Six take a test ....
Six student are sitting in a row of seating with an aisle on each end. The professor has instructed that students may leave as soon as they finish the examination.
If they finish in random order ... what is the probability that a student will have to pass by one or more students when exiting the row of seating?
If they finish in random order ... what is the probability that a student will have to pass by one or more students when exiting the row of seating?
posted by Zaux at
12:29 PM
61 Comments:
When the first student finishes, probability is 4/6.
Second student: 3/5
Third student: 2/4
Fourth student: 1/3
Fifth and sixth students: 0
Each situation occurs 1/6 of the time.
So the required probability is 1/6 * (4/6 + 3/5 + 2/4 + 1/3) = 7/20.
Hi Wiz ....
does not agree with wizards who set the problem ...
Depends a lot on the interpretation of a row of seating with an aisle on each end. I would interpret this as a single "row" oriented as such:
1 2 3 4 5 6
then students 1, 2, and 3 would exit to the "left", while 4, 5, and 6 would exit to the right. (Similarly, if the "row" were vertical they would exit to the front or back.)
Anyway, I'd guess 1/36?
-h
Six take a test ....
Assume students must take route with no students if available.
Staring on the left hand side.
Analyze chance of leaving being blocked by student.
Analyze 1st chair
-No matter what can exit by left aisle passing no students, 0%
Average=0%
Analyze 2nd chair:
1st: 100% leaving by left or right=100% combined
2nd: 1-1/5=80% chance of leaving by left. 100% chance leaving by right=80% combined
3rd:1-2/5 = 60% chance of leaving by left. 100% chance of leaving by right=60% combined
4th:1-3/5= 40% chance of leaving by left. 100% chance by leaving by right.=40% combined
5th:1-4/5=20% chance leaving by left. 1-4/5*3/4*2/3*1/2=80% leaving by right=16% combined
6th: 0% chance leaving by left or right.=0% combined
Average=1/6*(100+80+60+40+16+0)=49.33%
Analyze 3rd chair:
1st: 100% leaving by left . 100% chance leaving by right=100% combined
2nd: 100% chance of leaving by left. 100% chance leaving by right=100% combined
3rd:1-2/5*1/4 = 90% chance leaving by left. 100% chance of leaving by right=90% combined
4th:1-3/5*2/4= 70% chance leaving by left. 1-3/5*2/4*1/3=90% chance by leaving by right.=63% combined
5th:1-4/5*3/4=40% chance leaving by left. 1-4/5*3/4*2/3=60% leaving by right=24% combined
6th: 0% chance leaving by left or right.=0% combined
Average=1/6*(100+100+90+63+24+0)=62.8333%
4th chair is a mirror image of 3rd chair
5th chair is a mirror image of 2nd chair
6th chair is a mirror image of 1st chair
Average probability=1/3*(0%+49.33%+62.8333%)=37.388%
Answer:
37.388%
Cam
There is a flaw in my solution...
namely if the second chair was 5th he would have to be able to exit free by at least one route as only #6 can block him.....
Back to drawing board.....
Cam
Six take a test ....
Corrected solution !
Assume students must take route with no students if available.
Staring on the left hand side.
Analyze chance of leaving being blocked by student.
Analyze 1st chair
-No matter what can exit by left aisle passing no students, 0%
Average=0%
Analyze 2nd chair:
1st: 100% leaving by left or right=100% combined
2nd: 1-1/5=80% chance of leaving by left. 100% chance leaving by right=80% combined
3rd:1-2/5 = 60% chance of leaving by left. 100% chance of leaving by right=60% combined
4th:1-3/5= 40% chance of leaving by left. 100% chance by leaving by right.=40% combined
5th: only 6 can block so, 0% combined
6th: 0% chance leaving by left or right.=0% combined
Average=1/6*(100+80+60+40+0+0)=46.667%
Analyze 3rd chair:
1st: 100% leaving by left . 100% chance leaving by right=100% combined
2nd: 100% chance of leaving by left. 100% chance leaving by right=100% combined
3rd:1-2/5*1/4 = 90% chance leaving by left. 100% chance leaving by right=90% combined
4th: only 5 and 6 can block if one is on the left and the other on the right.
Let's count the ways....
