An Introduction to the Geometric Distribution

  Рет қаралды 311,999

jbstatistics

jbstatistics

10 жыл бұрын

An introduction to the geometric distribution. I discuss the underlying assumptions that result in a geometric distribution, the formula, and the mean and variance of the distribution. I work through an example of the calculations and then discuss the cumulative distribution function.
For those using R, here is the R code for the example in this video:
NB R uses a different definition of the random variable than I do here. I define the random variable X to be the number of trials required to get the first success. R defines the random variable to be the number of failures before getting the first success (let's call this Y). Then Y = X - 1, and we'll have to make this adjustment when using dgeom, pgeom, or rgeom. Some might find this confusing, and if you do, don't use these functions.
Sampling from a large population where 30% have CPR training until we get the first person with CPR training.
Finding the probability that it happens on the sixth person sampled:
(.3)*(.7)^5
[1] 0.050421
or
dgeom(6-1,.3)
[1] 0.050421
Finding the probability that it happens on or before the third person sampled:
.3+.3*.7+.3*.7^2
[1] 0.657
or
1-.7^3
[1] 0.657
or
pgeom(3-1,.3)
[1] 0.657

Пікірлер: 187
@GMan-um1pi
@GMan-um1pi 7 жыл бұрын
I go to one of the best universities in the world and you taught this infinitely better than my professor, thank you.
@jbstatistics
@jbstatistics 7 жыл бұрын
You are very welcome. I'm glad I could be of help!
@sharathnair1702
@sharathnair1702 6 жыл бұрын
which university
@qwertytwerqy2440
@qwertytwerqy2440 Жыл бұрын
@@sharathnair1702 asking the real questions
@keanaleong7745
@keanaleong7745 4 жыл бұрын
WAAAAAAAY better than my lecturer - You deserve my tuition fees! Thank you for explaining in such a simplified manner!
@tanvirkaisar7245
@tanvirkaisar7245 8 жыл бұрын
You are the best statistics teacher i have ever found!
@shengchuangfeng227
@shengchuangfeng227 6 жыл бұрын
Agree.
@mohammedelghitany9153
@mohammedelghitany9153 4 жыл бұрын
I agree, both mathematically and intuitively
@alexcipriani6003
@alexcipriani6003 4 жыл бұрын
@@theethatanuraksoontorn2517 if JB is not intuitive you have an intuition problem
@yencyperez8362
@yencyperez8362 3 жыл бұрын
Completely agree!
@Elfajito
@Elfajito 8 ай бұрын
we can agree with a 95% of confidence that jb is the best statistics teacher based on a sample of 5 comments
@george4746
@george4746 4 жыл бұрын
Your words make perfect sense to me. I wonder why professors and books like to make things overcomplicated than it really is.
@julieye5832
@julieye5832 8 жыл бұрын
It doesn't get any clearer than this. Thank you!!
@jbstatistics
@jbstatistics 8 жыл бұрын
+Julie Ye You are very welcome Julie! Thanks for the compliment!
@sunnywang7501
@sunnywang7501 3 жыл бұрын
I read my textbook three thousand times about geometric distribution and was still confused. 10 minutes video of yours is able to make me understand what my textbook has tried to explain to me for the past hour. Thanks so much. Honestly youtube videos like these have been such good friends of mine for years. They always explain concepts better than my profs.
@MrMarshmellowman22
@MrMarshmellowman22 5 жыл бұрын
jesus christ man, you make everything make sense. i didn't understand the hyper geometric and Geo metric formulas and you made them very simple to understand, love you work
@abhinavbichal8798
@abhinavbichal8798 2 жыл бұрын
You are the best statistics teacher I have ever found!
@hvsampad4554
@hvsampad4554 4 жыл бұрын
You are one of the best stats teacher ever seen sir. you make even critical concepts lucid!!!
@jeskow19
@jeskow19 4 жыл бұрын
These videos are better than any formal instruction I ever had in undergrad, masters, MBA....anything. Don't go to school kids. Just find the right youtube channel.
@JAlternative106
@JAlternative106 8 жыл бұрын
These are the greatest. I can't express how thankful i am to your clear explanations. Thank you so much
@jbstatistics
@jbstatistics 8 жыл бұрын
+Jbaker8390 Thanks! And you're very welcome!
