
Advanced Statistics Advanced Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
June 5th, 2015, 08:00 AM  #1 
Newbie Joined: Jun 2015 From: Booztwon Posts: 4 Thanks: 0  Difficult Probability Problem
Hi, I am struggling with a problem and I desperately need some help. "A real estate developer owns 5 flats, that are to be construted. Flat1, Flat2, Flat3, Flat4 and Flat5. The developer promises his best friend one of these flats. Moreover they agree that the friend chooses his flat after all flats have been constructed. The developer, however, wants to sell the other 4 flats before the end of the construction. The developer doesn't know,though, which flat his friend will pick but he guarantees him his first choice so he decides the following: He asks 4 potential buyers to give him two preferences regarding which flat they want to buy. E.g. "Potential buyer 1" will give him (Flat1, Flat2) or (Flat3, Flat5)... meaning that this buyer is indifferent between Flat1 and Flat2 (Flat3 and Flat 5) and either flat would satisfy his demand. Now, what is the probability that, after the friend has chosen his flat, all the other potential buyers end up getting one of the flats, they wanted?" (Accidentally, I've posted the question earlier in high school math) 
June 7th, 2015, 03:40 PM  #2 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 507 Thanks: 74 
There are 10 combinations for each of the four people choosing 2 flats and 5 permutations for the person who gets first choice. Therefore given enough time, you could write out 10 * 10 * 10 * 10 * 5 = 50,000 possibilities to solve it, but nobody wants to do that. For starters if the same combination of flats is chosen by three or more of the four people, not everybody can get his or her choice. For any one combination of 2 flats, the probability of 3 or more of the 4 people wanting it is 37/10,000. There are 5 flats, so if the combinations were independent, the probability of no combination being wanted by 3 or more of the 4 people would be ((10,000  37)/10,000)^5 = about 0.9816. However, since each flat is in exactly 4 of the combinations, I don't know if the combinations are independent. It won't calculate the probability, but I might do 100 random number simulations to estimate it. Last edited by EvanJ; June 7th, 2015 at 03:47 PM. 
June 9th, 2015, 06:12 AM  #3 
Newbie Joined: Jun 2015 From: Booztwon Posts: 4 Thanks: 0 
hey thanks for your answer. Some of those points were very helpful. I am not sure, though, how you got to 37/10000. For me the probability that 3 or 4 potential buyer end up having the same preference would be (4!/3!) * (1/10)^3 * (9/10) + 10* (1/10)^4 = 46/10000. And another issue is: Even if only 2 buyers had the same preferences it might be possible that things don't work out for the developer. Imagine: 2 potential buyers both want flat 1 and flat2 and the friend of the developer wants flat 2 also. Maybe you have accounted for that and I just didn't understand your argument properly? Anyways thanks for your help so far and don't be shy to correct my mistakes That is a very messy problem. 
June 9th, 2015, 11:48 AM  #4  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 507 Thanks: 74  Quote:
For the other issue, I know I didn't account for that. My random number simulations resulted in a probability of 216/500 = 0.432. The denominator is 500 because I did 100 simulations for the 4 people and then looked at if each simulation was compatible with the 4 people and with the person who gets first choice of 5 flats. This is not the exact probability, which I don't know how to calculate.  
June 12th, 2015, 03:03 AM  #5 
Newbie Joined: Jun 2015 From: Booztwon Posts: 4 Thanks: 0 
Hey thanks for the ballpark figure! Gives a good sense of what to expect!!! Problem itself is probably only solvable the hard way. Generally I would agree that there is only one way for four people to all want the same pair. However, it might be the case that all 4 want (Flat1, Flat2) or that all 4 want (Flat2, Flat3). So don't we ignore those events if we don't account for the 10 possible combinations? Many thanks! 
June 13th, 2015, 08:43 AM  #6  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 507 Thanks: 74  Quote:
 
November 18th, 2016, 08:02 PM  #7 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 507 Thanks: 74 
I'm going to try to do this problem in a spreadsheet. I'm going to do a little at a time and I may not post an answer for weeks.

December 30th, 2016, 08:25 AM  #8 
Newbie Joined: Dec 2016 From: Jordan Posts: 2 Thanks: 0 
Hi I need help to write code using mathmatica Can help me ??? 

Tags 
difficult, probability, problem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Difficult Probability Problem  lOWIq  Probability and Statistics  0  June 5th, 2015 07:49 AM 
a difficult conditional probability  20824  Advanced Statistics  1  January 29th, 2014 01:29 PM 
Interesting but very difficult probability problem  alpha_century  Probability and Statistics  0  May 5th, 2013 11:22 AM 
Difficult problem  MATHS FRIEND  Number Theory  3  August 17th, 2012 11:42 AM 
Probability a bit to difficult  mxmadman_44  Advanced Statistics  0  November 16th, 2010 05:56 PM 