6th in chair 1, 5th in 4=(1*3)*(1*2*1)=6
6th in chair 1, 5th in 5=(1*3)*(2*1*1)=6
6th in chair 1, 5th in 6=(1*3)*(2*1*1)=6
6th in chair 2, 5th in 4=(3*1)*(1*2*1)=6
6th in chair 2, 5th in 5=(3*1)*(2*1*1)=6
6th in chair 2, 5th in 6=(3*1)*(2*1*1)=6
Double for swapping 5 and 6=2*6*6=72
Total ways to arrange=5*4*3*2*1=120
72/120=60%
Thus 4th is 60% combined
5th: only 6 can block so, 0% combined
6th: 0% chance leaving by left or right.=0% combined
Average=1/6*(100+100+90+60+0+0)=58.33333%
4th chair is a mirror image of 3rd chair
5th chair is a mirror image of 2nd chair
6th chair is a mirror image of 1st chair
Average probability=1/3*(0%+46.667%+58.333%)=35%
Answer:
35%
Wait a minute......
35% is the same as what Wiz got !
I ran a simulation to confirm numbers for each chair and position and they confirm my results.
Wondering if the solution assumes something different from the students must leave by the path where no students are.Perhaps it assumes chairs 1-3 must go left and 4-6 must go right, which would change the answer significantly.
Cam
I've only partially read what's been posted. If the question was
intended to mean to find the probability of at least one student
crossing another, then that's 1-P(no student crossing another)
P not crossing: 1st 2/6, 2nd 2/5, 3rd 2/4, 4th 2/3, 5th & 6th 1.
So desired probability = 1 - (2/6)(2/5)(2/4)(2/3) ≈ 95.55%
I'll post soon with case of exactly one student crossing another.
good job Chris....
you got it ...
Case if exactly 1 student crosses another: then let p(n) be the
probability that the next student to leave crosses another,
when there are n students. Let Q(n) be the cumulative
probability that no student crosses another when there are n
students. Let P(n) be the cumulative probability that exactly 1
student crosses another when there are n students; we want P(6).
In words, the probability of exactly 1 student crossing another
is the probability of the next student to cross another times
the probabilty that no more of the remaining students do so,
plus the probability the the next student doesn't cross another
but exactly one of the remaining students does. This yields
the recursion:
P(n) = p(n)*Q(n-1) + (1-p(n))*P(n-1)
p(n) = (n-2)/n = 1-2/n and Q(n) = (1-p(n))(1-p(n-1))...(1-p(3))
=> Q(n) = (2/n)(2/(n-1))...(2/3) = 2^(n-1)/n!
=> P(n) = ((n-2)/n)*2^(n-1)/n! + (2/n)P(n-1) (n > 2)
P(3) = 1/3 (easily seen by straightforward inspection).
Plugging that into the recursion =>
P(4) = 1/3 ≈ 33.33%, P[5] = 14/75 ≈ 18.67%,
P(6) = 52/675 ≈ 7.70%
P(7) ≈ 2.56%
P(8) ≈ 0.72%
I've been trying to figure out how Cam's and my answer of 35% could be so far away from Chris's "correct" answer of 95.55%.
The way I see it, Cam and I treated each student's departure as separate events, each occurring with 1/6 probability, so we then in effect took an AVERAGE of all students' probabilities of having to pass another student. That is how we catered for students finishing "in random order" as the question stated. (I hope I'm doing justice to Cam's approach).
Chris, on the other hand, worked out the overall probability that ANY student would pass another student at ANY stage over the time period in question. Thus, "random order" of finishing doesn't come into it.
It's amazing how different interpretations of the question can give such widely (I was going to say wildly) different results.
Wiz. I don't know what you meant by, "each situation occurs
1/6 of the time". That's makes no sense.
Cam. LOL, that's got to be the worst analysis that I've ever
seen from you. What's all this probability leaving by the left
or the right? You should have taken more notice of what Wiz did.
The only ambiguity I see is is it exactly one or is it at least one student that crosses another.
That a student by an aisle doesn't cross another seems pretty obvious as an assumption. Otherwise, I'd go for it's 50-50 as to which way they leave. But I think that that is already going too far - you wouldn't do that in real life.
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Hi Wiz. I've just seen your 10:52 PM post. As I said in my last
post, I don't understand your 1/6 of the time statement at all.
And what Cam has done has completley baffled me - he got himself
stuck in a (wrong) rut.
When you have n students, then the probability of any one of them
leaving is 1/n. If it's an outside student, s/he doesn't have to
cross anyone, otherwise, s/he has got to cross another student.
The probability of it being an outside student is 2/n (as they are
independent events), and the probability of it being an inside
student is (n-2)/n. I know you understand that.