@zingg7203
@zingg7203 8 жыл бұрын
This saves me from cancer. Thank you!
@jbstatistics
@jbstatistics 8 жыл бұрын
You are welcome!
@abdallahelshinawy5536
@abdallahelshinawy5536 7 жыл бұрын
how this saved you from cancer ?!
@dhidhi1000
@dhidhi1000 7 жыл бұрын
relying on crappy probability teachers that use crappy books gives you cancer, this is scientifically proven.
@epsilonoclon7371
@epsilonoclon7371 7 жыл бұрын
i can sustain this theory myself!
@GuppyPal
@GuppyPal 2 жыл бұрын
Your videos are so simple and clear... You are a great educator! Thank you!
@bobcavanagh1482
@bobcavanagh1482 2 жыл бұрын
Just discovered this video. Concise, precise, excellent. I'm adding it to my A-Level scheme of work.
@probono2876
@probono2876 8 жыл бұрын
What a great series on statistics, fantastic contents and presentation. Many thanks for that.
@jbstatistics
@jbstatistics 8 жыл бұрын
+pro bono What a nice compliment! Thanks!
@akiyoshii1757
@akiyoshii1757 7 жыл бұрын
Thank you. I've just learned it less than 15mins, I have my presentation tomorrow about this, thanks for this
@zakariasaidy7783
@zakariasaidy7783 7 жыл бұрын
I never comment on videos but you just saved my life, I was having trouble with binomial, negative binomial and geometric distribution but now its all clear thanks to you. Thank you a million times. God bless you.
@jbstatistics
@jbstatistics 7 жыл бұрын
You are very welcome!
@zli-eo8xg
@zli-eo8xg 7 жыл бұрын
excellent explanation, 100 times clearer than the instruction book
@christophergonzalez1464
@christophergonzalez1464 5 жыл бұрын
5 years later and you still have students coming to your videos for help. Thank you sir
@jbstatistics
@jbstatistics 5 жыл бұрын
You are very welcome. I tried to make them stand the test of time :)
@sg5sd
@sg5sd 3 жыл бұрын
Make it 6 years.
@aymenechchalim4654
@aymenechchalim4654 Жыл бұрын
@@sg5sd make it 9 mam
@user-nv7dz9cp3u
@user-nv7dz9cp3u 4 ай бұрын
@@aymenechchalim4654 make it 10, @jbstatistics 🤩🥰
@AtlasMvm
@AtlasMvm 7 жыл бұрын
one of the best instructional math videos I have ever seen
@jbstatistics
@jbstatistics 7 жыл бұрын
Thanks! That's quite the compliment!
@achuthadivine
@achuthadivine 9 жыл бұрын
Amazing Playlist :) You made my day :)
@rishabhnarula1999
@rishabhnarula1999 10 ай бұрын
great explanation sir, and very well presented, really cleared up my doubts and confusions regarding this topic.
@kuldeeplakheshwar7811
@kuldeeplakheshwar7811 3 жыл бұрын
I don't have words to express my thankfulness 😌
@chitralarora2012
@chitralarora2012 5 жыл бұрын
bro you are the best never ever got so much clarity about distributions
@jbstatistics
@jbstatistics 5 жыл бұрын
Thanks! I'm glad to be of help!
@Daniel-aaaaa
@Daniel-aaaaa 2 жыл бұрын
You helped me pass my probability course. Just wish there was more stuff like chebyshev's inequality, but the videos explaining the common distributions are all golden.
@jbstatistics
@jbstatistics 2 жыл бұрын
I'm glad to be of help! I hope to add more videos in the near future.
@AshrafulAlam-
@AshrafulAlam- 3 жыл бұрын
Sir, you are doing really great. We, students in statistics really thankful to you. Please do more tutorials on statistics. (From Bangladesh)
@alexisbader5189
@alexisbader5189 7 жыл бұрын
This was absolutely fabulous, thank you for a wonderful clear explanation of the geometric distribution
@jbstatistics
@jbstatistics 7 жыл бұрын
Thanks for the very nice compliment! I'm very glad I could be of help.
@kajolandheriya7917
@kajolandheriya7917 Жыл бұрын
Thank you for such nice explanation, I clearly understood now, please try to make such more videos on data science concepts.
