George the silver fish

[BMAD]

There are M gold fish and K silver fish in a lake. They are caught and eaten one at a time at random until only one color of fish remains in the lake. One of the silver fish is named George. Find the probability George is not eaten.

[gavinksong]

Suppose there are M gold fish and K silver fish and W red herrings in a lake. The gold and silver fish are caught and eaten one at a time until only two colors of fish remain in the lake. One of the silver fish is named George. One of the red herrings is named Harold. Find the probability George is not eaten.

 

I am finding this problem very hard. :\

Edited April 8, 2015 by gavinksong

[bonanova]

x fish remain when their population becomes monochromatic.

George's survival chances are simply x / (M+K).

All we need is x.

 

x = M/(K+1) + K/(M+1)

[gavinksong]

x fish remain when their population becomes monochromatic.

George's survival chances are simply x / (M+K).

All we need is x.

 

x = M/(K+1) + K/(M+1)

 

Your answer makes it sound as if it doesn't matter whether George is a gold fish or a silver fish.

But it does. Suppose George is the only silver fish in the lake. The only chance he has of surviving is if all the other fish get eaten, whereas a gold fish just has to avoid being eaten before George does. The odds aren't exactly in George's favor.

 

For the case where there are two gold fish and one silver fish. The silver fish will survive only if the other two fish are eaten first: (2/3) (1/2) = 1/3. However, bonanova's analysis yields 4/9, which is too optimistic for George.

Edited April 8, 2015 by gavinksong

[gavinksong]

I'm gonna go out on a limb here, but based partially from a trend I have noticed and the formula that bonanova cited...

1/(M+1)

[gavinksong]

From George's perspective, except for George, the other silver fish are completely irrelevant - red herrings, if you will. For George to survive, he simply has to stay alive longer than all of the gold fish. If a different silver fish is caught and eaten, it doesn't affect George's ultimate chances of survival at all. Essentially, it's a "please roll again" scenario.

 

So take the silver fish out of the equation: we have M+1 fish remaining. Now, pretend that the game doesn't end until one fish remains. Again, for George, the situation is still exactly the same. If he survives, he is the last surviving fish. If he dies, it doesn't matter what happens afterwards; the game might as well go on until one fish remains. So in this modified version of the game, among the remaining M+1 fish, the probability that any one of them survives is simply 1/(M+1) due to symmetry. Thus, George's chances of surviving is also 1/(M+1).

Edited April 9, 2015 by gavinksong

[bonanova]

Brain fart in previous post.

Expected number surviving silver fish is K/(M+1).

Chance that any of the survivors is George is 1/K.

Chance that George is a survivor is 1/(M+1)

[bonanova]

From George's perspective, except for George, the other silver fish are completely irrelevant - red herrings, if you will. For George to survive, he simply has to stay alive longer than all of the gold fish. If a different silver fish is caught and eaten, it doesn't affect George's ultimate chances of survival at all. Essentially, it's a "please roll again" scenario.

 

So take the silver fish out of the equation: we have M+1 fish remaining. Now, pretend that the game doesn't end until one fish remains. Again, for George, the situation is still exactly the same. If he survives, he is the last surviving fish. If he dies, it doesn't matter what happens afterwards; the game might as well go on until one fish remains. So in this modified version of the game, among the remaining M+1 fish, the probability that any one of them survives is simply 1/(M+1) due to symmetry. Thus, George's chances of surviving is also 1/(M+1).

 

Agree. In fewer words:

George, accompanied by other silver fish or not, occupies one of M+1 equally likely spaces between the M gold fish.

If and only if he follows all the gold fish, he survives.

[gavinksong]

Incidentally, this explanation provides a much simpler way to arrive at the solution to your aha​ problem, bonanova.

 

Edit - I just realized that there was an equally elegant words-only solution further along the aha thread. And it has a slightly different angle than the explanation we have above. Man, this is what I love about math.

Edited April 9, 2015 by gavinksong