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Old 07-25-2005, 07:38 PM
BruceZ BruceZ is offline
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Default Another theatre line problem

10 people stand in a single file line in front of a ticket window. The 10 people have 10 different heights. The ticket clerk can see a person if and only if no one taller stands in front of that person. How many people can the ticket clerk see on average?
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Old 07-25-2005, 08:00 PM
SheetWise SheetWise is offline
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Default Re: Another theatre line problem

<font color="white">4 (I get 4.5)</font>
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Old 07-25-2005, 08:40 PM
zerosum zerosum is offline
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Default Re: Another theatre line problem

[ QUOTE ]
10 people stand in a single file line in front of a ticket window. The 10 people have 10 different heights. The ticket clerk can see a person if and only if no one taller stands in front of that person. How many people can the ticket clerk see on average?

[/ QUOTE ]

Do patrons wearing hats count? [img]/images/graemlins/grin.gif[/img]
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  #4  
Old 07-25-2005, 08:45 PM
BruceZ BruceZ is offline
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Default Re: Another theatre line problem

[ QUOTE ]
The ticket clerk can see a person if and only if no one taller stands in front of that person.

[/ QUOTE ]
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Old 07-25-2005, 08:49 PM
irchans irchans is offline
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Default Re: Another theatre line problem

Cute. My answer is below in white

<font color="white"> 2.93 on average </font>
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  #6  
Old 07-25-2005, 08:55 PM
kyro kyro is offline
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Default Re: Another theatre line problem

<font color="white"> I haven't done the math yet, but my instinct tells me it is lower than this</font>
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  #7  
Old 07-25-2005, 08:58 PM
bobman0330 bobman0330 is offline
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Default Re: Another theatre line problem

<font color="white"> The answer seems like it should be the 10th partial sum of the harmonic series.
The ticket taker can always see the first in line. There's a probability of .5 that the second person will be taller than the first. There's a probability of 1/3 that the third person will be taller than either of the first two, etc., etc.
1+1/2 + 1/3 + 1/4 + ... +1/10 = 2.929 </font>
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Old 07-25-2005, 09:47 PM
PairTheBoard PairTheBoard is offline
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Default Re: Another theatre line problem

[ QUOTE ]
10 people stand in a single file line in front of a ticket window. The 10 people have 10 different heights. The ticket clerk can see a person if and only if no one taller stands in front of that person. How many people can the ticket clerk see on average?

[/ QUOTE ]

Begin Aborted Attempt
<font color="white">
There are 10! ways they can line up.

First arrange them in order 1,2...,9,10 to see all 10. One way to do this.

Number of ways to see exactly 9:
Choose any of the first 9 and move him back one or more places. So, 9+8+7+6+5+4+3+2+1 = 9(9+1)/2 = 45

Number of ways to see exactly 8:
Choose any two of the first 9 and move them back one or more places.

Yikes. There's got to be an easier way.
</font>
End Aborted Attempt

PairTheBoard
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  #9  
Old 07-25-2005, 09:54 PM
KenProspero KenProspero is offline
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Default Re: Another theatre line problem

LOL, I started down the same way, and came to the same conclusion
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  #10  
Old 07-26-2005, 12:11 AM
PairTheBoard PairTheBoard is offline
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Default Re: Another theatre line problem

I think I've got it.

PairTheBoard
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