This was the last question in a Mathematics Competition for Uni students. Nobody could do it and I'm still struggling to complete it, days after the exam. Any help would be appreciated.
"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?"
This is the only method and I can think of and it doesn't provide an accurate answer:
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. I'm legitimately stumped.
Let me name the flats $A,B,C,D,\; and\; E$
It doesn't matter which flat the favored customer chooses (say, $E$)
Assuming that each equi-probable combo of two flats is independently and randomly chosen by the others, there will be a total of $[\binom52]^4 = 10,000$ possible choices of combos.
Suppose the $4$ "ordinary" customers $1 - 4$ get one of the $24$ possible permutations of $A,B,C,D$.
They each have $4$ options for their other choice, thus favorable ways = $24\cdot4^4$,
and $Pr = \dfrac{24\cdot4^4}{10,000} \approx 39 \%$
Note: It is possible that in rare cases, a particular choice of combos could yield more than one feasible allocation, in which case the figure will be marginally less.