Geometric Distribution Puns

Say I enter a lottery that had 1/15 chance of winning per ticket. If I want to be 95% certain of winning, how many lottery tickets would I need to buy? Is this the math:

(14/15)^k= 0.95 and solve for k?

I’ve spent so long trying to figure out what the name is. If it even exists. Basically trying to calculate how many trials required for x amount of successes/failures. For example you have a given P, and want to know the average amount of trials until (10,20,1,000) successes.

Geometric random variable is the closest I’ve found, but it stops after a single success. With the other being negative binomial distribution, but needs a given amount of trials. Am I just overlooking something and making this harder than it should be? Or is there a specific name for this?

Edit: found a website with a calculator for what I’m looking for, not entirely sure what the equation is supposed to equal... https://www.anesi.com/negativebinomial.htm?p=0.5&r=20

9709 mj 2021 paper 51

Sharma knows that she has 3 tins of carrots, 2 tins of peas and 2 tins of sweetcorn in her cupboard. All the tins are the same shape and size, but the labels have all been removed, so Sharma does not know what each tin contains. Sharma wants carrots for her meal, and she starts opening the tins one at a time, chosen randomly, until she opens a tin of carrots. The random variable X is the number of tins that she needs to open.

Show that P(X = 3) = 6 /35

.

so the rules of geometric distribution are that an experiment is repeated over and over until the success is achieved and hence shouldnt the formula here used be: (4/7)^2 x (3/7) ??

can someone tell me why geometric distribution won't be applied in this question? its killing me, please answer me!!!!

Hi all!

Today I've been pondering about the geometric and exponential distributions. They're the two main memoryless distributions, while one is discrete and the other continuous (on the positive reals). So I had a pretty simple idea: You can "approximate" an exponential distribution by partitioning the positive reals as follows:

Let $E \tilde Exp(\lambda ) , \lambda > 0$. Choose $\epsilon > 0$ and define $ X = \lbrace i\epsilon \mid i \in \mathbb{N} \rbrace$. Now we can define a geometric distribution on this set. The only thing that remains to be chosen is the transition probability $p$. We choose it, such that we mimic the exponential distribution: We use the simple fact that

$P(E < (i+1)\epsilon \vert E > i\epsilon) = 1 - P(E > (i+1)\epsilon \vert E > i\epsilon) = 1-e^{-\lambda\epsilon}$

Thus we define the Variable $G_{\epsilon}$ according to the geometric distribution $Geo(X,1-e^{-\lambda\epsilon}$ .

Now, is it true that $P(G\in [a,b]) \rightarrow P(E\in [a,b])$ as $\epsilon$ goes to 0? I think it is, it's pretty intuitive this far. At least for $[a,b] = [k\epsilon ,l\epsilon]$ we have

$ P(G \in [k\epsilon , l\epsilon] ) = e^{-\lamba\epsilon l} - e^{\lambda\epsilon (k+1) = F_{E}(\epsilon l) - F_{E}(\epsilon (k+1)) $ where $F_{E}$ is the CDF of $E$.

If so, we can think of a Poisson-Process $P_t$ where the interarrival times are i.i.d. exponentially distributed. Could we define any interesting Stochastic Processes (maybe a Markov-Chain) using $G_{\epsilon}$ that relate in any way to the Poisson-Process (or it's embedded Markov-Chain)?

Thanks in advance!

Hey all, new here so I hope this is an appropriate question for this subreddit.

I am currently researching replay reviews in the NBA. Specifically, I wish to look at possible changes to the coaches challenge rule.

To summarise, currently the rule is that both teams get only 1 challenge per game. However there has been a push to change this from fans to allow teams to get an additional challenge if they successfully overturn the call being challenged. Basically, a team can continue to challenge until they are unsuccessful once, and then they lose their ability to challenge altogether.

The problem with this change however is that this would add additional challenges to a game that arguably has too many. But I want to go further than just saying there are more, I want to quantify how many more on average.