@andrewmackechnie6594
@andrewmackechnie6594 3 жыл бұрын
So helpful! Thank you for making these videos
@davidmungai6048
@davidmungai6048 5 жыл бұрын
Very clear and precise explanation
@harisrg92
@harisrg92 7 жыл бұрын
I am not surprised this video has 0 dislikes. YOU ARE AWESOME MAN!!! THANK YOU SO MUCH!!
@jbstatistics
@jbstatistics 7 жыл бұрын
Thanks for the compliment! I'm sure the dislikers will come out of the woodwork eventually :) For now I'll be content with the 435:0 ratio.
@hellodarknessmyoldfriend2976
@hellodarknessmyoldfriend2976 7 жыл бұрын
jbstatistics make that a 492:0 like to dislike ratio
@clintonpambayi893
@clintonpambayi893 3 жыл бұрын
Danki Sir you nailed it!!!!!!!!!!!!! You are the best
@jean-francoisgirouard1133
@jean-francoisgirouard1133 9 жыл бұрын
Thank you! You explain at least 4012395719875 times better than my teacher ;)
@mothusitamajasi1094
@mothusitamajasi1094 2 жыл бұрын
Great explanations, you are the best
@muskp
@muskp Жыл бұрын
Short, Crisp, excellent explanation
@tenzin8773
@tenzin8773 6 жыл бұрын
Great video! Clear and concise. Thank you very much!
@jbstatistics
@jbstatistics 6 жыл бұрын
You are very welcome! Thanks for the compliment!
@flaviabeqiri
@flaviabeqiri 6 жыл бұрын
Fantastic Job! Thank you very much!!!
@kiranthota5137
@kiranthota5137 2 ай бұрын
Great explanation, while reading in wiki i realized that we have 2 diff type of Geometric distributions, 1. random variable is no. of trials for 1st success 2. random var is no. of failures to see 1st success. which is very important while conducting the experiment, which i think missed in this current video lecture. thanks.
@jbstatistics
@jbstatistics 2 ай бұрын
I think bringing that up in an introductory video on the geometric distribution does more harm than good. It's an extra layer of confusion that people don't need at first. The difference in the random variables, the difference in the means, describing why the variances are the same...it just takes away from the big picture of what the geometric distribution does for us. Sure, I bring it up elsewhere, especially as I use R in my courses and R uses the other definition of the r.v., but I think it would cause more confusion than it's worth in an intro video. Once one is understood, the other comes naturally.
@yevseldev
@yevseldev 2 жыл бұрын
Wow. Im really impressed. Thanks alot.
@yazanziad6718
@yazanziad6718 Ай бұрын
THANK YOU SO MUCH from Jordan 🇯🇴 🌹
@annalam8624
@annalam8624 7 жыл бұрын
it helps me a lot! thank you!
@blahmonster1234
@blahmonster1234 9 жыл бұрын
Excellent videos!
@thestupidsofheaven2042
@thestupidsofheaven2042 7 жыл бұрын
every time I fail my exam, I watch these videos..! what a series of wonderful video tutorials..!
@jbstatistics
@jbstatistics 7 жыл бұрын
Thanks for the compliment! I'm glad I could be of help!
@natemcclintock2258
@natemcclintock2258 8 жыл бұрын
Thanks for this, I couldn't understand my textbook but this was so helpful!
@jbstatistics
@jbstatistics 8 жыл бұрын
+Nate McClintock You're welcome Nate! I'm glad you found it helpful!
@marziehdo6788
@marziehdo6788 8 жыл бұрын
I really couldn't get why P(X>x) is calculated like that, Now it is really clear to me.Thanks alot, well explained
@jbstatistics
@jbstatistics 8 жыл бұрын
You are very welcome.
@thepegu8287
@thepegu8287 4 жыл бұрын
Thanks JB🔥. wish my professors taught like u😔
@LNCMD2023
@LNCMD2023 3 жыл бұрын
This is better than the statistics book I am using now. It only gives the formula but does not explain how it was derived.
@salvatoregiordano6816
@salvatoregiordano6816 5 жыл бұрын
Thank you sir!
@Ceilvia
@Ceilvia 5 жыл бұрын
Thank you I feel educated
@evandevapriya9653
@evandevapriya9653 10 жыл бұрын
Thanks, helped a lot :)
@YashwithQuantumDots
@YashwithQuantumDots 2 жыл бұрын
You the best!!