The probabilty should have be calculated as:
(4/6) + (1-(4/6))((3/5) + (1-(3/5))((2/4)+(1-(2/4))(1/3))))
Getting rid of unnecessary ()s for readability =>
4/6 + (1 - 4/6)(3/5 + (1 - 3/5)(2/4 + (1 - 2/4) 1/3))
= 2/3 + 1/3 (3/5 + 2/5(1/2 + 1/2 1/3)) ≈ 99.95%
Sorry that's a bit hard to read. Compare with probability of throwing a head in 3 throws:
= 0.5 + (1-0.5)(0.5 + (1-0.5)0.5) = 0.5+0.5(0.5+0.5*0.5)
= 0.5 + 0.5*0.75 = 0.875 = 7/8.
You're more likely to have done that as 0.5 + 0.5² + 0.5³ = 7/8.
99.95% above is a typo. It should have been 95.55%
Hi Chris,
My aside "each occurring with 1/6 probability" in my earlier post was badly put. What I meant to say was that we have 6 separate events, i.e. each student's departure, each with its own probability of passing another student, and that we take the average of these probabilities.
Since the judges have handed down their verdict in your favour I have to concede that my approach is probably invalid. I'm still not quite sure why.
Wiz.I suspect that you plucked 1/6 out of the air to make sure
that you didn't get a probability > 1.
Let me try another way. Let p(n) be the probability that the next
student to leave crosses another when there are n students
remaining, and let P(n) = probability that at least one student
has crossed another when all n have left. Note that (1-p(n)) is
the probability that a student doesn't cross another when there
are n student's left. Then P(n) = p(n) + (1-p(n))*P(n-1), and is
a recursion. Hence those deep ()s in my previous post, I had recursed in situ. P(2) = 0, P(3) = 1/3 by direct thought. or
P(3) = p(3) + (1-p(3))*P(2) = 1/3 + (1- 1/3) * 0 = 1/3.
Then P(4) = p(4) + (1-p(4))*P(3) = 2/4 + (1 - 2/4)*1/3 = 2/3
P(5) = p(5) + (1-p(5))*P(4) = 3/5 + (1 - 3/5)*2/3 = 13/15
P(6) = p(6)+(1-p(6))*P(5) = 4/6 +(1- 4/6)*13/15 = 43/45 ≈ 95.56%
P(7) = 311/315, P(8) = 314/315
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I made a slip in my 9:50 PM post. It should have read:
P(n) = ((n-2)*2^(n-2)/n! + (2/n)P(n-1), for n ≥ 2
P(3) = 1/3 ≈ 33.33% (easily seen by straightforward inspection).
Plugging that into the recursion =>
P(4) = (4-2)*2^(4-2)/4! + 2/4 * 1/3 = 1/2 = 50%
P[5] = (5-2)*2^(5-2) + 2/5 * 1/2 = 2/5 = 40%
P(6) = (6-2)*2(6-2) + 2/6 * 2/5 = 2/9 ≈ 22.22%
For the question of:
If they finish in random order ... what is the probability that a student will have to pass by one or more students when exiting the row of seating?
If one performed n trials
One could arrange the chairs by columns and the order that the student left by rows
Each position would occur 1/36 of the time.
Then one could record the % of times that position was blocked.
The expected probability of a student being blocked would be:
the sum of the #of times a student was blocked / (the total number of trials*6 )
OR
the probability for each chair and order * 1/36
OR
1/6*(1/6*the sum of probabilities for each row)
OR
1/6*(1/6*the sum of probabilities for each column)
Wiz formulated his answer by row, My answer was formulated by column. The answers confirmed each other with the answer of 35%.
Given a student, the chance that they will be blocked is 35%.
Chris answered a different question:
If they finish in random order ... what is the probability that AT LEAST ONE student will have to pass by one or more students when exiting the row of seating?
For this question the answer is ~=95.55%
Apparently, this is the intended question and answer from the source.
That being said.....
I'm somewhat disappointed that the following line of reasoning was made:
I don't understand your method = The method and it's author is subject to being mocked
I would suggest the following alternative approaches:
I don't understand your method= I will make further inquiry to see if the method is sound
I don't understand your method=I will specifically describe why the method must yield an invalid result
Just my thoughts...
Cam
Thanks for coming to my defence, Cam, but I never thought that Chris was being unfair in his criticism. As I said earlier I had phrased my argument badly. I did need to explain myself better.
I thought your summation of the 35% result was excellent and I now feel better about my effort.
That said, Chris's answer is absolutely correct on the intended interpretation of the question.
Peace be to all mankind . . .
Cam, I've just realised that you might have been defending yourself, not me.
Oh well, peace anyway . . .
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Cam, I'm stunned by much of your Feb 15, 1:24 PM comment.
When I say that I don't understand something, that's precisely
what I mean.
I believe I have also answered the question, "what is the
probability that at least one student is blocked?" What is it
that you (and Wiz) have calculated the probability of?
