Probability of Flowers

Let's start the generalization. Assume that the total number of roses be $n$ and the number of roses to be observed be $k$ . Then, arrange the flower according to their degree of beautiful (I know beautiful is different to everyone, but just focus on the mathematical side!). We will have the set $S=\{a_1,a_2,a_3,\dots,a_n\}$ where $a_1$ is the most beautiful flower, $a_2$ is the second most beautiful flower, $a_n$ is the ugliest flower.

Note that, $a_1$ cannot be in the first $k$ roses because the student will not pick them, so we only consider the case $a_1$ on the position $d$ where $k+1\leq d\leq n$ . Furthermore, we need to know the chance $a_1$ appear on every position is $\dfrac1n$ , so every case with a value $d$ will have a base probability $B=\dfrac1n$ .

Let's discuss if $a_1$ is on the position $d$ , then consider if we put the first $k$ ugliest roses on the first $k$ position which are the roses in the set $\{a_{n-k+1},a_{n-k+2},\dots,a_n\}$ , then the student must pick up the $k+1^{th}$ roses because it is the most beautiful compared to the first $k$ . It isn't what we want. So there is a bound where we need to decide to put the roses in the first $k$ position.

There are exactly $d-1$ roses in front of $a_1$ , so the first $d-1$ ugliest roses must at front of $a_1$ as the minimum condition. To calculate this, we need to know the $d-1^{th}$ ugliest rose is $a_{n-d+2}$ . At the minimum case, $a_{n-d+2}$ must be the most beautiful rose in the first $d-1$ roses and must place at position inside the first $k$ roses, to ensure the student will pick the rose at the position $d$ , which is what we want.

Before calculating these messy things, we first define the set $A_i$ to be the valid permutation of roses when the most beautiful rose in the first $k$ roses is $a_i$ . To make it clear, every element in the set $A_i$ is a string contains all of the flowers $a_1$ to $a_n$ , but with different permutations. Then, the student must pick up the actual most beautiful rose by the permutation of these elements, which is called "valid permutation".

Then, let the sample set $SP$ be all of the permutation when $a_1$ fixed at the position $d$ , so $n(SP)=(n-1)!$

After the setting process, we now count the number of elements aka the valid permutations of roses in the set $A_{n-d+2}$ . $a_{n-d+2}$ must in the first $k$ position, there are $k$ choices for it to be planted. then all of the roses uglier then $a_{n-d+2}$ must place in front of position $d$ , that is $(d-2)!$ . Then, all of the roses more beautiful than $a_{n-d+2}$ must place after position $d$ , that is $(n-d)!$ . So $n(A_{n-d+2})=k[(d-2)!][(n-d)!]$

Then, we consider the case for the set $A_{n-d+1}$ . We put $a_{n-d+1}$ in first $k$ position, which is $k$ choices. Then there are $n-d-1$ roses more beautiful than it (not $n-d$ , already excluded $a_1$ ), and we gonna put it after position $d$ , which is $n-d$ slots, so it is $\dfrac{(n-d)!}{1!}=(n-d)!$ . After that, there are $d-1$ roses left, all of them are uglier than $a_{n-d+1}$ , so the number of permutation of them will be $(d-1)!$ . Combining all of these, we get $n(A_{n-d+1})=k[(d-1)!][\dfrac{(n-d)!}{1!}]$

Then, the case $A_{n-d}$ will lead to the result $n(A_{n-d})=k[d!][\dfrac{(n-d)!}{2!}]$ .

We just need to loop through the case until $A_2$ , not $A_1$ , because the most beautiful rose will not place on the first $k$ position, so $n(A_2)=k[(n-2)!][\dfrac{(n-d)!}{(n-d)!}]$

Let the probability the student successfully pick up $a_1$ when it is on the $d$ position be $P(d)$ , then it can be expressed as $\begin{aligned}P(d)&=\frac{n(A_{n-d+2})}{n(SP)}+\frac{n(A_{n-d+1})}{n(SP)}+\frac{n(A_{n-d})}{n(SP)}+\dots+\frac{n(A_2)}{n(SP)}\\&=\frac{k[(n-d)!]}{(n-1)!}\bigg[(d-2)!+\frac{(d-1)!}{1!}+\frac{d!}{2!}+\dots+\frac{(n-2)!}{(n-d)!}\bigg]\end{aligned}$