@MrAlbashiri
@MrAlbashiri 8 жыл бұрын
you are the best. thank you for the video
@jbstatistics
@jbstatistics 8 жыл бұрын
+Mohammed Al-Bashiri You are very welcome!
@smartdesignengineering
@smartdesignengineering 5 жыл бұрын
Your videos are too awesome.
@jbstatistics
@jbstatistics 5 жыл бұрын
Thanks!
@GOODBOY-vt1cf
@GOODBOY-vt1cf 4 жыл бұрын
thank you so much
@muhammadkhan-lb8rx
@muhammadkhan-lb8rx 8 жыл бұрын
Fantastic .
@thomasjefferson6225
@thomasjefferson6225 2 жыл бұрын
Anyone know where to find a good video about this using geometric series mathematically? I gotta answer a question on it, and this guy is an amazing teacher.
@sahabkausar3097
@sahabkausar3097 8 жыл бұрын
Tnxxx for giving me a such ideas of geometric distribution but can u plz tell me how I can find the Mgf of Hyper_Geometric distribution..
@Mohammadalhashash
@Mohammadalhashash 2 жыл бұрын
Very helpful
@shikhagupta6409
@shikhagupta6409 8 жыл бұрын
Thanks a lot :-)
@bhavikbitspilani1186
@bhavikbitspilani1186 5 жыл бұрын
One of the best
@jbstatistics
@jbstatistics 5 жыл бұрын
Thanks!
@MarkGingrass
@MarkGingrass 7 жыл бұрын
Very useful explanation.
@jbstatistics
@jbstatistics 7 жыл бұрын
Thanks!
@kanikabagree1084
@kanikabagree1084 4 жыл бұрын
Hey thankyou somuch for this video helped alot but can you please explain the difference between when to use geometric and negative binomial distributions
@amaa3619
@amaa3619 8 жыл бұрын
Thanks so much. Your videos are a life saver 😀😀😀
@amaa3619
@amaa3619 8 жыл бұрын
Your presentation is so much easier to understand than the textbooks
@jbstatistics
@jbstatistics 8 жыл бұрын
+Ama opokua Asomani-Adem Thanks! And you are very welcome!
@DaiMoscv
@DaiMoscv Жыл бұрын
Crystal 🔮 clear
@eliasbiral
@eliasbiral 9 жыл бұрын
Hi dude your classes are great! What software do you use to make the slides and write on them as you make a video? Thanks
@claudiamesaaparicio8517
@claudiamesaaparicio8517 4 жыл бұрын
bravo!
@marketcycles2399
@marketcycles2399 8 жыл бұрын
thanks a lot sir
@caribbeankpoplover
@caribbeankpoplover 6 жыл бұрын
Ugh who even had the time to come up with all of this?! Anyways, thank you so much for your help!
@jbstatistics
@jbstatistics 6 жыл бұрын
You're welcome. The geometric distribution comes up frequently in theory and practice -- it's not just an obscure abstract notion.
@doukas21original
@doukas21original 3 жыл бұрын
Hello, is the text used in the videos available anywhere? I would like to use it as notes
@Darieee
@Darieee 5 жыл бұрын
Great explanation
@jbstatistics
@jbstatistics 5 жыл бұрын
Thanks!
@harshilandhariya7799
@harshilandhariya7799 6 ай бұрын
Thank you👍👍
@jbstatistics
@jbstatistics 6 ай бұрын
You are very welcome!
@covidnineteen5249
@covidnineteen5249 4 жыл бұрын
Thanks bro
@bhavadeepbhukya5926
@bhavadeepbhukya5926 2 жыл бұрын
thankyou sir
@rafatshaikh8516
@rafatshaikh8516 5 жыл бұрын
I want the formula for calculating the quartile deviation of geometric distribution
@messynkocierh1917
@messynkocierh1917 6 жыл бұрын
nice one bruh
@adminenetzone6330
@adminenetzone6330 9 ай бұрын
I couldn't understand why p is the parameter of the geometric's PMF, while it's a constant. Can you explain that to me please... Thanks for the video
@tajammalnawaz5552
@tajammalnawaz5552 7 жыл бұрын
one of precious play list for my type student who fail to understand in class
@jbstatistics
@jbstatistics 7 жыл бұрын
I'm glad to be of help.
@triassic995
@triassic995 10 жыл бұрын
Hi, could geometric distribution be applied to 3 outcomes? Somewhat like how multinomial distribution is an extension of binomial distribution. Thanks!
