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  #1  
Old 08-22-2002, 10:08 PM
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Default 2 Dark Room Questions



I have two questions. I am not sure if the first question has a good answer. I know the answer to the second question, and it makes me think that I know the answer to the first question. Try to think about the first question a while before reading the second.


1. Tom is the 5th person to have entered a darkened room. He is asked to estimate the number of people in the room, but the only thing he knows is that he was the 5th person to have entered the (large) dark room and no one has exited. He guesses that there are 25 or fewer people in the room. What is the probability that he is correct? Does the answer change if you replace 5th and 25 with nth and 5*n respectively?


2. N people enter a room. As they enter, they are told their entry number. (i.e. the first person is told that he is number 1. The second person is told that he is number 2. ...) If every person in the room says, "There are five times my entry number or less people in the room.", what percentage of the people will be correct. (For simplicity, assume that N is a multiple of 10.)
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  #2  
Old 08-23-2002, 01:58 AM
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Default Re: 2 Dark Room Questions



Okay, Im going to have a stab at this, please remember I wasn't trained in mathspeak so this will be a) probably wrong and b) badly worded :P


To try to explain away the paradox Im going to change the problem a bit.

Say the maximum number of people in the room can be 1000 and any number from 1-1000 could be the number of people in the room.


Q2.The number is selected as random(random(1000))

these people dont know their numbers so the random outside the bracket is all they are working on, the odds that of the people selected, their number comes up in the top 4/5ths of available numbers is the 80% required.


Q1. The number is selected as 5(random(1000)

so we know our number is 5 and there are 1-1000 people in the room, we dont like this bet.

We cannot rely on the second random element to help our guess now, we know we are 5 from what we have to assume is 500.


What I am saying is that in my opinion the two things are not the same question , whether there is a way of using infinite series to work out what our is for an "infinite" room I wouldnt like to guess.


Something on a similar note. I believe that 30% of the time the number you are assigned will end in the digit "1" but cannot remember how to show this, anyone out there willing to put that to rest for me and solve this too?


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Old 08-23-2002, 03:40 AM
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Default ooops



Thats why i couldnt remember how, I meant there is a 30% chance that the FIRST digit is a "1".

Sorry for any brain meltdowns caused


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Old 08-23-2002, 11:02 AM
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Default FIRST digit is a \"1\"



> 30% chance that the FIRST digit is a "1".


I remember that the first digit of numbers from accounting tend to be distributed something like


distr(i) = Log( (i+1)/i ) / Log(10)


which gives


1, 30.1%

2, 17.6%

3, 12.5%

4, 9.7%

5, 7.9%

6, 6.7%

7, 5.8%

8, 5.1%

9, 4.6%


I don't remember who figured that out.
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  #5  
Old 08-23-2002, 11:04 AM
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Default Answer to Q2



lorinda,


You are correct. 80% of the people will be correct as long as N is a multiple of 5. I still don't know the correct answer to Q1, if there is a correct answer.



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  #6  
Old 08-24-2002, 04:54 AM
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Default Re: 2 Dark Room Questions



2. 80% of the people +1 will be correct.


1. This is what we would call an ill-posed problem, because we have no information as to how the number of people in the room is selected. We can work this problem if we assume that this number is chosen randomly to lie between 0 and some N, with all numbers equally likely (uniform distribution). Now, the equating of a uniform distribution with total lack of knowledge is a topic of philosophical debate, and has led philosophers to some rather strange conclusions since Plato. Nevertheless, here is the result of that assumpion in this case:


From the definition of conditional probability:


P(A|B) = P(AB)/P(B) = P(B|A)P(A)/P(B)


where P(A|B) means the probability of A given B, and P(AB) means the probability of A AND B.


Let A be 25 people or less

Let B be you are 5th


P(A) = 25/N

P(B) = 1/N

P(B|A) = 1/N(1/5 + 1/6 + 1/7 + ... + 1/N)


that is, the probability that there are any number of people n in the room is 1/N, and the probability that you are 5th for that n is 1/n.


P(A|B) = (25/N)sum[n=5 to N](1/n)


Now I ask:


3. If you had to guess how many people were in the room, what guess would have the highest probability of being correct.


4. What guess would on average come the closest.


5. What guess would on average minimize the mean squared error (average of the square of the error).


Make the uniform assumption for 3-5.


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  #7  
Old 08-24-2002, 05:25 AM
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Default Ignore above, read this



2. 80% of the people +1 will be correct.


1. This is what we would call an ill-posed problem, because we have no information as to how the number of people in the room is selected. We can work this problem if we assume that this number is chosen randomly to lie between 0 and some N, with all numbers equally likely (uniform distribution). Now, the equating of a uniform distribution with total lack of knowledge is a topic of philosophical debate, and has led philosophers to some rather strange conclusions since Plato. Nevertheless, here is the result of that assumpion in this case:


From the definition of conditional probability:


P(A|B) = P(AB)/P(B) = P(B|A)P(A)/P(B)


where P(A|B) means the probability of A given B, and P(AB) means the probability of A AND B.


Let A be 25 people or less

Let B be you are 5th


P(A) = 25/N

P(B) = 1/N(1/5 + 1/6 + 1/7 + ... + 1/N)


that is, the probability that there are any number of people n in the room is 1/N, and the probability that you are 5th for that n is 1/n.


P(B|A) = 1/25(1/5 + 1/6 + 1/7 + ... + 1/25)


P(A|B) = sum[n=5 to 25](1/n)/sum[n=5 to N](1/n)


Now I ask:


3. If you had to guess how many people were in the room, what guess would have the highest probability of being correct.


4. What guess would on average come the closest.


5. What guess would on average minimize the mean squared error (average of the square of the error).


Make the uniform assumption for 3-5.



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  #8  
Old 08-24-2002, 06:22 AM
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Default minor correction



The above derivation holds for a uniform distribution from 1 to N rather than 0 to N, though the result would be the same for 0 to N as well.
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  #9  
Old 08-30-2002, 12:13 AM
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Default Re: minor correction



The correct answer, I believe, to the 2nd question is:

[(n/5) - 1] / n = % of wrong people
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  #10  
Old 08-30-2002, 09:40 AM
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Default Benford\'s Law.



it was frank benford in 1938 - it's called benford's law in his honour


it applies to a fairly arbitrary piece of real data - the lengths of various rivers; the individual payments received in a month by a mortgage lender; the population of a collection of towns; or similar


the only restriction is that it should be neither systematic, nor deliberately random: no lists of telephone numbers, or of prime numbers, or the output from a computer's random number generator


whatever the source of the data focus on the first significant digit - whether the data value is 81 or 0.81 or 0.081 that digit is 8


the probability of that digit being 1 is 30.1%



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