Re: Another theatre line problem
 

Begin; I think I've got it Solution
<font color="white">
This is another nifty application of the Sum of Expected Values that Bruce has shown us before.

Let the 10 people be in order of height, A1,A2,...,A10 with A1 the shortest and A10 the tallest.

For k=1...10 define the random variables Ik by:

Ik = 1 if Ak can be seen.
Ik = 0 if Ak cannot be seen.

Let N be the number of people in line that can be seen. We want to know E[N]. By the definitions,

N=I1+I2+...+In

So, E[N]=E[I1]+...+E[I10]

But E[Ik]=P[Ik=1] which we can calculate for each k.

P[I1=1] = 1/10
P[I2=1] = 1/10 + (1/10)(1/9)
P[I3=1] = 1/10 + (1/10)(2/9) + (1/10)(C(2,2)/C(9,2))
P[I4=1] = 1/10 + (1/10)(3/9) + (1/10)(C(3,2)/C(9,2)) + (1/10)(C(3,3)/C(9,3))
P[I5=1] = 1/10 + (1/10)(4/9) + (1/10)(C(4,2)/C(9,2)) + (1/10)(C(4,3)/C(9,3)) + (1/10)(C(4,4)/C(9,4))
...
P[I10=1] = 1/10 + (1/10)(9/9) + (1/10)(C(9,2)/C(9,2)) + ... + (1/10)(C(9,8)/C(9,8)) + (1/10)(C(9,9)/C(9,9))

Summing these as expectations and rearranging terms we get
E[I1]+...+E[I10] =
(1/10)10 +
(1/10)[(1+2+3+...+9)/9] +
(1/10)[C(2,2)+C(3,2)+...+C(9,2)]/C(9,2) +
(1/10)[C(3,3)+C(4,3)+...+C(9,3)]/C(9,3) +
(1/10)[C(4,4)+...+C(9,4)]/C(9,4) +
... +
(1/10)[C(8,8)+C(9,8)]/C(9,8) +
(1/10)C(9,9)/C(9,9)

= (1/10)(10 + 5 + 3.4 + 2.5 + 2 + 1.7 + 1.4 + 1.2 + 1.0 + 1)

= (1/10)(29.2)

or about 2.9 people.

</font>
End; I think I've got it Solution

Seems kind of small. Oh well.


PairTheBoard

Re: Another theatre line problem
 

Wow. Bobman's solution is so much nicer.
White
<font color="white">
Let Ik = 1 if the kth person in line can be seen.
</font>
End White

PairTheBoard

Re: Another theatre line problem
 

I've no idea how to 'do it' exactly, so I tried a little simulation in Excel. Just thought it'd be interesting to compare my results with the numbers the mathematicians are coming up with.

Using a million simulations (actually, twenty simulations, each of fifty thousand queues) I got an average visible patrons of 2.928935.

Yeah, I know - small sample size.

Thanks to OP - interesting puzzle.

Re: Another theatre line problem
 

Ok, I think I went about this the hard way and had to look up a sequence to get the answer.

So I computed the answers for N = 1, 2, 3, and 4.

N = 1, I got 1 / 1.
N = 2, I got 3 / 2.
N = 3, I got 11 / 6.
N = 4, I got 50 / 24.

So I have the denominators as factorials, and the numerator was this sequence that I didn't know off the top of my head. So, I looked up the sequence 1, 3, 11, 50. And I found that this is the sequence of Stirling Numbers (of the 1st kind)... some important combinatorial sequence that I vaguely learning a little bit about years ago.

So, I looked up the 10th Stirling number (indexed from 1), which was 10628640. And, 10! = 3628800.

So, we have 10628640 / 3628800 = 2.929.

-RMJ

Re: Another theatre line problem
 

That is truly a Stirling solution. Kind of amazing too.

PairTheBoard

Re: Another theatre line problem
 

[ QUOTE ]
Ok, I think I went about this the hard way and had to look up a sequence to get the answer.

So I computed the answers for N = 1, 2, 3, and 4.

N = 1, I got 1 / 1.
N = 2, I got 3 / 2.
N = 3, I got 11 / 6.
N = 4, I got 50 / 24.

So I have the denominators as factorials, and the numerator was this sequence that I didn't know off the top of my head. So, I looked up the sequence 1, 3, 11, 50. And I found that this is the sequence of Stirling Numbers (of the 1st kind)... some important combinatorial sequence that I vaguely learning a little bit about years ago.

So, I looked up the 10th Stirling number (indexed from 1), which was 10628640. And, 10! = 3628800.

So, we have 10628640 / 3628800 = 2.929.

-RMJ

[/ QUOTE ]

I went the same way as you.... I had never heard of Stirling numbers. I went to N=5 for 274/120 and then found the relationship with the Harmonic series when I was trying to figure out the numerators. The numerator (Stirling Number) is just Harmonic Number(n) * n!