Sounds like your typical geometric distribution right? Well not really, cause you aren’t necessarily ending with a success (being an unsuccessful challenge in this case). Basically, there’s 4 cases:

Case 1: Team never challenges anything

Case 2: Team fails their first challenge

Case 3: Team successfully challenge some number of plays but never unsuccessfully challenged (i.e. still had the ability to challenge at the end of the game)

Case 4: Team successfully challenged some number of plays but also unsuccessfully challenges a play (causing them to lose their ability to challenge)

At first I decided to just assume teams would always fail a challenge at some point, but teams only ever used their challenge roughly 25% of the time when they only had one, so that indicates to me this isn’t really a realistic assumption (right?)

My next thought was maybe somehow adding a probability (corresponding to the chance they use their challenge if they have it) that each trial occurs? But not really sure how to go about this or if it’s even possible.

I also tried splitting into a sum of two random variables but this similarly didn’t go too far…

Any ideas as to how I go about this? I’m fine with making assumptions given they are somewhat reasonable (I’m just making a video for fun not to be highly accurate).

Happy to clarify anything further if needed as I know people might not be familiar with the NBA and it’s rules.

I'm looking at the proof for this distribution and there is a step that I don't understand:

-p d/dp ( sum_y=1 ^infty (1-p)^y ) = -p d/dp (1-p)/p

I'm not sure how it goes from sum (1-p)^y to (1-p)/p

Is there a property that I don't know about?

Edit: I figured it out! You can use the a+ar+ar^2+ar^3+... = a/(a-r) property if anyone was wondering.

I have a probability question about independent random variables on some probability space using geometric distribution and I am stuck on how to solve it. The question is as follows

"let p,q be in (0,1), p+q=1. X and Y are the independent random variables on some probability space with X and Y having a geometric distribution with parameters p and q, meaning: P(X=n) = P(Y=n) (q^n)*p, n in the Natural Numbers including 0.

Find P(X=Y)"

Any help in answering this would be massively appreciated! Thanks so much

Quick question about Geometric Distributions:

Are they used to find the probability that the first success will occur on exactly the Yth trial with (Y-1) failures prior

OR

Are they used to find the probability that the first success will occur in Y trials (so it could occur on any trial 1 to Y).

The geometric probability density function builds upon what we have learned from the binomial distribution. In this case the experiment continues until either a success or a failure occurs rather than for a set number of trials. There are three main characteristics of a geometric experiment.

1. There are one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a “success” so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP.
2. In theory, the number of trials could go on forever.
3. The probability, p, of a success and the probability, q, of a failure is the same for each trial. p + q = 1 and q = 1 − p. For example, the probability of rolling a three when you throw one fair die is 1\6. This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three.
4. X = the number of independent trials until the first success.

There are three main characteristics of a geometric experiment.

1. There are one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a “success” so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP.
2. In theory, the number of trials could go on forever. There must be at least one trial.
3. The probability, p, of a success and the probability, q, of a failure is the same for each trial. p+q=1p+q=1 and q=1−pq=1−p. For example, the probability of rolling a three when you throw one fair die is 1616, the probability of a failure. The probability of getting a three on the fifth roll is X=X= the number of independent trials until the first success.

Hello, on my quest about learning probability I have found an exercise which I believe to have solved but no solution was provided for me to double check.

It goes like this:

1. Out of 3 cards, one is chosen uniformly at random and their values are: 5,6,7.

2. There is a box with 6 white balls.

3. The value of the card drawn represents how many black balls are put into the box which contains white balls. Then, balls are drawn WITH REPLACEMENT.

I am asked to find the PMF of T=First time a black ball is drawn.

REASONING AND SOLUTION?

Since I have to find the probability of first success of independent events (there is replacement), I thought that the Geometric distribution was the right tool.

I thought that T would be like this:

P(T=k)=P(T=k|5)*P(5)+P(T=k|6)P(6)+P(T=k|7)P(7)

which leads to: (1/3)(5/11)(6/11)^(k-1) + (1/3)(6/12)(6/12)^(k-1) + (1/3)(7/13)(6/13)^(k-1) = P(T=k)

Is this correct? Thanks.

what type of questions would they ask for those w/o calculator use

How can I show that? Thanks.

http://imgur.com/gallery/TxulT6S

I did the easiest one.

