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?

Re: Another theatre line problem
 

<font color="white">4 (I get 4.5)</font>

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]

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 ]

Re: Another theatre line problem
 

Cute. My answer is below in white

<font color="white"> 2.93 on average </font>

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>

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>

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

Re: Another theatre line problem
 

LOL, I started down the same way, and came to the same conclusion

Re: Another theatre line problem
 

I think I've got it.

PairTheBoard