### Coupon collector problem simulation

Assuming that the mouse is initially placed in Area 1, what is the expected number of moves in order for the mouse to find food?

## Coupon Collector’s Problem

Problem 8 Consider the maze that is described in Problem 6. Assuming that the mouse is initially placed in Area 1, what is the probability that the first area of food reached by the mouse is Area 9? Problem 1 and Problem 2 has a natural interpretation as a Markov chain. As the coin is tossed repeatedly, the chain moves through these 5 states. One characteristic of a Markov chain is that the probability of moving from one state to the next depends the current state the last state visited but not on the states prior to the current state, i.

Before any toss is made, the state is 0. Suppose the first toss is Head.

The second toss could be Head the Markov chain then moves to state HH or could be Tail the chain then moves to state 0. Suppose the second toss turns out to be a Tail. Then the next toss could be a Head the chain then moves to the state H or could be a Tail the chain continues to stay at state 0.

• bt broadband best deals existing customers;
• Coupon collector’s problem | Level Up Lunch?
• my pink vanity coupon code 2019.
• macys coupons wallets.

A good framework is to describe such movements in a matrix called transition probability matrix. To answer Problem 2, just raise this matrix to the 9th power see here. Problem 1 is more involved see Example 2 in this post. Another advantage of using Markov chains for these problems is that the method scales up quite easily.

For example, for the occupancy problem Problems 3, 4 and 5 , if the number of cells is higher than 6, it is quite easy and natural to scale up the transition probability matrix to include additional states. Then proceed with the same method. There is a learning curve at first. But once the concept of Markov chains is understood, the probability problems described here or other similar problems can be tackled quite easily and routinely.

For a basic discussion of Markov chains, see the early posts in this companion blog. The problem is usually stated as a coupon collector trying to collect the entire set of coupons. For example, each time the coupon collector buys a product e. Suppose that there are different types of coupons prizes and that the coupon collector wishes to collect the entire set.

How many units of the product does the coupon collector have to buy in order to collect the entire set?

## Coupon collector's problem

A simplified discussion of the coupon collector problem is found in another blog post. This post is more detailed discussion. This blog post in another blog discusses the coupon collector problem from a simulation perspective. As shown below, if there are 5 different coupons, it takes 12 purchases on average to get all coupons.

If there are 10 different types of coupons, on average it would take 30 purchases. If there are 50 different types of coupons, it would take on average purchases collect the entire set. The first few coupons are obtained fairly quickly. As more and more coupons are accumulated, it is harder to get the remaining coupons.

For example, for the coupon case, after the 49 coupons are obtained, it takes on average 50 purchases to get the last coupon. Suppose that the coupon collector does not want to collect the entire set and only wishes to collect distinct coupons where.

It turns out that this special case only requires a minor tweak to the case of collecting the entire set. Our strategy then is to focus on the main case. The special case will be discussed at the end of the post.

1. Associated Data!
4. coupon reduction amazonia.
5. Mathematics > Probability?
6. american girl coupons feb 2019;
7. We first consider the main case that the coupon collector wishes to collect the entire set. The problem can be cast as a random sampling from the population. Selecting a number at random from with replacement is equivalent to the coupon collector receiving a coupon. Let be the minimum number of selections such that each number in is picked at least once. In this post we discuss the probability function of as well as its mean and variance. Another interpretation of the problem is that it can be viewed as an occupancy problem, which involves throwing balls into cells at random.

The random quantity of interest is the number of balls that are required to be thrown such that each cell has at least one ball.

### The problem

Clearly this formulation is identical to the coupon interpretation and the random sampling interpretation. The angle of occupancy problem is useful since we can leverage the formulas developed in this previous post. A description of the occupancy problem is given here. Regardless of the interpretation, the goal is obtain information on the random variable , the minimum number of random selections from in order to have the complete set of distinct values represented in the sample.

The mean and variance of are easier to obtain. So that is where we begin. The key is to break up into a sum as follows:. For example, is the number of random selections to get a number that is distinct from the two distinct numbers obtained up to that point. Note that each involves repeated sampling until some criterion is reached, thus resembling a geometric random variable.

Indeed they are. As the sampling continues and as more distinct values are obtained, it is not as easy to obtain a new number. After distinct numbers have been obtained, the probability of drawing a new distinct number is. As geometric random variables, each has the following mean and variance. Note that the random variables are independent. The value of does not depend on how many trials it takes to draw the previous distinct numbers. The following gives the mean and variance of. We can work this out, using the formal definition of the variance.

The great thing about the Markov property is that the probability of moving to a given state depends only on the previous state. For this simple model the probability of moving up or down is actually independent of state. In other words, the formula turns out to be. After one hundred time steps the variance is one hundred.

## The coupon collector urn model with unequal probabilities in ecology and evolution

In fact over time, the probability of returning to the origin diminishes. Imagine instead if the coin had a slight amount of bias i. We can see the impact of a slight bias by varying the probability of a win in the simulation below. The more we increase the probability the smaller the variance becomes. This is the original algorithm used by Google to rank web pages. A more in depth look can be found here.

When it arrives at a web page, it finds all the links and then picks one at random to click on. Each time the walker passes through a page we record this by adding a point to the page. After a long time the pages can be ranked by how many times the walker has visited it. One issue is that the rank number for each page keeps increasing. We can simulate the process on a small random network below.

The colour denotes how recently the walker had visited that page and the size denotes its PageRank. In reality there are far quicker ways to calculate the PageRank than to just simulate it. Of course you cannot find a solution per se, but can only make an estimate--and there's no reasonable upper bound to that estimate, because there could be arbitrarily many objects having vanishingly small probabilities. What, then, do you mean by "a way forward" exactly? Completely glossed over that detail. For instance, perhaps each unobserved object only occurs once in an infinitely large population.