420chan now has a web-based IRC client available, right here
Leave these fields empty (spam trap):
You can leave this blank to post anonymously, or you can create a Tripcode by using the float Name#Password
A subject is required when posting a new thread
[*]Italic Text[/*]
[**]Bold Text[/**]
[~]Taimapedia Article[/~]
[%]Spoiler Text[/%]
>Highlight/Quote Text
[pre]Preformatted & Monospace text[/pre]
1. Numbered lists become ordered lists
* Bulleted lists become unordered lists


Community Updates

420chan now supports HTTPS! If you find any issues, you may report them in this thread
Time by Shower thought - Sat, 01 Aug 2015 17:37:45 EST ID:eAsqVIz1 No.14841 Ignore Report Reply Quick Reply
File: 1438465065277.jpg -(32969B / 32.20KB, 342x392) Thumbnail displayed, click image for full size. 32969
I have just stumbled upon a question while in the shower, what is the speed of time?
21 posts and 4 images omitted. Click Reply to view.
Polly Pittgold - Fri, 25 Sep 2015 03:48:03 EST ID:i84x+n57 No.14918 Ignore Report Quick Reply
Planck time is around 5.39*10^-44 seconds. The Planck scale is where quantum gravitational effects become significant.

I swear, you guys are way overthinking OP's question.
Augustus Gubblesare - Fri, 25 Sep 2015 19:08:49 EST ID:GKhdNJGT No.14919 Ignore Report Quick Reply
>around 5.39

AH that's better.
Fanny Wivingdale - Wed, 21 Oct 2015 15:16:21 EST ID:bgR+wrEg No.14946 Ignore Report Quick Reply

the poster just rounded 5.39e-44 to 1e-43. i wouldnt use that in a calculation though
Cornelius Gorryban - Thu, 22 Oct 2015 22:46:02 EST ID:GKhdNJGT No.14948 Ignore Report Quick Reply
I know, but it's kind of like me rounding my height up to 10 feet tall when I'm really only about 5 and a half feet tall. It wouldn't have bugged me so much if he had at least said "approximately 1e-43", even though it's barely even half that amount. I mean, If you're going to round to zero digits of precision, why not make it 5e-44? Plus in the next sentence, he used EIGHT digits of precision!

It's cool, though. I'm over it now.
Barnaby Blommerstudge - Mon, 26 Oct 2015 16:15:26 EST ID:sT1kFNdT No.14955 Ignore Report Quick Reply

thought you do have a point in terms of the psychologcal there is an actual determinator which directly shows the value of what OP speaks, but though it seems that rather because of the psychologi involved there is beleived for a long time to be a fluxuation in this, and tough it is true in certain " frames of visioN" there is not a trace of

i want the fommula to a shitload vs an assload by Phineas Bunforth - Sat, 03 Oct 2015 02:04:23 EST ID:tiSYuoE5 No.14925 Ignore Report Reply Quick Reply
File: 1443852263874.jpg -(9330B / 9.11KB, 480x360) Thumbnail displayed, click image for full size. 9330
can any of you meet my challenge that i am challenging here today

it is monday monday gotta get down on monday!
Lillian Shittingwater - Sat, 03 Oct 2015 02:46:17 EST ID:i84x+n57 No.14926 Ignore Report Quick Reply
According to circlejerk:

buttload * 10 = 1 butt ton
butt ton * 10 = 1 assload
assload * 10 = 1 asston
asston * 10 = 1 shitload
shitload * 10 = 1 shitton
shitton * 10 = 1 fuckload
fuckload * 10 = 1 fuckton
William Firrylack - Sat, 03 Oct 2015 14:10:01 EST ID:GKhdNJGT No.14927 Ignore Report Quick Reply