Someone else already done normal(or binomial).

Who will go for a harder one like log-normal or student-t?

An event has a 1/420 chance of happening each trial for the first 420 trials and then it doubles, becoming 1/210.

I thought the probability function for the number of trials needed to succeed would be (1/420)(419/420)^(x-1) for x=1,2,...,420 and then (419/420)^(420)(1/210)(209/210)^(x-1) for x=421, 422,..., n,....

I threw in some random values for x and they seemed to make sense, but when I went to calculate the expected value and it was weirdly low. Then I tried summing up just the probabilities and saw they don't even sum up to 1, so that's not even a probability function.

What am I missing here?

question

I’ve done parts 1 to 4 using Geometric and Binomial Distribution and got them right but I don’t understand how I’m meant to do part 5. The answer in the mark scheme is 0.0466 but I don’t get their method (9 x 0.88^8 x 0.12^2). Thanks to any1 that helps.

Hi, I'm calculating the expectation of the geometric distribution and I'm confused why my intuition doesn't work out.

E(X) = Σ(n=0 to ∞) * n * q^n * p where q = (1-p) = p * Σ(n=0 to ∞) * n * q^n

This (Σ(n=0 to ∞) * n * q^n) to me looks like the geometric series p * (n/(1-q)), but this implies E(X) = n which it's not. What's going wrong here? Is it because n appears both as the base and the multiple?

The derivations I've seen resort to using the derivative which results in the true E(X) = q/p.

I was looking at the proof of the memoryless property for geometric distributions, and I ran into a line which says,

P(X>=x) = (1-p)^x for x = 0,1,2,...

Can someone explain why that is true. Thanks

Hello, I've been trying to prove the mean/expected value of the geometric distribution (it should be 1/p), and was wondering if anyone could help me with the last step?

I start of with

E(X) = ∑ x*P(X=x)

which for a geometric distribution gives

E(X) = ∑ x*(p(1-p)^((x-1)))

=p∑x*(1-p)^((x-1))

The issue is I have know idea how to evaluate the summation, I tried writing out a few terms

1 + 2(1-p) + 3(1-p)^(2) + 4(1-p)^(3) + 5(1-p)^(4)...

I tried some substituting some values for p and it seemed to tend to 1/p^(2) which is what would be required for the proof, but this might just be a coincidence, so I was wondering if anyone knew how to evaluate this series.

Alternatively if someone knows a better way to prove E(X) for a geometric distribution, that would be great too!

**NOTE** - I'M USING SIGMA TO MEAN THE SUM TO INFINITY RATHER THAN JUST SUM AS I DON'T KNOW HOW TO WRITE THAT IN REDDIT.

Thank you

Hi all,

This is my second post regarding statistics in the SWCT app. The first (and probably more useful) post on the binomial distribution along with how you can calculate your specific probabilities for a particular insert is found here https://redd.it/37mvo8. So there are still a lot of posts by people who were devastated in not receiving an insert after X number of packs and the counterpoint to that of people pulling in their first or second pack. Hopefully I can shed some light on that for those who want to learn a bit about stats and have the time to read this.

Thought experiment:

In this example we will use the series 2 vintage 1 insert. 2500 count with 1/80 probability of pulling. Let’s say I tell 2500 people to open packs until they get one vintage insert and then stop, assuming they have unlimited credits (The rules here are strict- they must stop after getting an insert). So by the end every person will have one insert. How would the numbers break down? Think about this by yourself for a second before reading on: How many people would be ‘first pack’ pullers? How many people would pull on exactly the 80th pack? 100th? Do you have a better chance of getting your first insert on the first pack or exactly on the 80th?

