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I scatter 5 pellets in a tank of 5 goldfish. Assuming all pellets are eaten and each goldfish has equal probability of eating each pellet (and that goldfish don't eat less as they eat more), what is the probability that at least one goldfish eats more than 1 pellet?

I had originally thought the answer was $\frac{1}{5^5}$, but then I realized that the denominator was wrong because the pellets are not replaceable.

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up vote 2 down vote accepted

Think of it as rolling five $5$-sided dice: if pellet $1$ is eaten by fish $3$, die $1$ has come up $3$. In these terms you’re asking for the probability that the five dice all show different numbers. There are $5^5$ different ways they can come up; how many of those give the desired result?

Added: Here’s another way to attack it. Imagine that no two pellets are eaten at exactly the same moment. Label them $P_1,\dots,P_5$ in the order in which they’re eaten. The probability that $P_1$ is eaten be a fish that hasn’t already swallowed a pellet is obviously $1$. The probability that $P_2$ is eaten by a different fish is $\frac45$. Keep going to find the probabilities that $P_3,P_4$, and $P_5$ are eaten by fish that haven’t already swallowed a pellet. What do you do with these probabilities to get the final result?

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your reputation has almost a flat histogram. cool. –  Seyhmus Güngören Sep 10 '12 at 23:13
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Oh! The number of numbers available or fish available for picking so that the picked elements remain distict decreases by 1 each time. Therefore, the probability of no fish ever eating more than 1 pellet is $\frac{5!}{5^5}$, right? Therefore, the probability that at least 1 fish eats more than 1 pellet is $1 - \frac{5!}{5^5}$. –  David Faux Sep 10 '12 at 23:17
    
@David: Yes, that’s exactly right. –  Brian M. Scott Sep 10 '12 at 23:18
    
Ah, thanks! I appreciate you mentioning the two different paradigms of looking at the problem. Indeed, the counting rule would dictate the second method to also yield $1 - \frac{5!}{5^5}$. –  David Faux Sep 10 '12 at 23:50
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