# Probability - The fox and the rabbits

#### Robin04

Problem Statement
One day the fox decided to hunt for rabbits. As he was very hungry, he made a plan: he'll go the a rabbit hole at a random time and will wait there for time $t$ (let's call this strategy $A$) hoping that he'll encounter a rabbit. But that night he saw the devil in his dreams who said to him: "Fox! I know that a rabbit will show up at exaclty 11 o'clock." But the devil knew that the rabbit is very fast and it'll show up at time $\epsilon$ before 11. The next day the fox arrived at exactly 11 but he was late for the rabbit so he waited for time $t$ (let's call this strategy $B$).

Which strategy is better for the fox?

In average two rabbits show up every 45 minutes and we say that strategy $A$ is better than strategy $B$ if it is true that with strategy $A$ the fox had a higher chance to catch at least one rabbit.
Relevant Equations
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So if 2 rabbits show up every 45 minutes then trivially one rabbit shows up every 22.5 minutes. With strategy $A$ the probability of catching a rabbit is $\frac{t}{22.5}$. With strategy $B$ it is $\frac{t}{22.5-\epsilon}$, so strategy $B$ is always better. Am I correct or I'm missing something? This is posted as a harder problem in my assignements (however it turned ot with other problems that they're not that difficult either...).

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#### PeroK

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I think you are modeling rabbits like trains with a regular timetable. I'm not sure that is the idea.

#### Robin04

I think you are modeling rabbits like trains with a regular timetable. I'm not sure that is the idea.
How else could I model them?

#### lomidrevo

Try to see whether the appearance of rabbits could be modeled by some of the well-know distributions.

#### Robin04

Try to see whether the appearance of rabbits could be modeled by some of the well-know distributions.
Well, it could be Poisson distribution maybe. But how do I decide what distribution can be applied? I don't think there's any information about this in the text.

#### PeroK

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How else could I model them?
Have you heard of a Poisson distribution?
However, that might not be the issue here.

#### Robin04

What distribution did I assume in my solution?

#### PeroK

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What distribution did I assume in my solution?
The point is that if events happen with a certain frequency, then the probability of getting $n$ events in a given time interval is a Poisson distribution.

The question asks the probability of the fox catching at least one rabbit. You actually calculated the expected number of rabbits. If you rephrase the question, then that it what you are given:

The expected number of rabbits in any 45 minute period is 2.

The main problem with your solution is that you have confused expected number and probability of at least one.

That said, the question really only asks you to compare the strategies. Perhaps you could simply look at that. As I understand the question when the fox turns up he has only just missed a rabbit and the the question is what difference does that make?

#### lomidrevo

Well, it could be Poisson distribution maybe. But how do I decide what distribution can be applied? I don't think there's any information about this in the text.
Nice guess. The info is there, you just have to "decipher" the problem statement. Let me quote form this link:
Poisson Distribution
The Poisson distribution is one of the most widely used probability distributions. It is usually used in scenarios where we are counting the occurrences of certain events in an interval of time or space.
Can you see the similarities with this particular problem?

#### lomidrevo

I would suggest you to first focus on strategy A. Try to compute the parameter $\lambda$ of the Poisson distribution. Hint: all the values necessary for that are given.

#### Robin04

So, according to my lecture notes the Poisson distribution is: $\frac{(\mu t)^k}{k!}e^{-\mu t}$, where $\mu$ is the frequency of the event, in our case $\frac{2}{45}$, and $k=1$ as we're only looking for the probability of catching one rabbit. This is strategy $A$.

For strategy $B$ $\mu$ has to express the difference. My guess would be $\mu = \frac{1}{45-\epsilon}$. Is this correct?

#### PeroK

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For strategy $B$ $\mu$ has to express the difference. My guess would be $\mu = \frac{1}{45-\epsilon}$. Is this correct?
Why would that be?

#### Robin04

Why would that be?
Because one rabbit is already gone (that's for the numerator), and that was $\epsilon$ time ago so that should compensate a bit for the frenquency dropping to its half. Or I'm not sure anymore...
Edit: if $\epsilon$ was 22.5 that would mean that $\mu$ would remain constant, which seems intuitive to me.

#### PeroK

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Because one rabbit is already gone (that's for the numerator), and that was $\epsilon$ time ago so that should compensate a bit for the frenquency dropping to its half. Or I'm not sure anymore...
What happens if you see a second rabbit a few minutes later? Does that mean there will be no more rabbits for 40 minutes or so?

#### lomidrevo

where μμ\mu is the frequency of the event, in our case 245245\frac{2}{45}
Don't you think you are missing somewhere the time spent by the fox near the rabbit hole? I.e. the period of "observation"

k=1k=1k=1 as we're only looking for the probability of catching one rabbit
Not correct. You are looking for probability of catching at least one rabbit.

#### lomidrevo

Don't you think you are missing somewhere the time spent by the fox near the rabbit hole? I.e. the period of "observation"
sorry, I noticed just right now that you are using slightly different definiton of the distribution as I am used to... forget this comment.... you have $t$ already there

#### Robin04

What happens if you see a second rabbit a few minutes later? Does that mean there will be no more rabbits for 40 minutes or so?
I think yes. Because then there was two rabbits for a time of $\epsilon$ + a few minutes and considering their frequency (2 rabbits / 45 minutes) it's unlikely that we'll see a third one soon.

