Probability Help (1 Viewer)

phamtom44 · Jul 6, 2017

Hey guys, I ran into this question today:

Catch any pokemon every 5 minutes on average, 42 different pokemon total, how long before you have caught every kind of pokemon? you can use whatever program you want, need to explain assumptions.

I was thinking of using a Poisson process/Exponential distribution but I'm stuck.

InteGrand · Jul 6, 2017

phamtom44 said:
Hey guys, I ran into this question today:

Catch any pokemon every 5 minutes on average, 42 different pokemon total, how long before you have caught every kind of pokemon? you can use whatever program you want, need to explain assumptions.

I was thinking of using a Poisson process/Exponential distribution but I'm stuck.

$\noindent We can find the distribution of the desired time in the following way. Let $n$ be the total types of Pok\'emon (in your case, $n = 42$). I will assume Pok\'emon arrive according to a Poisson process with mean interarrival time 5 minutes and conditional on an arrival of a Pok\'emon, any of the $n$ types of Pok\'emon are equally likely to be the one that appears, independently of the arrival process. I will also assume that when a Pok\'emon appears, it is always successfully caught and the catching takes no time.$

$\noindent Let $\lambda$ be the rate of arrival of Pok\'emon per minute. For your case, $\lambda = \frac{1}{5}$. In general, it is the inverse of the mean time between arrivals. Due to our assumptions, we have a ``thinned'' Poisson process: it can be shown that the arrivals of each Pok\'emon type form independent Poisson processes with rate parameter $\frac{\lambda}{n}$ (I'll give a link to some information about thinned Poisson processes at the end for further reading). Therefore, if we let $T^{(i)}$ be the random variable denoting the time of first arrival of the $i$-th type of Pok\'emon ($i = 1,2,\ldots, n$), the $T^{(i)}$ are iid rate $\frac{\lambda}{n}$ exponential random variables, with CDF $F(t) = 1-e^{-\frac{\lambda}{n} t}$, $t \geq 0$.$

$\noindent Now, the time at which we have caught all $n$ types of Pok\'emon is the random variable$

$M= \max \left\{T^{(1)},\ldots ,T^{(n)}\right\}.$

$\noindent Therefore, the support of $M$ is $[0,\infty)$ and the CDF of $M$ is $F_{M}(t) \equiv \mathbb{P} (M\leq t) = \left(1 -e^{-\frac{\lambda}{n} t}\right)^{n}$, $t \geq 0$. This gives us the distribution of $M$.$

• Poisson thinning: http://www.math.uah.edu/stat/poisson/Splitting.html

InteGrand · Jul 6, 2017

InteGrand · Jul 6, 2017

$\noindent Alternatively, if you only care about the expected time to catch all types of Pok\'emon, you can note that $\mathbb{E}[M] = \frac{n}{\lambda}\sum_{k = 1}^{n}\frac{1}{k}$. One way to show this is to consider the time to arrival of a new type of Pok\'emon at each stage that you just catch a Pok\'emon type that hadn't been caught before. It is explained here:$

http://www.stat.berkeley.edu/~mlugo/stat134-f11/exponential-maximum.pdf

$\noindent and is essentially a continuous time version of the coupon collector problem:$

https://en.wikipedia.org/wiki/Coupon_collector's_problem .

InteGrand · Jul 6, 2017

So for your example (42 total types of Pokémon and 5 minute mean arrival time), E[M] comes out to be about 908.6, so about 15 hours, 8 minutes and 36 seconds: http://www.wolframalpha.com/input/?x=0&y=0&i=42*5*(sum+from+k+=+1+to+42+of+1/k) .

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phamtom44

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