Get that metric shit out of here.
Reuben Gigglepot - Sun, 25 Oct 2015 19:34:19 EST ID:JsLKELFF No.14953 Ignore Report Quick Reply
take a pen

take a pencil

grab a pen

grab a pencil

paper or write on you will do


discover what you seek

foremost you have done

"Good at math" by Archie Gennerlark - Thu, 01 Oct 2015 01:08:46 EST ID:M30RiRC5 No.14920 Ignore Report Reply Quick Reply
File: 1443676126033.jpg -(109223B / 106.66KB, 640x465) Thumbnail displayed, click image for full size. 109223
What does it mean to be "good at math?"
3 posts omitted. Click Reply to view.
Archie Snodfield - Thu, 01 Oct 2015 20:23:53 EST ID:i84x+n57 No.14924 Ignore Report Quick Reply
To me it means someone excels at using abstract reasoning to solve problems. To most people it means someone can follow a formulaic procedure to solve a particular type of problem.
Doris Goodville - Wed, 21 Oct 2015 15:43:01 EST ID:d7QflcOd No.14947 Ignore Report Quick Reply
applying the concepts derived form it, and also the feeling of knwoing that time spent in actualy organizing the several different patterns of thinking that revolves around the " eqaution " gives, which is very important because after a while what noramlly takes 9 hours to solve, becomes well, 8.9 hours and with repetition the number decreases and more insight to these new patterns of thinking and unifying the relation between different symbols becomes apperant, i think anyways
Basil Cricklehune - Sat, 24 Oct 2015 22:42:32 EST ID:G+WyMdI9 No.14950 Ignore Report Quick Reply
like woah
Barnaby Clullerfoot - Sun, 25 Oct 2015 04:15:05 EST ID:AuBkeUkZ No.14951 Ignore Report Quick Reply
When you can sign Brian McKnight - Back At One
Jarvis Clayson - Sun, 01 Nov 2015 13:12:36 EST ID:ryc+8ox4 No.14962 Ignore Report Quick Reply
what do you mean? nb

Physics question about elasticity by Rasclot - Sat, 17 Oct 2015 18:55:11 EST ID:gpkCKW1u No.14941 Ignore Report Reply Quick Reply
File: 1445122511224.jpg -(81568B / 79.66KB, 800x800) Thumbnail displayed, click image for full size. 81568
Yo, so I should preface this by saying this is a physics question, but it is quite mathematical.
My issue right now with computational physics research is that I am trying to prove a 2d simplification of a 3d model of a vesicle has an elastic response to applied forces. My 2d simplification of a 3d vesicle, a sphere of lipids containing a fixed volume of water, is a loop containing a fixed area. It is a discrete model in which I move the vertices representing the loop in order to reduce line-tension, but however never does so in a manner which changes area (to linear order). The issue is that when I apply an outward gaussian force to this bitch, for small magnitude forces, it converges to a new length. My best fits of this regime look like newlength=k(fapplied)^2+lengthoriginal. While this is interesting, and I am glad that my loop isn't diverging in these cases, I am not sure whether that relation actually describes an elastic surface/the elastic modulus of such a surface. For high forces it doesn't converge at all.

Anyhoo, how would you go about calculating the elastic relationship between the length of an elastic loop and the force applied to it?

Any help would be much appreciated.
A human.
John Hennerkut - Mon, 19 Oct 2015 10:09:39 EST ID:i84x+n57 No.14943 Ignore Report Quick Reply
>Anyhoo, how would you go about calculating the elastic relationship between the length of an elastic loop and the force applied to it?

That depends on how the force is applied. Is it two parallel lines squeezing the loop from opposite sides?
Cedric Guckleworth - Fri, 23 Oct 2015 18:58:36 EST ID:gpkCKW1u No.14949 Ignore Report Quick Reply
1445641116924.gif -(190352B / 185.89KB, 2000x2000) Thumbnail displayed, click image for full size.
it's an elastic loop, think rubber band.

