you want) as (k+1) seats, then the (k+1)-passenger as 50% of chances to get his (k+1)-seat.

(k+1) case:

a) Probability of pass 1 to take seat 1 = 1/(k+1)

b) Probability of pass 1 NOT to take seat 1 or seat(k+1) = (k-1)/(k+1).

Now part b) assume that pass 1 as taken a seat between 2 and k, (say seat 7)… then for the passengers 2 to 6 there is no problem… but when pass 7 arrive his seat is occupy… then he do not have a seat… think carefully… if pass 7 takes seat 1, then the situation is fixed… but he does not now it… so he is some kind of “crazy”, so at this moment the probability of the(k+1) pass to get his (k+1) seat is 1/2. Because, from the math point of view this situation is equivalent to having a small plane where 7 is the crazy, and 8,9,…. to (k+1) are the rest of the passengers…. and the assumption says that this probability is 1/2… you should understand the Math Induction, to agree with me at this point.

Now do the calculations:

Prob(pass1 take seat1) +

P(pass1 did not take seat 1 or seat (k+1) ) x P(last pass take his seat, given that pass 1 did not took seat 1 or (k+1).

1/(k+1) + (k-1)/(k+1) * (1/2) = (k+1) / (2(k+1)) = 1/2… as we expected !!!

0July 15, 2003 at 8:32 pm #87979

phaedrusParticipant@phaedrus**Include @phaedrus in your post and this person will**

be notified via email.let n be number of seats:

after first passenger sits down, prob last seat is open = (n-1)/n

after second passenger sits down, last seat is still open if his seat is open (prob=(n-2)/(n-1)) and the next passenger can take his designated seat or if the second passenger’s seat is occupied but he doesn’t sit in the last passenger’s seat (prob=1/(n-1)*(n-2)/(n-1)). So prob last seat is open after two have entered plane is (n-1)/n*((n-2)/(n-1)+1/(n-1)*(n-2)/(n-1)),

simplifying (n-1)/n*((n-1)/n+1/n*(n-2)/(n-1))=(n-2)/(n-1)

after third passenger sits down, last seat is still open if the third passenger’s seat is open (prob=(n-3)/(n-2)) or if the third passenger’s seat is occupied but he doesn’t sit in the last seat (prob=1/(n-3)*(n-3)/(n-2)). so prob last seat is open after three have entered plane is (n-2)/(n-1)*((n-3)/(n-2)+1/(n-2)*(n-3)/(n-2)),

simplifying (n-2)/(n-1)*((n-3)/(n-2)+1/(n-2)*(n-3)/(n-2))=(n-3)/(n-2)

…..after m passengers have taken their seats, the prob that any one of the remaining passengers’ seats is empty is (n-m)/(n-m+1), for the last passenger, m=n-1, and the prob =1/2.

the beauty of this solution is that it also answers the more general question of what is the probability of the next passenger’s seat being open after m passengers have entered for any value of m, not just m=n-1.

how do we interpret the beautiful result that after m passengers have taken their seats, the prob of the next passenger’s seats being empty is (n-m)/(n-m+1)? it means the prob that his seat is occupied is 1/(k+1), where k (=n-m) is the number of empty seats. it is the same prob as if the passenger just before him were the first passenger to find his seat occupied. how does this make sense? i’ll have to think about it some more to see whether there is a simpler way to view the problem.

0October 2, 2004 at 9:31 pm #108470Assume 10 people on a 10seater airplane.

P1 takes a randomized seat- a 50/50 shot he takes his own- either he does, or he takes someone elses. So, off the bat, if he takes is proper seat, then person 10 gets his seat. If he does not:

1/9 chance of a particular seat- or 1/(q-1) assuming q=#of seats.

So he takes seat #4. (Just because)

Person 2 takes seat 2, p3=seat 3, and now p4 has a choice of taking another seat. He has a 50/50 shot of taking seat 1 or some other seat- if he takes seat 1, then we see that p5 takes seat 5, p6=s6, etc, and again, P10 gets his seat. So if he does not take s1, he will take another one- say, s7.

Ok…. so p5=s5, p6=s6, p7=uh oh…same thing…so randomly, he takes s9, so then p8=s8, p9=again, we have a annoyed passenger.

Well he has a choice- now we have a whopping 2 seats left, seats 1 and 10. All the other seats are taken. So passenger 10 has a fifty shot that p9 will take seat 1!!!!!!! So that means are bud p10 has a 50-50 shot at getting his own seat! This will work for any number q.

Passenger 100 has a 0.5 chance at getting his seat= 1/2!

You might be wondering what if p4, took randomly seat 10. Well, its the same logic as him taking seat 1- he either takes it or not. If he takes it, it is over for p10. Keep in mind that this detail doesnt really change the probability- because each time a passenger has to sit in a different seat he will have, among others, the choice of seat 1 and seat 10. The passenger chosing s1 basically ‘neutralizes’ his taking s10- they average out to 1/2!

Again, 1/2 is the final answer.0October 4, 2004 at 11:01 am #108488The only possibility for 100th person to sit on his seat is that passanger 1 sits on seat 1. If he changes his seat this will disturb whole of the sitting plan.

