Basket with Lilies and Roses. Find the number of roses.

Question

Ques: A covered basket of flowers has some lilies and roses. In search of rose, Sweety and Shweta alternately pick up a flower from the basket but put it back if it is not a rose. Sweety is 3 times more likely to be the first one to pick a rose. If sweety begin this 'rose hunt' and if there are 60 lilies in the basket, find the number of roses in the basket.

Let the roses be x and lilies be y. Since they find all of the x roses finally $$ x!/(x+y)! + y/(x+y) * x!/(x+y)! \cdots = 1 $$ So to be straight forward, I do not have any good idea how to deal with this.

Answer 1

Assume that the probability to pick a rose is $r$. Then - if $p$ denotes the probability that Sweety (who starts picking) will be the first to pick a rose - we find: $$p=r+(1-r)^2p\tag1$$

Here $r$ is the probability that her first pick is succesful and $(1-r)^2$ is the probability that both are unsuccesful at their first pick so that the process starts over with again a probability $p$ that Sweety will be the first one with success.

Next to that we also have:$$p=3(1-p)$$or equivalently:$$p=\frac34$$and this makes it possible to solve $(1)$.

One solution is: $r=\frac23$ telling us that next to $60$ lillies there are $120$ roses.

Another solution is $r=0$ telling us that there are no roses at all. Then Sweety and Shweta are still picking at the moment, both having probability $0$ to succeed once. Observe that $0=3\cdot0$ so that indeed the probability of Sweety to be the first is $3$ times the probability of Shweta to be the first.

Actually this second solution could be excluded on base of the info that the basket contains lillies and roses (so $r>0$).

Answer 2

Let $x$ be the number of roses in the basket.

It's given that the first person to choose is $3$ times more likely to get the first rose than the second person, hence the first person gets the first rose with probability $\frac{3}{4}$.

With probability ${\large{\frac{x}{x+60}}}$, Sweety wins on the first try.

With probability ${\large{\frac{60}{x+60}}}$, Sweety does not win on the first try. In that case, Shweta is effectively the first player in a new game. Hence, starting at that point, Shweta has probability $\frac{3}{4}$ to win, and Sweety has probability $\frac{1}{4}$ to win.

This yields the equation $$\frac{3}{4}=\frac{x}{x+60}+\left(\frac{60}{x+60}\right)\left(\frac{1}{4}\right)$$

Solving for $x$ yields $x = 120$.