> A theater has 15 seats in the first row. Patrons enter
> in single file, taking each consecutive seat, as they
> arrive. In this particular case, there are 8 boys, 7 girls.
>
> If a boy and girl occupy adjacent seats, we have a candidate
> match. If boy sees girl on each side, that's two matches.
> Clearly, any possible arrangement will contain 1...14
> candidate matches.
>
> What's the average number of matches, for this row?

--
Rich

On Sat, 4 Sep 2021 16:07:23 -0700 (PDT), RichD
<r_delaney2001@yahoo.com> wrote:

>> A theater has 15 seats in the first row. Patrons enter
>> in single file, taking each consecutive seat, as they
>> arrive. In this particular case, there are 8 boys, 7 girls.
>>
>> If a boy and girl occupy adjacent seats, we have a candidate
>> match. If boy sees girl on each side, that's two matches.
>> Clearly, any possible arrangement will contain 1...14
>> candidate matches.
>>
>> What's the average number of matches, for this row?
>

Are we to assume that the eight boys are indistinguishable from one
another (ditto for the girls)? That is to say, the arrangement
BBBBBBBBGGGGGGG is single arrangement regardless of whether the last
boy is Tom, Dick, or Harry?

Since the problem was presented as an entry sequence, can we assume
that mirror images are distinct? That is, the sequence
BGB BBBBBBGGGGGG
is distinct from the sequence
GGGGGGBBBBBB BGB
?

Given the above, there are 6,435 distinct arrangements, determined by
simply counting the number of 15-bit integers that happen to have
exactly seven bits set to one. (The smallest integer is 127 and the
largest is 32,512.)

--
Remove del for email

On 9/4/2021 6:07 PM, RichD wrote:
> A theater has 15 seats in the first row. Patrons enter
> in single file, taking each consecutive seat, as they
> arrive. In this particular case, there are 8 boys, 7 girls.

For a single pair :
[BG + GB ]/ total: [BG + GB + BB + GG]
[(8*7) + 7*8]/ [8*7 + 7*8 + 8*7 + 7*6] = 112/210

Then, 14 pairs: 14 *[112/210]

On 05/09/2021 00:07, RichD wrote:
>> A theater has 15 seats in the first row. Patrons enter
>> in single file, taking each consecutive seat, as they
>> arrive. In this particular case, there are 8 boys, 7 girls.
>>
>> If a boy and girl occupy adjacent seats, we have a candidate
>> match. If boy sees girl on each side, that's two matches.
>> Clearly, any possible arrangement will contain 1...14
>> candidate matches.
>>
>> What's the average number of matches, for this row?
>
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Assuming the ordering of people in seats is random:

Let I_n = { 1 if match on seats (n,n+1)
{ 0 if no match

Then E(#matches) = E(I_1 + ... + I_14)
= E(I_1) + ... + E(I_14)
= 14 * E(I_1) [symmetry]

and E(I_1) = P(seats 1,2 are BG or GB)
= 2 * P(seats 1,2 are BG)
= 2 * 8/15 * 7/14
= 8/15

So E(#matches) = 14 * 8/15
= 112/15 [approx 7.467]

Mike.

Mike Terry ... wrote:
> On 05/09/2021 00:07, RichD wrote:
>>> A theater has 15 seats in the first row. Patrons enter
>>> in single file, taking each consecutive seat, as they
>>> arrive. In this particular case, there are 8 boys, 7 girls.
>>>
>>> If a boy and girl occupy adjacent seats, we have a candidate
>>> match. If boy sees girl on each side, that's two matches.
>>> Clearly, any possible arrangement will contain 1...14
>>> candidate matches.
>>>
>>> What's the average number of matches, for this row?
>>
> .
> .
> s
> .
> p
> .
> o
> .
> i
> .
> l
> .
> e
> .
> r
> .
> .
> .
>
> Assuming the ordering of people in seats is random:
>
> Let I_n = { 1 if match on seats (n,n+1)
> { 0 if no match
>
> Then E(#matches) = E(I_1 + ... + I_14)
> = E(I_1) + ... + E(I_14)
> = 14 * E(I_1) [symmetry]
>
> and E(I_1) = P(seats 1,2 are BG or GB)
> = 2 * P(seats 1,2 are BG)
> = 2 * 8/15 * 7/14
> = 8/15
>
> So E(#matches) = 14 * 8/15
> = 112/15 [approx 7.467]

That matches up with Serg io's answer, although he didn't provide
motivation for his calculation. Anyhow, in a slightly more general
case, if there are b-1 G's and b B's, the expected number of matches
is 2*(b-1)*b/(2*b-1).

On September 5, Mike Terry wrote:
>> A theater has 15 seats in the first row. Patrons enter
>> in single file, taking each consecutive seat, as they
>> arrive. In this particular case, there are 8 boys, 7 girls.
>> If a boy and girl occupy adjacent seats, we have a candidate
>> match. If boy sees girl on each side, that's two matches.
>> Clearly, any possible arrangement will contain 1...14
>> candidate matches.
>> What's the average number of matches, for this row?

> Let I_n = { 1 if match on seats (n,n+1)
> { 0 if no match
>
> Then E(#matches) = E(I_1 + ... + I_14)
> = E(I_1) + ... + E(I_14)
> = 14 * E(I_1) [symmetry]
>
> and E(I_1) = P(seats 1,2 are BG or GB)
> = 2 * P(seats 1,2 are BG)
> = 2 * 8/15 * 7/14 = 8/15
>
> So E(#matches) = 14 * 8/15 = 112/15

I found this problem in a book. The author's approach matches yours -
compute the mean for one pair of seats, then multiply by 14. Thereby
avoiding the brute force combinatorics.