I ran a single simulation of the experiment. Then I counted the number of people who pull in one pack, the number of people who pull in two packs, in three packs, and so on until the end into a histogram shown here: http://i.imgur.com/4rekhLt.jpg So for example, if you wanted to find the number of people pulling in exactly 200 packs, go along the horizontal axis until 200, and you will see about 5 people in that group. The number of people pulling in one pack? About 35. And then there are people that take 400-600 packs, though a lot fewer. Don’t focus on the specific details but notice the general graph trends.

Notice how a lot of people pull the insert early! This number drops off and continuously drops off for each additional pack until you get to those poor people who need 500-600 packs to pull a 1 in 80 insert. So how does this make sense that there are actually more people pulling it in exactly one pack than people in exactly 80 and so on? Everything can be explained by the geometric distribution (https://en.wikipedia.org/wiki/Geometric_distribution). Think about it like this- the chance you pull on the first pack is 1/80. In the example, with 2500 people pulling, there should be 2500/80 or 35 people getting it first pa

Title, but I am a bit too tired to enter in calculations, come up with a spreadsheet, but i NEEDED to post this to the subreddit so that another fellow statistics major can pick up the slack where i fail (sorry, midterm tomorrow)

This is my logic. based on the assumption that you will have 1 leg in 400 blood shards (based off of 5 shard pieces) thats a leg item ~ every 80 items that you buy from kadala.

We will use this 1/80 to calculate expected amount of blood shards spent

If you are searching for the elusive Cindercoat: this is a calculation that you will use.

Geometriccdf (CDF because you dont care the probability that it will happen on specifically any trial, just that you get it in n trials) probability of success is .0625, now you just have to select your N legendaries note, when you select N legendaries, its 1 legendary per 400 shards, ie, you should expect 5 legendaries in ~ 2k shards

say i am willing to gamble 4k shards for a cindercoat, thats 10 legendaries.

Geometriccdf(probability, lower bound, upper bound) where you have probability = .0625, lower bound = 1 (need at least 1 legendary to find a cindercoat), upper bound = 10, (im only willing to spend 4k shards, which is ~10legs) Geometriccdf(.0625, 1, 10) is what you can put into a calculator if you have one with the capability / you have the know how

This yields a probability of .47554... so you have a 47% chance of actually acquiring a cinder coat given you gambled 10 legendaries from your 4k blood shards. I hope that someone can use this info to help build a table suitable for the community, love you guys.

Source for leg drop rate on cindercoat goes to the spreadsheet i acquired from /u/oTradeMark when he made this post: http://www.reddit.com/r/Diablo/comments/24qhi9/sortable_spreadsheet_for_new_legendary_items_drop/

Thanks for your time, I must sleep

TL;DR: 100k shards looking for wand of woh yields about a 25% chance of finding it, based off of my assumptions. 47% chance of finding a cindercoat in 4k shards. I implore more people to post more numbers or ask me questions to calculate their chances with a given number of shards spent. Good night!

EDIT: Not going to do all the math, but if you want a wand of woh from her, dont fucking gamble it- im fairly certain you would spend on the order of ~100k shards before you found one because you are gambling 15 shards at a time for a 1h weapon

EDIT2: did the math (1/4 weapons would be a wand, 1/66 will be a wand

There are 3 balls numbered from 1 to 3, we take the balls out without returning them one after the other. We'll mark X as the number of the first ball, and Y as the number of the second ball. Calculate E(XY).

I don't understand why the answer is 3.66 and not 4.

Thanks a lot in advance !

Sharma knows that she has 3 tins of carrots, 2 tins of peas and 2 tins of sweetcorn in her cupboard. All the tins are the same shape and size, but the labels have all been removed, so Sharma does not know what each tin contains. Sharma wants carrots for her meal, and she starts opening the tins one at a time, chosen randomly, until she opens a tin of carrots. The random variable X is the number of tins that she needs to open.

Show that P(X = 3) = 6 /35

.

so the rules of geometric distribution are that an experiment is repeated over and over until the success is achieved and hence shouldnt the formula here used be: (4/7)^2 x (3/7) ??

can someone tell me why geometric distribution won't be applied in this question? its killing me, please answer me!!!!