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Topic: 9 jars and 9 labels (Read 1141 times) |
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Kozo Morimoto
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9 jars and 9 labels
« on: Dec 4th, 2002, 4:17am » |
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I have 9 opaque jars/tins of paint each with a different colour paint.
I also have 9 labels with the colour written on it.
Without opening the jars/tins, I place the labels on the jars/tins.
What's the expected number of labels that I would have correctly placed?
My guess is 1 from looking at n=3 and n=4, but I don't know how I need to go about finding a solution to this problem.
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| « Last Edit: May 1st, 2004, 7:35pm by Icarus » |
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towr
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Re: help needed: 9 jars and 9 labels
« Reply #1 on: Dec 4th, 2002, 9:14am » |
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Simulation would be my easy answer.. just try it out by computer several thousand times.. (actually since it's just nine, you can try all possibilities once)
Probability theory would be a better way.. But I'd have to think about that first..
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| « Last Edit: Dec 4th, 2002, 9:46am by towr » |
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James Fingas
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Re: help needed: 9 jars and 9 labels
« Reply #3 on: Dec 4th, 2002, 12:34pm » |
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Kozo,
Maybe a proof by induction would work here. I'm thinking along the lines of:
1) Assuming label 9 ends up on jar 9 (p = 1/9), then our expectation is one more than the expectation for 8 jars and 8 labels.
2) Assuming label 1 ends up on jar 9 (p = 1/9), then we get the expectation for 8 jars and 8 labels, but we subtract the probability that the 1st label that was correct (which we get by symmetry of the labels 1 to 8).
3) The same logic applies to all labels from 2 to 8 ending up on jar 9.
I haven't worked through it, but I imagine that it will work nicely.
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| « Last Edit: Jan 22nd, 2003, 3:05pm by william wu » |
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TimMann
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Re: help needed: 9 jars and 9 labels
« Reply #4 on: Dec 4th, 2002, 7:27pm » |
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What's the probability that exactly one label ends up on the wrong jar? 
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Kozo Morimoto
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Re: help needed: 9 jars and 9 labels
« Reply #5 on: Dec 4th, 2002, 11:33pm » |
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What's the probability that exactly one label ends up on the wrong jar?
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That's a good one.
But working on the problem, 9! = 362,880
no. of 9 correct labels = 1
no. of 8 correct labels = 0
no. of 7 correct labels = COMBIN(9,7) = 36 ?
for each 7 correct labels, you can only have 1 combination of 2 wrong labels...
but how do I calc the next step with no. of 6 correct labels?
COMBIN(9,6) = 84 and FACT(3)=6, but some of the 6 wrongs include the situation where you may have 7 or 9 correct labels. So out of the 6, 1 would be when 9 labels are correct and 3 of then are when 7 labels are correct which only leaves us with 2?
So,
no. of 6 correct labels = 168 ?
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Nigel_Parsons
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Re: help needed: 9 jars and 9 labels
« Reply #6 on: Apr 11th, 2004, 8:45am » |
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To continue this thought process, and it gets complex.
For 6 correct there are indeed 168 ways to select 6 correct from 9, including the fact that the 3 'wrong' labels can be selected 2 ways. i.e. if labels 4-9 are correct, to have 6 correct & only 6 then labels 1 2 & 3 must all be wrong. The 6 arrangements of labels for 1 2 & 3 give:
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 & 3 2 1
of which only 2 3 1 & 3 1 2 are totally incorrect.
The next level (5 correct) gets tougher. There are 126 arrangements of 5 correct. i.e. 9!/(5! * 4!)
However, these can appear with any arrangement of 4 incorrect labels, and there are 9 ways to get 4 wrong labels.
Thus there are 9*126 or 1134 ways to get 5 correct labels
Back to the drawing board for the next level (4 correct) I have a feeling the complexity will increase
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Barukh
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Re: help needed: 9 jars and 9 labels
« Reply #7 on: Apr 11th, 2004, 10:32am » |
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Check the Umbrela makers' convention thread - it discusses the same problem.
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Grimbal
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Re: help needed: 9 jars and 9 labels
« Reply #8 on: May 1st, 2004, 5:17pm » |
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Thinking of just trying all the combinations.
Imagine all the jars in a circle. For each possible arrangement of the labels, you can get 8 other arrangements by rotating all the labels. Over these 9 arrangements, each label is right exactly 1 time out of 9. That is an average of 1 correct label per arrangement. The set of all arangments is the sum of all such sets of 9. So the average is 1 overall.
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| « Last Edit: May 1st, 2004, 5:19pm by Grimbal » |
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rmsgrey
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Re: 9 jars and 9 labels
« Reply #9 on: May 2nd, 2004, 6:22am » |
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Very nice grimbal. Makes all our messing around with conditional probabilities and case counting look messy and inefficient! 
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yadayada
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Here is one way to look at it.
Say there are n cans.
let X(i) be 1 if you label the ith jar correctly.
Let it be 0 othewise.
Now Expected value of X(i) is 1/n.
The number of correct labels is X = X(1) + X(2) + ...+ X(n)
By linearity of expectation,
Expected value of X = sum of Expected values of X(i)'s
= n * 1/n = 1
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