I found a hint at the answer here:http://www.wired.com/2012/08/do-rubber-bands-act-like-springs/
I also found an APS journal article about it that is more rigorous American Journal of Physics 31, 938 (1963); doi: 10.1119/1.1969212

Solving Recurrence Relations by Wesley Tillingbury - Fri, 16 Oct 2015 21:23:55 EST ID:pCKCgraP No.14939 Ignore Report Reply Quick Reply
File: 1445045035508.png -(14284B / 13.95KB, 269x235) Thumbnail displayed, click image for full size. 14284
I did discrete math last year but I've forgotten a lot about how to solve recurrence relations. In the solutions to one of my assigned homeworks, I don't understand the process the professor used to find the explicit formula. For instance, in the pic related example:

C(n) = 3C(n-1) + 4; C(0) = 4

Why aren't the "+ 4"s being multiplied by the 3 as the relation is backwards substituted? Why is it exempt?
Cyril Blavingstutch - Sat, 17 Oct 2015 01:50:54 EST ID:i84x+n57 No.14940 Ignore Report Quick Reply
1445061054900.jpg -(2828662B / 2.70MB, 3504x2336) Thumbnail displayed, click image for full size.
>Why aren't the "+ 4"s being multiplied by the 3 as the relation is backwards substituted? Why is it exempt?

You're right; they should be multiplied by the 3s. Whoever made your pic fucked up hardcore and has no business teaching math. I hope for your sake that it wasn't your professor.

Line 3 of the red text is where they first fucked up. Should be:

C_n = 3*[3*[3C_(n-3) + 4] + 4] +4 = 3*3*3*C_(n-3) + 4*(1 + 3 + 3*3)

This means:

C_n = 3^k*C_(n-k) + 4*(summation from i = 0 to k-1 of 3^i)

The summation is just a geometric series (https://en.wikipedia.org/wiki/Geometric_series#Formula):

C_n = 3^k*C_(n-k) + 4*(1 - 3^k)/(1 - 3) = 3^k*C_(n-k) + 2*3^k - 2
Comment too long. Click here to view the full text.
Fucking Clurrykire - Sat, 17 Oct 2015 20:27:30 EST ID:i84x+n57 No.14942 Ignore Report Quick Reply
>and verifying with the original recurrence relation
and checking against the original recurrence relation

I'll fix the link too: https://en.wikipedia.org/wiki/Geometric_series#Formula

Phyllis Pemmerhut - Mon, 19 Oct 2015 18:39:21 EST ID:xJK33YUo No.14944 Ignore Report Quick Reply
Thanks a lot for your help, I'm glad that I wasn't missing something huge. It was my professor so I'll talk to her today. I actually need some more help if you're willing, with this problem:

"In class we showed that the recurrence relation for divide and conquer algorithms such as the quick and merge sorts is C_n = 2*C_n/2 + n, where C_1 = 0. Using this recurrence relation, the analysis of these algorithms is Theta(n*lg(n)). Solve the recurrence relation, using a different base case: C_1 = 15, to demonstrate whether changing the base case results in the same (or different) Big Theta result."

Now I'm not sure if you're familiar with computer science, but I don't think you need to be to solve this problem. I'm just unsure of how to approach 'solving' the relation, and how/if a different base case would change the analysis (which we perform using the Master Theorem https://en.wikipedia.org/wiki/Master_theorem#Generic_form ). Any guidance?
Shit Bloblingwater - Tue, 20 Oct 2015 02:32:41 EST ID:i84x+n57 No.14945 Ignore Report Quick Reply
I've only ever taken one CS course. But if I understand correctly, the base case should be irrelevant with regard to using the master theorem. The recurrence relation in question falls under case 2 mentioned the the Wikipedia article, and accordingly is theta(n log n). This result is independent of the base so should be true of C_1 = 15.

For solving the relation, just use the same backward substitution argument as before. Doing that gives you:

C_n = 2^k*C_(n/2^k) + k*n

Set 2^k = n:

C_n = n*C_1 + n*log_2(n)

For C_1 = 15:

C_n = 15n + n*log_2(n) = theta(n log n)

So it checks out.