#### PeroK

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Because one rabbit is already gone (that's for the numerator), and that was $\epsilon$ time ago so that should compensate a bit for the frenquency dropping to its half. Or I'm not sure anymore...
Edit: if $\epsilon$ was 22.5 that would mean that $\mu$ would remain constant, which seems intuitive to me.
You need to be careful. These questions depend on the events being independent. The rabbits aren't lined up ready to to come forward, one ever 22.5 minutes. You have to imagine a large number of rabbits who turn up with the overall frequency of one ever 22.5 minutes.

#### PeroK

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I think yes. Because then there was two rabbits for a time of $\epsilon$ + a few minutes and considering their frequency (2 rabbits / 45 minutes) it's unlikely that we'll see a third one soon.
As I said above this is not correct. Let's use a different an example. A typical soccer game has a goal every 30 minutes, say. So, if it's 2-1 at half-time, does that mean that there is:

a) the usual probability of 1 goal every 30 minutes in the second half?

b) less chance of a goal in the second half, because the game already has its three goals?

#### Robin04

Not correct. You are looking for probability of catching at least one rabbit.
Hmm, maybe I have to add to it a second term where $k=2$?

You need to be careful. These questions depend on the events being independent. The rabbits aren't lined up ready to to come forward, one ever 22.5 minutes. You have to imagine a large number of rabbits who turn up with the overall frequency of one ever 22.5 minutes.
But the difference in strategy $B$ is still in $\mu$, right?

#### lomidrevo

Hmm, maybe I have to add to it a second term where k=2k=2k=2?
And what about $k = 3, 4, 5...$ ? Even though the probability is rapidly decreasing with $k$ increasing, you should not ignore it. Thankfully you don't need calculate the sum of these probabilities... There is much easier way. Hint: how do you calculate probability of the complement of an event?

#### Robin04

As I said above this is not correct. Let's use a different an example. A typical soccer game has a goal every 30 minutes, say. So, if it's 2-1 at half-time, does that mean that there is:

a) the usual probability of 1 goal every 30 minutes in the second half?

b) less chance of a goal in the second half, because the game already has its three goals?
I think I'm starting to understand what you're saying. But I'm still confused about how to express this strategy B.

I think that knowing that a rabbit has ran away $\epsilon$ time ago decreases the probability of catching another one soon. But now I'm not even sure about this one. Ah, complete confusion

And what about $k = 3, 4, 5...$ ? Even though the probability is rapidly decreasing with $k$ increasing, you should not ignore it. Thankfully you don't need calculate the sum of these probabilities... There is much easier way. Hint: how do you calculate probability of the complement of an event?
Omm, $1-p|_{k=1}$?

#### PeroK

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I think I'm starting to understand what you're saying. But I'm still confused about how to express this strategy B.

I think that knowing that a rabbit has ran away $\epsilon$ time ago decreases the probability of catching another one soon. But now I'm not even sure about this one. Ah, complete confusion
If you think of the sports analogy again. If someone has just scored a goal, why is it less likely that another goal is scored in the next few minutes? Okay, everyone knows that a goal has just been scored. But, if that doesn't change the way the game is played, then there would need to be a physical difference in the next few minutes. But, if the players are playing normally, then the next few minutes is just like any other few minutes.
This is a critical point of probability theory.

If you just miss a train (assuming a normal timetable), then you will wait longer for the next train. But this is not the random process being described here with these rabbits. Unless, therefore, the rabbits are coordinated in some way, the probability of a rabbit in the next few minutes does not depend on whether there has just been a rabbit or there have been no rabbits for over a hour.

In fact, that is the fundamental assumption in this case.

#### lomidrevo

Omm, $1-p|_{k=1}$?
This one gives you probability that fox doesn't catch exactly one rabbit. Is the same as "at least one rabbit"?
Good formula, wrong $k$.

#### Robin04

If you think of the sports analogy again. If someone has just scored a goal, why is it less likely that another goal is scored in the next few minutes? Okay, everyone knows that a goal has just been scored. But, if that doesn't change the way the game is played, then there would need to be a physical difference in the next few minutes. But, if the players are playing normally, then the next few minutes is just like any other few minutes.
This is a critical point of probability theory.

If you just miss a train (assuming a normal timetable), then you will wait longer for the next train. But this is not the random process being described here with these rabbits. Unless, therefore, the rabbits are coordinated in some way, the probability of a rabbit in the next few minutes does not depend on whether there has just been a rabbit or there have been no rabbits for over a hour.

In fact, that is the fundamental assumption in this case.
Ah, I see. But then what is different if we know that a rabbit has just gone $\epsilon$ time ago?

This one gives you probability that fox doesn't catch exactly one rabbit. Is the same as "at least one rabbit"?
Good formula, wrong $k$.
Ah, it should be $1-p|_{k=0}$

"Probability - The fox and the rabbits"

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