If the first house is the best, tough luck, you’ll miss it, and the first M houses. So the likelihood of them is 0.

But if the newly refurbished Turner 6-share, with a pool for only $ 500 a week is the (M+1)th  you’re in luck.

As it must be better than all previous house:

The greater the N, the smaller this chance, so let’s consider the (M+2)th house. This means the (M+1)th didn’t have a higher rating than the previous M, and the probability of the (M+2)th being the best is (1/N) again:

Now, for the (M+3)th, we repeat the argument for the (M+2) th case, and get:

That there are finitely many houses in Canberra is quite a reasonable assumption to make, as there is only a population of 381,000. So we can keep enumerating out the argument until we get to the point where all but the final ( (N−1)th cases) , as for the final one. It doesn’t matter. You got the house with the neighbours who are learning how to yodel on one side and cook meth on the other.

This final case will be:

When we collate all this, we get:

pick your target

So the condition we want to place on P(M,N) is: P(M-1, N) < P(M,N)  and  P(M+1,N)< p(M,N).You set up the conditions we specify, do some algebra and find a rule. Firstly that: P(M-1, N) < P(M,N)  goes to:

And next, that P(M+1,N)< p(M,N) goes to:

We now need to find the M that satisfies both of these conditions.

example

So, you’ve filled in so many applications that you are now sick of your name and nonchalant about the lie regarding your financial stability.

And grad job applications and assignments have got on top of you so you could only apply for five houses, so we get the following:

We would get M =2, so the first 2 houses should be ignored. Then you accept the next best one.

knock ’em down

So we have our equation now:

And our problem is to find M where this is true:

It is clear we are building squares above and below the line (1/x) Firstly above (the highest point on the curve is where the top of the square is):

And below (lowest point on the curve is where the top of the square is):

With these equations we are making rectangles 1 unit apart, then multiplying them by their position on the line 1/x, so if we were looking at the rectangle(likelihood of choosing the right one is increased by having this here) at the house being between M+5 and M+6, we make our rectangle by multiplying the width: (M+6) –(M+5) =1. By the height of the line at the point we will depend on whether we do the top or bottom value, it will be: 1/(M+5) and 1/(M+6) respectively. Multiplies by one, they stay the same.

Now if we integrate (this adds up all our increases in additional chances) our term to solve it, this sums up:

While our other term goes to:

And as (N/M)≈(N−1)/M when N and M are large (just put some fake numbers in and try!). We can just work on the term for our lower bound, then we get:

So the best place to execute our plan, to select the next best house(the house before is M)  after all the ones we have seen is at around 0.37 multiplied by the total number of houses – this is just 37%.

Veranda and a picketfence

You’ve done the maths, your housemates thought you were too weird and left you and so you took to craigslist to find new housemates. But that doesn’t matter, by picking the next house better than all the first 37% you saw, you found your dream house.

Dave who is coming down from Perth for an undisclosed line of work might think shitting in the garden is fine as “it’s all dirt anyway”.

Jimmy might ruin your date night every week by bringing around a different group of friends and all taking dark web hallucinogens.

And Andy will be arriving tomorrow with his bond, backpay of rent and all the lounge room furniture until the end of the contract.

But at least you have your castle, and it was all because of the 37% rule.