Trig troubles by Molly Bingergold - Mon, 12 Oct 2015 00:12:03 EST ID:P4gPE0kz No.14933 Ignore Report Reply Quick Reply
File: 1444623123652.png -(4785B / 4.67KB, 455x54) Thumbnail displayed, click image for full size. 4785
I'm told I have to expand the left side using double angle identities but I'm not sure where to start. Can someone help me get started with this?
Doris Sommerdock - Mon, 12 Oct 2015 22:13:31 EST ID:i84x+n57 No.14935 Ignore Report Quick Reply
Start with cos(2x) = 1 - 2*sin^2(x). Solve this for sin^2(x). You should be able to figure out the rest from here.
Frederick Banningsire - Wed, 14 Oct 2015 19:26:09 EST ID:i84x+n57 No.14937 Ignore Report Quick Reply
Meh, I looked at the problem again, and it's a little more involved than I initially thought. Okay, I'll just work it out fully. Doing what I previously said gives:

sin^2(x) = 1/2 - 1/2*cos(2x)

square both sides: sin^4(x) = 1/4 -1/2*cos(2x) + 1/4*cos^2(2x) (call this equation 1)

start with another double angle identity: cos(2x) = 2*cos^2(x) - 1

solve for cos^2(x): cos^2(x) = 1/2*cos(2x) + 1/2

sub 2x for x: cos^2(2x) = 1/2*cos(4x) + 1/2 (equation 2)

plug eq. 2 into eq. 1: sin^4(x) = 1/4 - 1/2*cos(2x) + 1/4(1/2*cos(4x) + 1/2)
Comment too long. Click here to view the full text.

Help me to understand by Fuck Sottinghet - Tue, 13 Oct 2015 17:55:25 EST ID:43kF2tms No.14936 Ignore Report Reply Quick Reply
File: 1444773325462.gif -(688596B / 672.46KB, 205x160) Thumbnail displayed, click image for full size. 688596
I'm doing an assignment for college, can you explain the math by providing an example please ?
This assignment involves developing a primitive plagiarism detection application
We will use the following simple comparison metric. Let

= (

Ebenezer Bazzleway - Fri, 16 Oct 2015 21:17:28 EST ID:2a+jHHTF No.14938 Ignore Report Quick Reply
You need to tell us more. All you've told us is:
  • you need to develop a primitive plagiarism detection application
Is this to detect when a given text is plagiarism from another?

  • simple comparison metric
This is the metric used to determine similarity?

A=(a0; a1; an
Is this the entire metric? Did you forget to close parentheses or is there more? What are the variables involved?

Show sets are equinumerous by finding a specific bijection between the sets. by Phoebe Hennerlock - Tue, 22 Sep 2015 13:24:29 EST ID:SKWHYv68 No.14909 Ignore Report Reply Quick Reply
File: 1442942669344.png -(18538B / 18.10KB, 462x143) Thumbnail displayed, click image for full size. 18538
Hi /math/, looking for some help. Several problems for my advanced calculus homework follow this format, and I just don't understand how to describe the bijection. I understand bijections, and that finding one proves two sets are equinumerous, but again I don't know how. For instance, this first problem:
a) S = [0,1] and T = [1,4]
What does the answer to this look like, and how can I apply it to other problems of a similar format?
James Cledgewig - Tue, 22 Sep 2015 16:57:33 EST ID:i84x+n57 No.14910 Ignore Report Quick Reply
t = 3s + 1 where s is an element of S and t is an element of T
Emma Gipperhood - Tue, 22 Sep 2015 18:09:49 EST ID:RsLdN3g2 No.14911 Ignore Report Quick Reply
you want to find a function, f, from one set to the other which is injective (no two elements get sent to the same element) and surjective (each element in the "receiving set" has an associated element in the first set). More formally, injectivity is illustrated by: if f(a) = f(b) then a = b, surjectivity: For any b in the receiving set, there exists an element a in the first set such that f(a) = b.

So in the question all it wants you to do is find a function from the first set to the second set that satisfies the above two properties.