@jbstatistics
@jbstatistics 10 жыл бұрын
If you're asking about the distribution of the number of trials required to get a certain number of successes (e.g. the number of tosses required to get the fourth success), then that's the negative binomial distribution. I have a video on the negative binomial distribution here: kzfaq.info/get/bejne/eLacoM2hl9_RqKs.html
@samsonwong8121
@samsonwong8121 4 жыл бұрын
isnt that variance equals to p/(1-p)^2?
@naheen4628
@naheen4628 4 жыл бұрын
Anyone here after they introduced this to the cambridge as level syllabus this year?
@johnbennett8948
@johnbennett8948 5 жыл бұрын
This is for sampling with replacement. What distribution would we use if we sampled without replacement?
@jbstatistics
@jbstatistics 5 жыл бұрын
It's not so much sampling with replacement, as that the probability of success is staying constant from trial to trial. (We might not be sampling from a finite population -- e.g. we might be flipping a coin repeatedly.) If we are sampling repeatedly without replacement from a finite population that is made up of a certain number of successes and a certain number of failures, then the # of trials required to get the kth success has a negative hypergeometric distribution. What you are looking for is the negative hypergeometric distribution with k = 1. (With this distribution, as well as the "regular" geometric and negative binomial, you have to be careful, as different sources use different notation and different definitions of the random variable -- e.g. # of trials required to get the kth success, or the # of failures needed to get the first success, or using k to represent a different quantity than I am here.)
@johnbennett8948
@johnbennett8948 5 жыл бұрын
@@jbstatistics thanks.
@ender5296
@ender5296 Жыл бұрын
how we found E and V please help
@arminehayrapetyan3373
@arminehayrapetyan3373 5 жыл бұрын
Why P(x > 3) is 0.7^3?
@Kaysar777
@Kaysar777 6 жыл бұрын
Thank you for this great explanation! Isn't mean 1-p/p though?
@jbstatistics
@jbstatistics 6 жыл бұрын
No, not under the way I've defined the random variable in this video. I've defined X to represent the trial # on which the first success appears. Then E(X) = 1/p. Sometimes people define the random variable to be the # of failures before the first success appears. If Y represents the number of failures before the first success appears, then Y = X-1, and E(Y) = E(X) -1 = (1/p) -1 = (1-p)/p.
@Kaysar777
@Kaysar777 6 жыл бұрын
Aaa... Thank you for this great clarification! You're awesome! :D
@dandanny1081
@dandanny1081 5 жыл бұрын
Hello Do you have a video for the Uniform distribution(discrete) please ? thanks for the lovely video , your voice is a Superb by the way !
@jbstatistics
@jbstatistics 5 жыл бұрын
No, I do not yet have a video on the discrete uniform distribution. Thanks for the compliments!
@G_anon
@G_anon 10 жыл бұрын
I expect that this is an update to a previous video you had on introducing geometric distributions?
@jbstatistics
@jbstatistics 10 жыл бұрын
Yes, it's an updated intro to the geometric. (There were a couple of things I didn't like about the first one -- this one is better.) Cheers.
@abdourahman87
@abdourahman87 Жыл бұрын
Can't Thank you enough for saving my a$$
@patrickgomez5688
@patrickgomez5688 Жыл бұрын
🧡🧡🧡
@imonghose553
@imonghose553 4 ай бұрын
Does the geometric distribution hold the memoryless property ? Also, is exponential distribution a continuous version of the geometric distribution ?
@jbstatistics
@jbstatistics 4 ай бұрын
I don't think you'd ask those two questions in that way if you didn't know the answers to them. So my question to you is, why are you asking those questions? I'm pretty sure we both know the answers.
@Stichting_NoFap
@Stichting_NoFap 5 жыл бұрын
the probability that x>3, why isn't it also + 0,7^4*0,3 + 0,7^5*0,3 + 0,7^6*0,3 etc?