By using the identity (which will be prove at the bottom of the solution) $r!+\frac{(r+1)!}{1!}+\frac{(r+2)!}{2!}+\dots+\frac{n!}{(n-r)!}=r!\binom{n+1}{r+1}$ We can simplify $P(d)$ to $\begin{aligned}P(d)&=\frac{k[(n-d)!]}{(n-1)!}[(d-2)!]\binom{n-1}{d-1}\\&=\frac{k[(n-d)!]}{(n-1)!}[(d-2)!]\frac{(n-1)!}{[(d-1)!][(n-d)!]}\\&=\frac k{d-1}\end{aligned}$

Overall, the student pick up $a_1$ no matter what position it is $k+1\leq d\leq n$ will be $\begin{aligned}P&=B\sum_{d=k+1}^nP(d)\\&=\frac kn\bigg[\frac1k+\frac1{k+1}+\frac1{k+2}+\dots+\frac1{n-1}\bigg]\\&=\frac kn\sum_{d=k+1}^n\frac1{d-1}\end{aligned}$ Since the harmonic series can't be simplify anymore, this is the generalized form of the probability. Let me write down here: $P(n,k)=\frac kn\sum_{d=k+1}^n\frac1{d-1}$ Substituting $n=11$ and $k=5$ , we will get $P(11,5)=\dfrac5{11}\bigg[\dfrac15+\dfrac16+\dfrac17+\dfrac18+\dfrac19+\dfrac1{10}\bigg]=\dfrac{2131}{5544}$ , so the sum $a+b$ will equal to $2131+5544=\boxed{7675}$ .

You can see the calculation about finding the optimum ratio of $n$ and $k$ done by Michael Mendrin in the comment section below.

Below is the proof of the identity.

Prove that $r!+\frac{(r+1)!}{1!}+\frac{(r+2)!}{2!}+\dots+\frac{n!}{(n-r)!}=r!\binom{n+1}{r+1}$

Proved by Induction:

When $n=r$ , $LHS=r!$ , $RHS=r!\binom{r+1}{r+1}=r!$ , the equality holds.

Assume that the identity holds when $n=r+k$ , that is $\displaystyle r!+\dfrac{(r+1)!}{1!}+\dfrac{(r+2)!}{2!}+\dots+\dfrac{(r+k)!}{k!}=r!\binom{r+k+1}{r+1}$ . Then when $n=r+k+1$ , we have $\begin{aligned}LHS&=r!\binom{r+k+1}{r+1}+\frac{(r+k+1)!}{(k+1)!}\\&=\frac{r!(r+k+1)!}{k!(r+1)!}+\frac{(r+k+1)!}{(k+1)!}\\&=\frac{(r+k+1)!}{k!}\bigg[\frac1{r+1}+\frac1{k+1}\bigg]\\&=\frac{(r+k+1)!}{k!}\frac{r+k+2}{(r+1)(k+1)}\\&=\frac{(r+k+2)!}{(r+1)[(k+1)!]}\\&=r!\frac{(r+k+2)!}{(r+1)!(k+1)!}\\&=r!\binom{r+k+2}{r+1}\\&=RHS\end{aligned}$ This completes the proof.

$\begin{aligned} P(n,k) &= \sum_{i=1}^{n} P(\text{The }i\text{-th rose is picked})\times P(\text{The }i\text{-th rose is the best}) \\ &= \frac{1}{n} \left(\sum_{i=1}^{k}0 + \sum_{i=k+1}^{n}P(\text{The most beautiful rose in the first } (i-1)\text{ roses is in the first } k\text{ roses})\right) \\ &= \frac{1}{n} \sum_{i=k+1}^{n} \frac{k}{i-1} \\ &= \frac{k}{n} \sum_{i = k+1}^{n} \frac{1}{i-1} \\ P(11,5) &= \frac{2131}{5544} \end{aligned}$

See Michael's comment for the optimum ratio for large $n$