So for the first question, we could by inspection try the function f(x) = 3x + 1 like James gave. We now need to show that it's injective and surjective (hence bijective). For injection, suppose we have f(a) = f(b).
Then 3a + 1 = 3b + 1 which implies that a = b.

For surjectivity, if we take any element in T (say b), then we need to show that there exists an element in S such that f(a) = b. So a such that 3a + 1 = b. Because of the sets in question we can just take cases. If b = 4 then 1 works, if b = 1 then 0 works. Done.

Note that functions don't have to be so neat. So we could have taken the function f, defined by f(x) = 1 if x = 0 and f(x) = 4 if x is 1. Then proving it's a bijection is easy.

So the basic strategy is to first "guess" a function based on trial and error, intuition or calculation and then try to show that your guess is a bijection.
Shit Hinkinshaw - Tue, 22 Sep 2015 20:56:00 EST ID:BfM+rTDn No.14912 Ignore Report Quick Reply
So wait, maybe I'm just confused about the presentation. [1,4] doesn't mean {1,2,3,4}? Just {1,4}?
Priscilla Serringpit - Wed, 23 Sep 2015 04:07:47 EST ID:i84x+n57 No.14913 Ignore Report Quick Reply
It's a continuous interval - the segment of the real number line between 1 and 4, end points inclusive as indicated by the square brackets. Parentheses, on the other hand, indicate that an endpoint is not included in the set.
Priscilla Bruzzlesit - Wed, 23 Sep 2015 14:48:51 EST ID:crJsE1sF No.14915 Ignore Report Quick Reply

oh yeah, my bad (in my defense I was high - and still am). The first function still works.

2 easy and funny mathematics tricks by No_name - Mon, 21 Sep 2015 17:45:39 EST ID:6m2TPZ+N No.14906 Ignore Report Reply Quick Reply
File: 1442871939125.jpg -(66247B / 64.69KB, 640x640) Thumbnail displayed, click image for full size. 66247
2 easy and funny mathematics tricks



Nell Clonningdale - Mon, 21 Sep 2015 18:25:45 EST ID:2JYFlB0V No.14907 Ignore Report Quick Reply
113 (a-b concatenated with a+b).

customer pays 0.9*(1.1*price_item_1 + 1.1*price_item_2 + ... + 1.1*price_item_n) = 0.99*(price_item_1 + price_item_2 + ... + price_item_n) < (price_item_1 + price_item_2 + ... + price_item_n) which is the price he would have normally paid.

these aren't mental tricks though, they're puzzles.

Advice on interesting maths topics by Nicholas Blackcocke - Tue, 04 Aug 2015 15:10:07 EST ID:84oIcStQ No.14849 Ignore Report Reply Quick Reply
File: 1438715407888.png -(53664B / 52.41KB, 250x249) Thumbnail displayed, click image for full size. 53664
I'm a maths student going into my third year in September. I've already started looking at the possible modules I can take next year and I'm feeling a little overwhelmed; the number of modules available have increased dramatically and they all seem pretty interesting.

Measure theory, Introduction to Topology, Galois theory, Structures of Complex Systems, Commutative Algebra, Algebraic Topology, Brownian Motion, Geometric Group Theory, Algebraic Geometry, Set Theory, Markov Processes and Stochastic Analysis are the modules I'm considering but there are possibly many interesting modules that I passed up.

I'd be grateful if anyone who has studied any of these topics could let me know how they found them and if anyone's really bored they could have a quick scan of the modules available and let me know if I missed anything interesting:

3 posts omitted. Click Reply to view.
Fuck Worthingway - Wed, 26 Aug 2015 22:48:36 EST ID:Dk8yywxc No.14866 Ignore Report Quick Reply

