Another theatre line problem

[PairTheBoard]

Begin; I think I've got it Solution
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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.

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End; I think I've got it Solution

Seems kind of small. Oh well.


PairTheBoard