Find the probability that a particular fish is caught before two of the colors of fish are exhausted.

Question

In a particular pond there are L white, M red, and N yellow fish. One of the yellow fish is George. A fisherman catches one fish per day and stops only if two of the colors of fish are exhausted. Find the probability that George survives.

I came up with P(George survives) = ((L+M+N-1)/(L+M+N))(N/(L+M+N)).

I first tried to solve the problem by finding P(George survives not including 2 colors exhausting) = (L+M+N-1)/(L+M+N). I found this by using permutations.

Then I found probability that two colors exhausted. I figured that I only have to worry about white and red fish being exhausted, since the only way George will survive is both white and red fish are exhausted. So P(white and red fish are exhausted) = N/(L+M+N). I believe that this is correct but I am not sure.

Answer 1

Ignore the yellow fish other than George. Of the $(L+M+1)$ relevant fish, George must be the only one not caught. The probability of this should be quite clear:

$$\frac{1}{L+M+1}$$

Answer 2

It is true but inconclusive that George dies unless the red and white fish are exhausted (i.e. unless the yellow fish are never exhausted). The complication is that George might still die if he is caught before the red and white fish are exhausted.

Perhaps the clearest approach is to pretend that all $(L + M + N)$ fish will be caught, but that once two of the colors are exhausted, all fish caught up until that point will be killed, and all fish caught after that point will be saved by the policy of catch and release.

So the question becomes what are the chances that George does not appear in the sequence of fish snagged until after all of the $(L + M)$ red and white fish appear in this sequence.

Let $D$ denote $(L + M + N)!$ which means that $D$ denotes all of the various ways of ordering the fish to determine the order in which the fish are caught.

Focusing only on the yellow fish, George can be the $k^{\text{th}}$ fish caught, where $k$ is equally like to have any of the values in $\{1, 2, \cdots, N\}.$

Suppose that George is the $k^{\text{th}}$ fish caught, and George survives. Regard this as a successful outcome.

The # of sequences that represent this successful outcome are $f(k) = \binom{N-1}{k-1} \times (L+M + [k - 1])! \times [N-k]!$

Here, the 1st RHS factor is used to determine which of the $(N-1)$ other yellow fish are caught before George.

Consequently, the chance that George survives should be
$\frac{\sum_{k=1}^N \,f(k)}{D}.$

Addendum Demonstrating that my answer is equivalent to Daniel Mathias' answer.

Lemma 1 : $~ \displaystyle \forall ~n,T ~\in ~\mathbb{Z^+}, ~\sum_{k=0}^{n-1} \binom{T + k}{T} ~=~ \binom{T + n}{T+1}.$

Proof by induction:

$~ n = 1 ~: \displaystyle \binom{T + 0}{T} ~=~ 1 ~=~ \binom{T + 1}{T+1}.$

$~ n = N + 1 ~:$ $~\displaystyle \sum_{k=0}^{N} \binom{T + k}{T} ~=~ \left[\sum_{k=0}^{N-1} \binom{T + k}{T}\right] + \binom{T + N}{T}$

$=~$ [by inductive assumption]

$~\displaystyle \binom{T + N}{T+1} + \binom{T + N}{T} ~=~ \binom{T + [N+1]}{T+1}.$

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Let $T \equiv (L + M).$

By Lemma 1, $~ \displaystyle \sum_{k=0}^{N-1} ~\binom{L + M + k}{L + M} ~=~ \binom{L + M + N}{L + M + 1} ~\Rightarrow $

$~ \displaystyle (L+M)! \times (N-1)! \times \sum_{k=0}^{N-1} ~\binom{L + M + k}{L + M} ~=~ \frac{(L + M + N)!}{(L + M + 1)} ~\Rightarrow $

$~ \displaystyle (N-1)! \times \sum_{k=0}^{N-1} ~\frac{(L + M + k)!}{k!} ~=~ \frac{(L + M + N)!}{(L + M + 1)} ~\Rightarrow $

$~ \displaystyle (N-1)! \times \sum_{k=1}^{N} ~\frac{(L + M + [k-1])!}{[k-1]!} ~=~ \frac{(L + M + N)!}{(L + M + 1)} ~\Rightarrow $

$~ \displaystyle \sum_{k=1}^{N} \left\{~ \binom{[N-1]}{[k-1]} \times ~ (L + M + [k-1])! \times (N - k)! ~\right\} ~=~ \frac{(L + M + N)!}{(L + M + 1)} $

$~\Rightarrow$

$~ \displaystyle \sum_{k=1}^{N} ~f(k) ~=~ \frac{(L + M + N)!}{(L + M + 1)} ~=~ \frac{D}{(L + M + 1)} $.