The only reason that I even attempted the problem was to try to
understand what you and Wiz were doing. I haven't succeeded.
The only ambiguity that I can see is, is it exactly one or
at least one, student that gets blocked? At the moment I
believe that the exactly one case => 22.22% probability.
Wiz. Thank you for appreciating that I was simply making fair
comment. I have far too much respect for you and Cam to
intentionally be scornful (especially as I'd be aware that my view
might be wrong). I might poke a jibe at you if I did it with
obvious humorous intent. I can be contacted using
alan.neuman@yahoo.com (that's a disposable email address and
that isn't my real name - my first name really is Christopher).
Your logic failed when you took the average of the probabilities.
It was that single fact that caused me to examine the problem in
detail - I knew that was wrong.
Cam. I apologise for the way that I wrote off your solutions.
Part of my problem is that I couldn't follow what you were doing.
I think I now do. But, I have the same problem that I have with
Wiz's logic, and that is the averaging of the probabilities.
My understanding of the problem is: there are 6 students sitting
in a row (name them 1,2,3,4,5,6). If 1 or 6 finish first, when
they leave, they don't have to pass another student. If 2,3,4 or 5
finish first, then they have to get pass another student(s).
The probability that any one of them finishes first is 1/6. The
probability that 1 or 6 finshed first is 2/6 = 1/3 and the
probability that 2,3,4 or 5 finishes first is 4/6 = 2/3. If the
problem stopped there, then the required probability would be 2/3.
This is already much higher than the 35% figure.
We now have the original problem, but with only 5 students. We
only need to consider the 5 student case if student 1 or 6
finished first (we can't simply add the probabilities together as
they are not mutually exclusive).
Explanation: If a and b are random events (a student finishing),
then P(a OR b) = P(a) + P(b) - P(a AND b).
For independent events, P(a AND b) = P(a)*P(b) =>
P(a OR b) = P(a) + P(b) - P(a)*P(b) = P(a) + (1-P(a))*P(b)
For a,b,c get P(a OR b OR c) = P(a OR (b OR c))
= P(a)+(1-P(a))*P(b OR c) = P(a) + (1-P(a)*[P(b) + (1-P(b))*P(c)]
Continuing, but substituting the appropriate probabilities for the
question => P(at least one blocked) =
(1)*[4/6] + (2/6)*[3/5] + (4/6)(2/5)*[2/4] + (2/6)(2/5)(2/4)*(1/3)
= 0.9555... I've not reduced 4/6 to 2/3 for clarity
I've used [] to indicate the individual (n-2)/n part.
The original probabilities have factors:
1, 2/6, (2/6)(2/5), (2/6)(2/5)(2/4), 0, 0
Cam. When I posted my LOL (Feb 14, 10:56PM), I really thought that you would see what was wrong with your solution. I hadn't expected to see you defend it. I was so confident that you'd spot the flaw, that I hadn't considered the possible ramifications of being wrong. I had expected something more like "Meh, I need more coffee" - but, again, I forgot to put a ;) at the end to make that clear. I apologise for that, and also for taking so long to getting round to saying so.
In my continued defence, maybe I'm not smart enough to give a more helpful approach. I've had to do a lot of thinking to recall all that stuff I've posted. I intuitively knew that I was doing the right math, but had to think hard about how to prove it. Then I rememberd Venn diagrams and the rest of the combined probability stuff came out.
But I remain bamboozled by your acknowledgement that 95.56% is right, when you also believe that 35% is also correct. I'm now simply asking you to help me to understand your interpretation of the problem. I mean, can you write a statement that is as easy to understand as "at least one student crosses another"?
I've just re-read my Feb 15, 1:07 AM comment. I can see why that one wouldn't have gone down well. I apologise for that too.
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Hi Wiz. I found the following from you:
"Chris, on the other hand, worked out the overall probability that ANY student would pass another student at ANY stage over the time period in question. Thus, "random order" of finishing doesn't come into it."
My solution totally depended on the the students finishing independently. It wasn't irrelevent, it was essential to my argument.
You also seem to be interpreting the question differently to me. I can't see another interpretation (except exactly one student is blocked).
As I was writing this, I suddenly realised how to demonstrate why your 1/6 average process is wrong: toss a coin 3 times, what's the probability of getting 3 heads? The argument that you and Cam seem to be using on is that the overall probability would be the average probability = (1/3)(1/2 + 1/2 + 1/2) = 1/2, and not 1/8. I'm sorry if I'm misinterpreting what you're doing, but I don't see how else to read it.
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I have found another interpretation (but it's stretching things) - what is the probability that the first student to finish crosses another? That's 2/3.