"The average of the sum is the sum of the averages." A basic theorem of
probability. The author hammers this again and again. However, there's an
assumption underlying this theorem, a pre-condition. What is that condition?
Do you see how it's violated, in this problem?

--
Rich

On 08/09/2021 01:00, RichD wrote:
> On September 5, Mike Terry wrote:
>>> A theater has 15 seats in the first row. Patrons enter
>>> in single file, taking each consecutive seat, as they
>>> arrive. In this particular case, there are 8 boys, 7 girls.
>>> If a boy and girl occupy adjacent seats, we have a candidate
>>> match. If boy sees girl on each side, that's two matches.
>>> Clearly, any possible arrangement will contain 1...14
>>> candidate matches.
>>> What's the average number of matches, for this row?
>
>> Let I_n = { 1 if match on seats (n,n+1)
>> { 0 if no match
>>
>> Then E(#matches) = E(I_1 + ... + I_14)
>> = E(I_1) + ... + E(I_14)
>> = 14 * E(I_1) [symmetry]
>>
>> and E(I_1) = P(seats 1,2 are BG or GB)
>> = 2 * P(seats 1,2 are BG)
>> = 2 * 8/15 * 7/14 = 8/15
>>
>> So E(#matches) = 14 * 8/15 = 112/15
>
> I found this problem in a book. The author's approach matches yours -
> compute the mean for one pair of seats, then multiply by 14. Thereby
> avoiding the brute force combinatorics.
>
> "The average of the sum is the sum of the averages." A basic theorem of
> probability. The author hammers this again and again. However, there's an
> assumption underlying this theorem, a pre-condition. What is that condition?
> Do you see how it's violated, in this problem?

I'm afraid not. Of course, there are assumptions that the boys and
girls are distributed "randomly" so any boy or girl is equally likely to
be in any seat, but if that isn't the case the problem would just be
silly so I don't think it's that.

If I had to guess what you're thinking of, I'd say it's that the I_1,
I_2,...I_14 indicator variables must be independent. But that is NOT a
requirement of the theorem! :)

Mike.

On 9/7/2021 7:00 PM, RichD wrote:
> On September 5, Mike Terry wrote:
>>> A theater has 15 seats in the first row. Patrons enter
>>> in single file, taking each consecutive seat, as they
>>> arrive. In this particular case, there are 8 boys, 7 girls.
>>> If a boy and girl occupy adjacent seats, we have a candidate
>>> match. If boy sees girl on each side, that's two matches.
>>> Clearly, any possible arrangement will contain 1...14
>>> candidate matches.
>>> What's the average number of matches, for this row?
>
>> Let I_n = { 1 if match on seats (n,n+1)
>> { 0 if no match
>>
>> Then E(#matches) = E(I_1 + ... + I_14)
>> = E(I_1) + ... + E(I_14)
>> = 14 * E(I_1) [symmetry]
>>
>> and E(I_1) = P(seats 1,2 are BG or GB)
>> = 2 * P(seats 1,2 are BG)
>> = 2 * 8/15 * 7/14 = 8/15
>>
>> So E(#matches) = 14 * 8/15 = 112/15
>
> I found this problem in a book. The author's approach matches yours -
> compute the mean for one pair of seats, then multiply by 14. Thereby
> avoiding the brute force combinatorics.
>
> "The average of the sum is the sum of the averages." A basic theorem of
> probability. The author hammers this again and again. However, there's an
> assumption underlying this theorem, a pre-condition. What is that condition?
> Do you see how it's violated, in this problem?
>
> --
> Rich
>
>

Generally no correct, it is only the same in specific cases.

ie. Generally

Sum(x) / Sum(y) not equal to Sum (x/y) / n

where n is the total entries x is row entries and y is column entries.

only true if all of the y's are equal

eg: (1/2 + 3/5)/2 = 11/20 (1+3)/(2+5) = 4/7

Where as if y is equal (1/7 + 4/7)/2 = 5/14 (1+4)/(7+7) = 5/14

On September 8, Mike Terry wrote:
>>>> In this particular case, there are 8 boys, 7 girls.
>>>> If a boy and girl occupy adjacent seats, we have a candidate
>>>> match.
>>>> Clearly, any possible arrangement will contain 1...14
>>>> candidate matches.
>>>> What's the average number of matches, for this row?
>
>>> Then E(#matches) = E(I_1 + ... + I_14)
>>> = E(I_1) + ... + E(I_14)
>>> = 2 * P(seats 1,2 are BG)
>>> = 2 * 8/15 * 7/14 = 8/15
>>> So E(#matches) = 14 * 8/15 = 112/15
>
>> I found this problem in a book. The author's approach matches yours -
>> compute the mean for one pair of seats, then multiply by 14.
>> "The average of the sum is the sum of the averages." A basic theorem of
>> probability. The author hammers this again and again. However, there's an
>> assumption underlying this theorem, a pre-condition. What is that condition?
>> Do you see how it's violated, in this problem?
>
> If I had to guess what you're thinking of, I'd say it's that the I_1,
> I_2,...I_14 indicator variables must be independent.

Right.

> But that is NOT a requirement of the theorem! :)

You're right.
For some reason, call it a brain bubble, I recalled that the variables
must be independent. Upon review, not so.

Anyway, you can see here, that adjacent pairs of seats in
this problem aren't independent. But apparently it doesn't matter.

--
Rich