# Probability query

• Jan 10th 2011, 01:34 AM
ravingcrew
Probability query
Last night my girlfriend posed a rhetorical question that I mistakenly answered leading to a several hour 'debate'.

Essentially she said "I was at a concert with my mum, and my brother's girlfriend, who bought her tickets independently of us, was sat next to me, what are the chances of that?" With the assumed capacity of the venue to ve 10,000 I answered approximately 1 in 10,000 (more accurately 1 in 9,998).

Several others have been bought into the debate, including some with a 1st Degree in Maths and statistics, but each side is unwilling to accept the others answer (with their answer being 1 in 10,000 multiplied by 1 in 9,999).

To answer this we have established that they both definately buy a ticket for the same event independently, so that's happened already, the question is then, now that they both have a ticket what is the chance of them sitting next to each other? Seeing as her mum was with her, there is only a free seat on one side of her. My logic is that she is definately going to sit somewhere, so thats a given, so the only thing we are working out the probability of is that her friend is in the one free seat next to her, there are 9,998 free seats left after her and her mum have taken 2 seats so I feel the answer is 1 in 9,998. The probability of this 1 event happening.

Their arguement is that there are 2 events, her getting one seat multiplied by her friend getting the one next to her.

So, having established that they are both definately going on the same night, what is the chance of them sitting next to each other??? Please help end the arguing!!!!

Thanks very much.
• Jan 10th 2011, 01:43 AM
pickslides
So what is the probability you sat next to either your mum or your brother's girlfriend given they were sitting next to each other in a 10,000 seat arena?

If so they are in a pair and you can sit either side of them. If this is true it acts as there is one less seat in the arena. Do you follow?

Now are there any rows that end? I.e. can one of these two people be on the end of a row, i.e. you will not be able to sit on one side of the pair?
• Jan 10th 2011, 01:50 AM
ravingcrew
It's the probability of the brother's girlfriend (let call her person B) sitting next to my girlfriend (person A).

So to summarise person A buys tickets for a concert and goes with her mum who she sits next to. Person B independently buys a ticket for the same concert. what is the probability that person B sits next to person A (remember the mother is sat on one side of her)? the capacity of the arena is 10,000. If we leave aisles out of it for simplicity for now. Or if somebody wants to bring that in then fine, but my answer has no aisles OR assumes that if an aisle is between the 2 this still counts as sitting next to each other.
• Jan 10th 2011, 08:03 AM
snowtea
Quote:

Originally Posted by ravingcrew
What is the probability that person B sits next to person A (remember the mother is sat on one side of her)? the capacity of the arena is 10,000. If we leave aisles out of it for simplicity for now.

Ok, no aisles (lets just assume one row of seats wrapped around the inside of a cylinder).
Your girlfriend and her mother can sit anywhere. Regardless, there is only 1 seat left in the auditorium next to her.
So I think you are right 1/9998.

This hand waving explanation might not seem satisfactory to nonbelievers, so lets count all possible seatings directly: Let M be mom, A B be person A and B respectively (assuming all seats are on one row).
Prob = (# ways to seat with MAB + # ways to seat with BAM) / (# ways to seat with MA + # ways to seat with AM)
= (9999*9997! + 9999*9997!) / (9999*9998! + 9999*9998!) = 1/ 9998

If this does not work. Use a smaller case where possibilities can be counted. How about 4 seats, 5 seats, 10 seats.