And another, what is the probability that a student would have to cross another, if the blocked ones, couldn't leave? That doesn't seem to fit, because the question makes it plain that a blocked student does exit. I think that situation is quite hard to analyse. I won't bother analysing it as the first student alone contributes 2/3 which is much higher than the 35%.
I have also realised when it is valid to average probabilities. That's when your measring them. You might have two groups making measurements and calculating a probability. Let group 1 calculate a probability p1 over t1 trials, and group 2 calculate p2 of t2 trials. The average probability is (p1*t1 + p2*t2)/(t1+t2), and is precisely what you'd get if you'd amalgamated the two groups data before calculating the overall probability.
Eureka! I know what the 35% is.
The average number of students leaving by passing another student is:
4/6 + 3/5 + 2/4 + 1/3 = 2.1. So the fraction of them leaving that way is 2.1/6 = 35%. This is not a probability. It is the average fraction of students passing another.
This depends on how you define probability.
One definition could be a prediction on how often a defined outcome occurs in repeated trials.
So, if we conducted repeated trials of counting students leaving from a row of 6 I am confident that much less than 95% of them would need to pass another student.
Cam and I think 35% in our scenario.
However, just to confuse matters, I now have another scenario.
If we count the number of cases where a student has to pass another versus the number where they don't, then we get 4 against 2 at the start, then 3/2, 2/2, 1/2, 0/2 and 0/1 as the students leave one by one.
So, adding these up, we get 10 cases where a student has to pass another and 11 where they don't.
That gives a prediction of 10/21 or 47.62%.
So, what do you think of that?
I go back to what I said earlier, that if probability is the prediction of outcomes, and we conducted repeated trials to verify our predictions, then this is what I think we would get.
My head is starting to spin, so I'll leave it there.
Where are you, Cam?
A trial in this problem, is 6 students finishing an exam, not one student leaving a row. You seem to be treating 1 student leaving a row as a trial.
If you toss a coin 3 times, the probabilty of getting at least one head is 7/8 (87.5%). The fraction of heads (per trial of 3 tosses) is (1/3)(1/2 + 1/2 + 1/2) = 50%. To interpret that as a probability, you must consider it as 3 trials of one toss, not 1 trial of 3 tosses.
Loking at you last comment ("So, if conducted...") - the probability of (at least 1) student crossing another in a (starting) row of 6 is 95.56%. 35% of them (on average) will cross snother. On average 2.1 students will cross another, and 3.9 students won't. The ratio of crossers to non-crossers is 2.1/3.9 = 53.85%.
For now, I'll pass on your 10 student (max) case as my dad has just arrived and wants to talk to me. But it corresponds to a fraction, not a probability.
Wiz. Your 10 student case. I was thought that you were going to deduce the number of students that were crossed by the exiting students. In the worst case, that would be 10 students (for first row if #2 finished first then he could pass by #3,#4,#5 and #6. etc. => 4+3+2+1+0 = 10. I don't know what your 11 not passing corresponds to. But you weren't meaning that. You are saying that even though there are only 6 students, 10 of them pass another (on average). After the first 6 have left (leaving 0), where did the other 4 come from? I'm sure that I haven't followed you argument properly, not that it is wrong.
There are 6! = 720, equally likely, different ways that the students could have left the exam. In only 32 (=2*2*2*2*2*1) of these cases does no student cross another. That leave 688 cases where a student does cross another. 688/720 = 95.56%
Please don't think I'm having a pop at you. Obviously I do want you to see the light, but I really am going on about it for my benefit. If it weren't for this site, I wouldn't have done anything with probability for at least 20 years. Thanks to ToM, I've now done quite a lot, and am thrilled at how much I've (re)learnt.
As there are a lot of aspects to this problem (we haven't started to consider the probability distribution function), I might knock up some code to simulate the many possible questions that could be asked.
PART 1 of 2
Apologies accepted Chris.
Anyhow, hopefully this post will put the whole thing to rest.
Classical Definition: probability of an event to occur is defined as number of cases favorable for the event, over the number of total outcomes possible in an equiprobable sample space.
Example 1:
I have 100 lotto tickets.
75 of these tickets are winners.
25 of these tickets are losers.
I put all of the tickets in a bag 0. I then select a lotto ticket.
The probability of the lotto ticket being a winner is :
75 favorable events
75+25=100 total possible outcomes
P=75/100 =75%
Example 2a:
I have 100 lotto tickets.
75 of these tickets are winners.
25 of these tickets are losers.
I put all of the winners in bag 1. I then select a lotto ticket from bag 1.
The probability of the lotto ticket being a winner is :
75 favorable events
75 total possible outcomes
P=75/75 =100%
Example 2b:
I put all the remaining tickets (losers) in bag 2.