So the probability that passanger 1 sits on seat 1 is 1/100

or 0.01

0December 17, 2004 at 12:53 am #112548u cant think of a good brain teaser think of a better one a email [email protected] and i will come to your site

0December 19, 2004 at 11:15 pm #112667the question was a bit mistated. if someone is sitting in your seat, you pick another random seat to sit in. so the answer isn’t .01.

i think you have to look at it recursively0December 20, 2004 at 3:11 am #112678great answer, totally right, here’s a follow on question that with your solution is really easy to answer. what is the expected number of people who will be sitting in the wrong seat on a plane with n seats?

my answer will be in the next post0December 20, 2004 at 3:16 am #112680expected number of people in wrong seat on an n seat plane will be the sum from i = 2 to n of 1/i, i.e. a truncated harmonic series. as the plane gets big, the number of people in the wrong seat will also get big, there is no asymptote.

another great question you might already know: $7.11 = a*b*c*d=a+b+c+d, where a, b, c, and d are all normal prices (i.e. $1.23, not 1.3333… or 1.234). what are a, b, c, and d?0January 28, 2005 at 12:07 am #114122I came up with a slightly different solution than the one already posted.Define p(n) to be the probability that the final passenger ends up in his seat, given that there are n seats. Now, if the crazy person ends up in his own seat (seat 1), then the final person will too. This has a probability of 1/n. If the crazy person ends up in seat 2, then person 2 “becomes” the crazy person (with only n-1 seats left), and his “correct” seat is now seat 1. Thus the probability of this occurring is 1/n * p(n-1). Similarly, if the crazy person ends up in seat 3, person 2 sits down in seat 2 and then person 3 “becomes” the crazy person (with only n-2 seats left), and seat 1 is his new correct seat. This has a probability of 1/n * p(n-2). Continuing the pattern, we see that:

p(n) = (1 + p(n-1) + p(n-2) + … + p(3) + p(2))/n

or

p(n) = 1/n * (1 + sum(p(i) as i goes from 2 to n-1))Now, if we use this recursive formula to write out p(n) for the first few numbers, we see that:

p(2) = 1/2, p(3) = 1/2, p(4) = 1/2, p(5) = 1/2It certainly seems that p(n) = 1/2 for all n >= 2. To prove this claim, we use mathematical induction:

Assume that p(k) = 1/2 for k = 2,3,…,n-1

Then:

p(n) = 1/n * (1 + sum(p(i) as i goes from 2 to n-1)

p(n) = 1/n * (1 + (1/2)*(n-2))

p(n) = 1/n * n/2

p(n) = 1/2

Q.E.D.= 2. To prove this claim, we use mathematical induction:

Assume that p(k) = 1/2 for k = 2,3,…,n-1

Then:

p(n) = 1/n * (1 + sum(p(i) as i goes from 2 to n-1)

p(n) = 1/n * (1 + (1/2)*(n-2))

p(n) = 1/n * n/2

p(n) = 1/2

Q.E.D.= 2. To prove this claim, we use mathematical induction:

Assume that p(k) = 1/2 for k = 2,3,…,n-1

Then:

p(n) = 1/n * (1 + sum(p(i) as i goes from 2 to n-1)

p(n) = 1/n * (1 + (1/2)*(n-2))

p(n) = 1/n * n/2

p(n) = 1/2

Q.E.D.Thus we see that for any number of passengers greater than or equal to 2, the probability that the final one is in his correct seat is 1/2.0January 28, 2005 at 12:12 am #114123Sorry about the previous post being hard to read – the editor decided to delete my carriage returns >:/

Lets try this again::/

Lets try this again::/

Lets try this again:p(n) = 1/n * (1 + sum(p(i) as i goes from 2 to n-1) p(n) = 1/n * (1 + (1/2)*(n-2))p(n) = 1/n * n/2p(n) = 1/2 Q.E.D.0February 19, 2005 at 11:21 pm #115117

Reality ManParticipant@Reality-Man**Include @Reality-Man in your post and this person will**

be notified via email.It’s ironic that Six Sigma would have so many incorrect replies, and so few–this one, for example–correct.

0April 26, 2005 at 10:35 am #118414

Vineet KumarMember@Vineet-Kumar**Include @Vineet-Kumar in your post and this person will**

be notified via email.This problem could be solved by taking following conditions into consideration.

The probability of 100th person sitting on 100th seat can be calculated as below:

The probability of 1st person not sitting on 100th Seat

* probability of 2nd person not sitting on 100th Seat (provided 1 person has already occupied some seat)

*probability of 3rd person not sitting on 100th Seat (provided 1st and 2nd persons have already occupied some seat) and so on.

This will be:

99/100 * 98/99 * 97/98 * ….. 2/3 * 1/2 = 1/100

Regards

Vineet

0February 1, 2006 at 1:58 pm #133204Send me more information about how to conduct interactive sessions like quiz, treasurehunt to spread 6 sigma information in a firm

0March 2, 2006 at 10:50 am #134511

AMIT KUMARParticipant@AMIT-KUMAR**Include @AMIT-KUMAR in your post and this person will**

be notified via email.It is not 0.01. What if the second person sits on seat1? Then everbody else will go to there respective seats.So in that case the problem is pretty simple. the whole problem depends on the seat occupied by 2nd person.if he goes to seat 3 then it depends on third person and like that…..what I mean to say that is if by any how seat 1 is occupied by any of persons then the 100 th person will go to his seat.In other words it is the probability of occupancy of seat 1.

Thanks0March 14, 2006 at 10:36 am #134996

WaskitaParticipant@ferry-waskita**Include @ferry-waskita in your post and this person will**

be notified via email.Easy question, tough answer ;p

The way to answer this is to work-out probability tree diagram.

This way, it would derive you to a conclusion that the probability of passanger#100 sits in his proper sit is :

= (1/100) + (99 x 99 x (1/factorial 100))

= 0.01 + 0.0000000000000…..

= approx 0.010May 30, 2006 at 12:18 pm #138359

ChockalingamParticipant@Chockalingam**Include @Chockalingam in your post and this person will**

be notified via email.The probability that 100th to sit in 100th place is the probability of the rest 99 people ignoring the one seat (100th one)…

Which equals

(1-1/100)*(1-1/99)…. = 99!/100! = 1/100 = 0.010July 20, 2006 at 3:33 pm #140714I see a 50% probability

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