@jbstatistics
@jbstatistics 5 жыл бұрын
First, note that you are missing a term: P(X>3) = P(X=4) + P(X=5) + ... = 0.7^3*0.3 + 0.7^4*0.3 + ... But, like I state in the video, this is equal to 0.7^3. Your way works, but you have to add up infinite terms. My way the answer is 0.7^3. I like my way a little better :)
@jibranimtiyaz4612
@jibranimtiyaz4612 4 жыл бұрын
this is nothing but geometric progression...use sum of infinite terms of a GP formula to compute P(X>x) in general...U will come up with (1-p)^x
@avocado.toast519
@avocado.toast519 5 жыл бұрын
Help!!!why is there different representation of mean on the internet,1/p and (1-p)/p
@jbstatistics
@jbstatistics 5 жыл бұрын
I've defined X to represent the trial # on which the first success appears. Then E(X) = 1/p. Sometimes people define the random variable to be the # of failures before the first success appears. If Y represents the number of failures before the first success appears, then Y = X-1, and E(Y) = E(X) -1 = (1/p) -1 = (1-p)/p.
@tvvt005
@tvvt005 4 ай бұрын
6:50 isn’t the mean (1-p)/p?
@socialdeveloper5570
@socialdeveloper5570 7 жыл бұрын
I cant stop watching these tutorials wtf
@jbstatistics
@jbstatistics 6 жыл бұрын
It's good stuff!
@ArshadAli-zk5kj
@ArshadAli-zk5kj 5 жыл бұрын
shouldn't the probability of getting success on the sixth trial be p if the trials are independent?
@jbstatistics
@jbstatistics 5 жыл бұрын
The probability of getting *a* success on the sixth trial is p, sure, but here we're looking for the probability that the *first* success occurs on the sixth trial.
@adiga202
@adiga202 8 жыл бұрын
hi jbstatistics ! can yo explain why p(X>3) = (0.7)^3 and not 1-(0.7)^3 ? thanks for your videos!
@thequiickbrownfox
@thequiickbrownfox Жыл бұрын
in Geometric Dist. we are trying to figure out the number of trials required to get the first success; since it is given that x>3, it means that the trial number that gave us success is more than , i.e., we had 3 failures. X=1 is failure 1, so x=3 will be (0.7)^3
@khaledadel7322
@khaledadel7322 5 жыл бұрын
first off, a million thanks to you man, your a savior for real ... second, at 8: 45 , when he says: the probability of the variable X taking a value greater than 3 is simply having 3 failiers in a row,, can someone please elaborate? because i feel that the probability of taking a value greater than 3 is "(3 fails in a row) or (4 fails in a row) ... etc " what am i getting wrong please?!
@jbstatistics
@jbstatistics 5 жыл бұрын
If the first 3 trials are failures, then the first success must come after the third trial. If you toss a coin repeatedly, and tosses #1, #2, and #3 are all tails, then the first time heads appears will be on trial #4 or later.
@JimbobFaz
@JimbobFaz 5 жыл бұрын
What really made it click for me was when I realised the statement you give at 9:10 is actually an equivalent statement. That is the converse to this statement is also true. Therefore the probabilities must to be the same. I wrote it out letting statement A = the first 3 trials are failures, B = more than 3 trials are needed to obtain a success. Then it is easy to see both A implies B and that conversely B implies A. Therefore A = B, hence P(A) = P(B). This was bugging me all day, glad I cleared it up.
@MCConfuz
@MCConfuz 3 жыл бұрын
at 7:48 i don't get how 0.7 to the zero power times 0.3 = success ... could you please explain for me? thanks!
@jbstatistics
@jbstatistics 3 жыл бұрын
That first term at 7:48 is just the probability the first trial is a success (P(X=1)). That is given as 0.3. But I wrote it as 0.7^0*0.3 so that it naturally fit with the other two terms (0.7^1*0.3 and 0.7^2*0.3).
@MCConfuz
@MCConfuz 3 жыл бұрын
@@jbstatistics thanks!
@feloria1862
@feloria1862 4 жыл бұрын
I'm confused about the shape of the distribution, especially for small probabilities of success, shouldn't we expect the probability to be highest at the mean number of trials instead of at 1.
@jokote
@jokote 4 жыл бұрын
No, this probability means "how big needs to be the distribution to get 1 success in a number x of trials" .... meaning that; if you want to get it in the first attempt then you need a big distribution.. if you incease the numer of trial, then you need a smaller distribution to get 1 success
@ugestacoolie5998
@ugestacoolie5998 11 ай бұрын
Reminds me of gacha game pulls whenever I do a geometric distribution
@jaimeeduardo159
@jaimeeduardo159 3 жыл бұрын
What if p=1?
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