I then select a lotto ticket from bag 2.
The probability of the lotto ticket being a winner is :
0 favorable events
25 total possible outcomes
P=0/25 =0%
Example 2c:
I put bag 1 and bag 2 in a bigger bag, called bag 3. The material from bags 1 and 2 magically dissolves.
I then select a lotto ticket from bag 3.
The probability of the lotto ticket being a winner is:
(Pbag1*nbag1+Pbag2*nbag2)/nbag3=(100%*75+0%*25)/(75+25)=75%
OR
same as the original bag 0. 75 winners. 100 total possible outcomes. P=75/100=75%
Example 3:
I take the tickets from bag 0 and seperate them into bags 1-5.
bag1 : 15 winners 5 losers. Pbag1=15/20=75%
bag2: 5 winners 15 losers Pbag2=5/20=25%
bag 3: 15 winners 5 losers Pbag3=15/20=75%
bag 4: 20 winners 0 losers Pbag4=20/20=100%
bag 5: 20 winners 0 loser Pbag5=20/20=100%
I then place all these bags into a bigger bag 6. The material from bags 1-5 magically dissolves. I then select a lotto ticket from bag 6.
The probability of the lotto ticket being a winner is:
Sum(Pbagn*nbagn)/Sum(nbagn)=(75%*20+25%*20+75%*20+100%*20+100%*20)/(20+20+20+20+20)
Sum(Pbagn*nbagn)/Sum(nbagn)=(15+5+15+20+20)/(100)=75/100=75%
Continued
Cam
PART 2 of 2
Example 4a:
I have a bag 0 with 360 tickets. 35% of them are winners. i.e. 126 winners
I take the tickets from bag 0 and seperate them into bags 1-6.
I put 1/6 of the 360 tickets in each bag. i.e each bag has 1/6*360=60 tickets
bag1: none of the tickets are winners. Pbag6=0%
bag2 : 4/6 of the tickets are winners. i.e40 winners Pbag2=4/6
bag3: 3/5 of the tickets are winners. i.e.36 winners Pbag3=3/5
bag 4: 1/2 of the tickets are winners.i.e. 30 winners Pbag4=1/2
bag 5: 1/3 of the tickets are winners.i.e. 20 winners Pbag 5=1/3
bag 6: none of the tickets are winners. Pbag6=0%
I then place all these bags into a bigger bag 7. The material from bags 1-6 magically dissolves. I then select a lotto ticket from bag 7.
The probability of the lotto ticket being a winner is:
Sum(Pbagn*nbagn)/Sum(nbagn)=(0*60+4/6*60+3/5*60+1/2*60+1/3*60+0*60)/(60+60+60+60+60+60)
Sum(Pbagn*nbagn)/Sum(nbagn)=(0+40+36+30+20+0)/(360)=126/360=35%
We may have noticed that if nbagn is the same for all the smaller bags then we can simplify to.
Sum(Pbagn)/n=(0+4/6+3/5+1/2+1/3+0)/6=2.1/6=35%
Example 4b:
I have a bag 0 with 6n tickets. 35% of them are winners. i.e. 35%*n winners
I take the tickets from bag 0 and seperate them into bags 1-6.
I put 1/6 of the 6n tickets in each bag. i.e each bag has n tickets
bag1: none of the tickets are winners. Pbag6=0%
bag2 : 4/6 of the tickets are winners. i.e 4/6*n winners Pbag2=4/6
bag3: 3/5 of the tickets are winners. i.e. 3/5*n winners Pbag3=3/5
bag 4: 1/2 of the tickets are winners.i.e. 1/2*n winners Pbag4=1/2
bag 5: 1/3 of the tickets are winners.i.e. 1/3*n winners Pbag 5=1/3
bag 6: none of the tickets are winners. Pbag6=0%
I then place all these bags into a bigger bag 7. The material from bags 1-6 magically dissolves. I then select a lotto ticket from bag 7.
The probability of the lotto ticket being a winner is:
Sum(Pbagn*nbagn)/Sum(nbagn)=(0*n+4/6*n+3/5*n+1/2*n+1/3*n+0*n)/(n+n+n+n+n+n)
Sum(Pbagn*nbagn)/Sum(nbagn)=(2.1*n)/(6*n)=2.1/6=35%
We may have noticed that if nbagn is the same for all the smaller bags then we can simplify to.
Sum(Pbagn)/n=(0+4/6+3/5+1/2+1/3+0)/6=2.1/6=35%
Now let's replace lottery ticket with student, and winning with blocked.
Example5:
I have a row with 6n students. 35% of them are blocked. i.e. 35%*6n students
I take the students from the row and seperate them into the random order that they leave.
