How this pattern arises?

Question

I am trying to solve this problem:
If I am offered ice-creams in n different sizes, and I don’t know what the sizes are, and I am offered one of each in succession, in random order. So, the strategy I used is, reject the first ice-cream I am offered and choose the first one thereafter that is bigger than their first one i was offered. If the first ice-cream is in fact the biggest one, then I have to put up with the last one.

Let Pn(k) be the probability I choose the Kth biggest ice-cream, where k=1 is the biggest and k=n is the smallest.
Find the expression for Pn(1). The actual answer is:
$\frac{1}{n} \sum_{r=1}^{n-1} \frac{1}{r} $ Could someone please explain to me how this pattern works, or is this just happen coincidentally.
Thank you very much for your reply.

Update: I knew how the formula works with the help of the answers.

However, I derived it again by myself, I found that there is a factorial, could you guys check for me?
My derivation:
There is n! way to choose the ice cream each time. So, we have:
$\frac{1}{n}\frac{1}{(n-1)}!$(this is the probability that chooses a certain size of ice cream, as you choose one of them, you have $\frac{1}{(n-1)}$ to choose, and choose another, you have $\frac{1}{(n-2)}$ to choose a certain ice cream, so it can be written into this form.

When n=2, we can choose things other than it self, so we have (n-1)! Way to choose from the ice creams in order to get 1. So, the probability for choosing 1 when n =2 is$\frac{(n-1)!}{(n-1)!}$.
When n=3, we can choose things other than itself and 2, so we have $\frac{(n-2)!}{(n-1)!}$ way to choose in order to get things. Observing this pattern:
$\frac{1}{n} \sum_{1}^{n-1}\frac{1}{(r-1)!}$ way to choose in order to get 1. Could you guys explain to me where things went wrong?

Answer 1

To put it simply, we start with what's the first ice-cream. There are $n$ choices for that, each equally probable. Let's check those cases.

If the first one is the largest, then you've already rejected it, so your chance of getting it is $0$.

If the first one is the $2$nd largest, then you'll definitely get the largest. So, your chance is $1$.

Now, if the first one is the $3$rd largest, there are two options. You either get the largest or the second largest, depending on which comes first. So, you have a chance of $1 \over 2$.

Similarly, for the kth largest, you have a chance of $1 \over k-1$.

Note that, all these cases have probability of $1 \over n$.

So, the total probability $P_n(k) = {1 \over n} \sum\limits_{r=1}^{n-1}{1 \over r}$.

Answer 2

You never choose first.

If you choose 2nd ice- cream and it is the biggest(first should not be biggest and second is biggest)= $\frac{n-1}{n}.\frac{1}{n-1}$(after that does not matter)

If you choose 3rd , 3rd is biggest but you also have to keep in mind that 1st and 2nd are given in decreasing order.First you choose 2 ice - creams from n-1 icecreams , once you chose their order are fixed which gives you ${n-1 \choose 2}$ cases out of n(n-1)(n-2) cases.Hence the probabillity $\frac{1}{2n}$.

If you chose 4th ice cream first u choose 3 icecreams out of n-1(again once you choose their order is fixed in which they can be given that is in decreasing order) and 4th is biggest which have only one way of happening. In this case ${n-1 \choose 3}$ out of n(n-1)(n-2)(n-3) cases hence the probability $\frac{1}{3!\times n}$.(so here it is where i was wrong , need not be in decreasing order as 3,1,2,4 is also valid, actually we will say the second place have $\frac{1}{2}$ of not being selected , and third one have $\frac{2}{3}$ of not being selected hence for 1st and 2nd not being selected simultaneously is $\frac{1}{3}$ .Of course seeing this above solution is better)

I think you forgot a factorial sign when you were writing the answer in general summation(there was no mistake).If you notice the good thing about this distribution is that you have maximum probability to eat the biggest ice cream compared to other ice creams