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SigmundFraud
Topic Author
Posts: 36
Joined: June 16th, 2012, 11:41 am

### Theatre Row Problem

The Theater Row: Eight eligible bachelors and seven beautiful models happen randomly to have purchased single seats in the same 15-seat row of a theatre. On the average, how many pairs of adjacent seats are ticketed for marriageable couples?

Posts: 23951
Joined: September 20th, 2002, 8:30 pm

### Theatre Row Problem

1. In today's era of gay marriage, the answer is 14. 2. The lower bound is 14*(1 - (7/15)^2 - (8/15)^2 ) = 6.96888...3. ... more later

EBal
Posts: 431
Joined: May 20th, 2005, 1:30 pm

### Theatre Row Problem

Let $a(w, m)$ is number of such pairs formed with w women and m men if a woman is in the leftmost seat, and $b(w, m)$ - number of pairs if a man is in the leftmost seat. Then we have $a(w,m)=\frac{(w+m-2)!}{(w-1)!(m-1)!}+b(w-1,m)+a(w-1,m)\,,\quad b(w,m)=\frac{(w+m-2)!}{(w-1)!(m-1)!}+a(w,m-1)+b(w,m-1)$. Introduce the transform$A(x,y)=\sum_{w=1}^\infty\sum_{m=1}^\infty a(w,m)x^wy^m$ and same for $B(x,y)$. Then $A, B$ satisfy$A=\frac{xy}{(1-x - y)}+xA+xB\,,\quad B=\frac{xy}{(1-x-y)}+yA+yB$. This can be solved, for the quantity of interest $A+B=\frac{2xy}{(1-x-y)^2}$. All that remains to do is to find the coefficient in front of $x^7y^8$.
Last edited by EBal on July 1st, 2014, 10:00 pm, edited 1 time in total.

EBal
Posts: 431
Joined: May 20th, 2005, 1:30 pm

### Theatre Row Problem

The expansion is actually not too hard, the result is $2\frac{(w+m-1)!}{(w-1)!(m-1)!}$ for the total number of pairs. The average is $\frac{2mw}{(m+w)}$. This is in agreement with Vanubis1 who posted correct answer before I could spot an error in my original recurrence relation. I find it hard to follow his logic though.
Last edited by EBal on July 1st, 2014, 10:00 pm, edited 1 time in total.

Vanubis1
Posts: 152
Joined: February 21st, 2011, 7:41 am

### Theatre Row Problem

I think you can do the average directly.A man has a probability of 1-1/(w+m) to have a left neighbour and in this case w/(w+m-1) it's a woman.Same number for right neighbour.Each man has an average number of women neighbours of 2*w/(w+m) By summing on all the men, we find 2*m*w/(w+m) = 7.46 with w=7 and m=8
Last edited by Vanubis1 on July 2nd, 2014, 10:00 pm, edited 1 time in total.

EBal
Posts: 431
Joined: May 20th, 2005, 1:30 pm

### Theatre Row Problem

Last edited by EBal on July 1st, 2014, 10:00 pm, edited 1 time in total.

chocolatemoney
Posts: 322
Joined: October 8th, 2008, 6:50 am

### Theatre Row Problem

katastrofa
Posts: 9332
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

### Theatre Row Problem

Can you solve this skipping the word "marriageable" (i.e. when one pair is classified as a couple, they cannot be married to somebody else)? E.g. female male female -> 1 couplea

Ultraviolet
Posts: 1655
Joined: August 15th, 2012, 9:46 am

### Theatre Row Problem

Previous solution - m * (m + n - 2) / (m + n) * n * (n - 1) / (n + m - 1) - analogous term for "beautiful models"?

katastrofa
Posts: 9332
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

nope...a