1/6 of the 6n students will belong to each i.e each has n students
Leave 6th: none of the students are blocked. Pblocked6=0%
Leave 1st : 4/6 of the students are blocked. i.e 4/6*n students. Pblocked1=4/6
Leave2nd: 3/5 of the students are blocked. i.e. 3/5*n students Pblocked2=3/5
Leave3rd: 1/2 of the students are blocked i.e. 1/2*n students Pblocked3=1/2
Leave4th: 1/3 of the students are blocked.i.e. 1/3*n students Pblocked4 =1/3
Leave5th: none of the students are blocked. Pblocked5=0%
I then place all these students into a test. I then select a student from the test.
The probability of the student being blocked is:
Sum(Pblockedn*nstudentsn)/Sum(nstudentsn)=(0*n+4/6*n+3/5*n+1/2*n+1/3*n+0*n)/(n+n+n+n+n+n)
Sum(Pblocked*nstudentsn)/Sum(nstudentsn)=(2.1*n)/(6*n)=2.1/6=35%
We may have noticed that if nstudentsn is the same for all the random orders of leaving then we can simplify to.
Sum(Pblocked)/6=(0+4/6+3/5+1/2+1/3+0)/6=2.1/6=35%
Note that these were either positive or negative events. They weren't affected by different values for each outcome. 35% is the probability that a randomly selected student will be blocked. It is not just an average. Although in this special case since each individual probability has the same number of events as the other probabilities, thus the average of the probabilities will yield this answer.
Cam
Hi Cam (and Wiz). I'll take (another) look at you Part 1 and 2 posts before I comment. I hate doing it with code, but it does have the advantage that it's easy to follow and check that there is no cheating etc.
Some code. If you have Excel, then try the code below. To insert it press F11, then Add a Module, and copy and paste the code. Turn the immediate window on. with the cursor in the code, press F5 to run it. I'm not used to using Excel (at all, yet alone for programming). I only know what to do because Ragknot told me.
Results are similar to:
Average number of students that crossed = 2.1001868998131 ( 35.003114996885 %)
% of exams in which at least one student crossed another = 95.5475044524955
% of exams in which exactly 0 student(s) crossed another = 4.45239554760445
% of exams in which exactly 1 student(s) crossed another = 22.2254777745222
% of exams in which exactly 2 student(s) crossed another = 38.8497611502388
% of exams in which exactly 3 student(s) crossed another = 27.7953722046278
% of exams in which exactly 4 student(s) crossed another = 6.67689332310668
% of exams in which exactly 5 student(s) crossed another = 0
% of exams in which exactly 6 student(s) crossed another = 0
-- The code
Sub RunTrials()
Dim nStudentsThatCrossed As Long
Dim nTrial As Long
Dim nStudentsRemaining As Long
Dim nStudent As Long
Dim nTrialsWithAtleastOneStudentCrossing As Long
Dim nTrialsWithNStudentsCrossing(0 To 6) As Long
Dim nAccumulatedNumberOfStudentsThatCrossed As Long
Randomize
For nTrial = 1 To 1000000
nStudentsThatCrossed = 0
For nStudentsRemaining = 6 To 1 Step -1
nStudent = Int(nStudentsRemaining * Rnd()) + 1
'check if is an aisle student
If nStudent <> 1 And nStudent <> nStudentsRemaining Then
nStudentsThatCrossed = 1 + nStudentsThatCrossed
End If
Next nStudentsRemaining
If nStudentsThatCrossed > 0 Then nTrialsWithAtleastOneStudentCrossing = 1 + nTrialsWithAtleastOneStudentCrossing
nTrialsWithNStudentsCrossing(nStudentsThatCrossed) = nTrialsWithNStudentsCrossing(nStudentsThatCrossed) + 1
nAccumulatedNumberOfStudentsThatCrossed = nAccumulatedNumberOfStudentsThatCrossed + nStudentsThatCrossed
Next nTrial
Debug.Print
Debug.Print "Average number of students that crossed ="; nAccumulatedNumberOfStudentsThatCrossed / nTrial; " ("; nAccumulatedNumberOfStudentsThatCrossed / nTrial * 100 / 6; "%)"
Debug.Print "% of exams in which at least one student crossed another ="; 100# * nTrialsWithAtleastOneStudentCrossing / nTrial
For nStudent = 0 To 6
Debug.Print "% of exams in which exactly"; nStudent; "student(s) crossed another ="; 100# * nTrialsWithNStudentsCrossing(nStudent) / nTrial
Next nStudent
End Sub
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Cam. In all my editing I accidentally deleted my thank you for accepting my apology.
I'm aware that you are very intelligent, and that if I contradict you, it is not with a feeling of smugness or to make you look foolish (I'm not that sort of person). But I'm sometimes insensitive to other people's feelings - it is rarely deliberate. If I'm abrupt, it's only because I'd fear that circumlocution could be seen as patronisation. The last thing I want to do is to fall out with you.
Much the same goes for Wiz too. I actually suggested the name (Wizard of Oz) to Wiz because I thought it very appropriate after seeing him practically trivialise some of the problems that I'd - I was very impressed (and still am).
I love this problem because it has forced me to find out/recall/sort out a lot of things that I had long forgotten. It's also caused me, for the first time, to write code to run under Excel.
My continued posting on the subject is not to beat you down; it is purely for the intellectual pleasure. Obviously I hope that you end up agreeing with my view ;)
Ooops. A typo. To insert the code you press ALT+F11.
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Cam. Oooops, I see that you have definded what you associate the probability with (in your last post). Your phrasing has hit me like a bolt of lightning. I'm gobsmacked and finally see what you and Wiz mean.
I've put to much effort into my last few posts to delete them. Besides, they'll serve as an object lesson in (my) stupidity. I've reread the question several times, and it is completely compatible with your interpretation. Despite having calculated the value a few times, I'd never thought to look at the problem from the point of view of a student, only of an onlooker.
I'm posting this now for safety - I hope you read it before commenting on my previous few comments. Think of it as an apology if you wish :)
As we all made a very reasonable interpretation of the problem, I declare Cam, Wiz and self as joint winners, LOL. I hope Zaux doesn't mind me taking a liberty with his problem.
I remain stunned at the enormous difference that being an onlooker or a participant makes to the interpretation of the problem, and that the question uses exactly the same words to [more than reasonably] mean two completely different things.
Wiz. Re your Feb 19 11:30 PM post.
I'd look at it like this. On average 2.1 students cross another per exam (from the point of view of an onlooker ;)) So on average, one student leaves by crossing another per 0.4762 (= 1/2.1) exams ;)
The ambiguity in the question makes it the most full-stength ToM that I can recall seeing. I'm going to email the question to my dad; I want to know how he interprets it.
We've had three different answers so far . . . it'll be interesting to see if your old man can come up with a fourth one.
Now I've had a day to get used to it, I realise that we've both modified the question.
I'd done: "... the probability that a[ny of the] student[s] will have to pass by one or more ...?"
Wiz and Cam have done: "... the probability that a [given] student will have to pass by one or more...?"
Yes, that's right.
Perhaps my latest effort (47.6%) could be phrased as "...the probability that the next student to leave will have to pass by one or more...".
Is your dad a professor of mathematical statistics? That's what we need.
Hi Wiz. I'll have a careful think about your 47.6%.
My dad is just a smart dude. About the first thing he did when he started a career, was to re-design the nosecone, the electrics power generator and a hysteresis motor for the Firestreak air-to-air missile.
Nice genes.
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Using the alternative interpretation of the problem (what is the prob that a particular student will cross..)
Here are two ways to do it.
Consider just the very first student to leave. On average 4/6 ths of a student will leave by passing another. Now just repeat for the remaining 5 students etc. You end up with the average number of student's to exit by crossing another is (4/6 + 3/5 + 2/4 + 1/3) = 2.1 students per exam. That's the same as 2.1 students out of every 6, which means that the probability of selecting a student that will cross another student is 2.1/6 = 35%.
---- other way
If I select a student first, then I get, the probability is:
SUM([P still in the exam]*{P of leaving}*(P of crossing)) =>
[1]{1/6}(4/6) +[1-1/6]{1/5}(3/5) +[1-1/6][1-1/5]{1/4}(2/4)
+[1-1/6][1-1/5][1-1/4]{1/3}(1/3) = 35%
NB I've used [] {} () to emphasize the roles of the different factors.
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Hi Chris, Sorry for the delay but I'm out in the desert helping out on a film shoot for the next month. So I'll largely be otherwise occupied till then.
I think we've probably taken this topic as far as we can but I'll keep an eye on it anyway.
Cheers, Wiz
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Hi Wiz. I'd noticed your absence. Topic definitely done to death, except I'm not going with your 47.6% argument ;)
Strangely enough, I recently did a problem Queue height, in which thinking directly as a probability (rather than as a fraction of a person) seems fine. But for this one my physical intuition just won't let me see it (your way); but I realise that's a limitation of mine. Obviously most of the stuff on this page is because I just didn't see the other interpretation of the problem. When I did see it, I was gob-smacked.
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another kick post
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