What classes did you end up going for? I'm taking intro to modern algebra, modern analysis, descriptive set theory, and mathematical logic
Cornelius Hammerway - Thu, 03 Sep 2015 20:34:25 EST ID:Ayxv8mCh No.14875 Ignore Report Quick Reply
I am a ph.d student studying lie groups and conservation laws. If I were you, I would take topology, measure theory, markov processes and stochastic analysis, and geometric group theory. My reasoning is because I had similar choices to make in undergrad and that is essentially what I picked. Do what interests you, anon, and be good at that.
Cornelius Hammerway - Thu, 03 Sep 2015 20:35:11 EST ID:Ayxv8mCh No.14876 Ignore Report Quick Reply
mathlogic? really? enjoy your topos theory and unemployment.
Frederick Pickford - Wed, 09 Sep 2015 17:36:03 EST ID:LzNXN+l+ No.14887 Ignore Report Quick Reply

I think what you're interested in is shit you elitist prick. You seriously think that your particular flavor of group theory or physics is going to put you leagues ahead of other people? Take your own advice, shut the fuck up, and go multiply some matrices rude boy.
Ernest Grimridge - Mon, 21 Sep 2015 17:44:25 EST ID:AdYmD9Mq No.14905 Ignore Report Quick Reply
I'm taking Measure Theory, Introduction to Topology, Functional Analysis I, Structure of Complex Systems, Galois Theory, General Relativity and Dynamical Systems.

I need to drop two or three of them though.

I'm taking Algebraic Topology for sure next term. Set Theory doesn't really lead on to anything and apparently is kind of easy so I'll do it next year when I'm doing my master's thesis.

I don't like random variables so I think I'd find stochastic analysis hard ... sucks as if I had taken it, I could possibly do my masters thesis on it with a fields medalist as my supervisor! :/

Geometric group theory's in term two so I might do it then.

btw, how did you decide to work on lie groups? Did you find it interesting at undergraduate level?
I know it's kind of early but I'm thinking about what I should do for my master's thesis next year however everything seems so interesting that I don't know how I'm going to specialize.

Freaking math man by Hugh Nickledale - Wed, 29 Jul 2015 15:09:16 EST ID:I+l45cST No.14831 Ignore Report Reply Quick Reply
File: 1438196956144.jpg -(22700B / 22.17KB, 560x247) Thumbnail displayed, click image for full size. 22700
So pics related, why is it possible that the original formula can't be solved for x=1, but the formula after applying a series of transformation rules can be solved for x=1.
I mean, it should be the same formula but only written differently, right? Am I "not getting" something extremely basic here?
1 posts omitted. Click Reply to view.
Caroline Blommleshit - Thu, 30 Jul 2015 03:51:15 EST ID:I+l45cST No.14834 Ignore Report Quick Reply
Yes, I get how it's done with the limit and all but I'm still confused why, why does factoring allow to solve it at value 1?
What I basically don't get is: if you simplify a formula (let's take the one in the OP, but backwards) it is initally solvable for x=1, but after factoring it isn't anymore. But if you rewrite a formula it should have the same outcome for every value of the variable, right? By factoring the result of that formula is changed for one value, so it basically isn't the same formula anymore?
I'm obviously not getting something fundamental here, I hope this isn't a completely retarded question. Thanks for the reply btw.
Caroline Blommleshit - Thu, 30 Jul 2015 03:57:01 EST ID:I+l45cST No.14835 Ignore Report Quick Reply
Or is the fact that the formula (x³-1)/(x²-1) isn't solvable for the value x=1 just a limitation of the 'excessive' factorization, and that's all there is to it?
Albert Blingerridge - Thu, 30 Jul 2015 13:29:45 EST ID:i84x+n57 No.14836 Ignore Report Quick Reply
>But if you rewrite a formula it should have the same outcome for every value of the variable, right? By factoring the result of that formula is changed for one value, so it basically isn't the same formula anymore?
Factoring doesn't change anything. It's the last step where the (x-1) terms are cancelled that changes things, and what you're left with is NOT equal to the original equation. Cancelling these terms is the same as division by zero (which of course is an illegal operation) UNLESS you specify x≠1. This means that the first and last functions are equal for all values of x except x=1. The last function essentially fills the "hole" in at x=1 with the limit i.e. 3/2.
Angus Sommlehood - Sun, 20 Sep 2015 00:19:57 EST ID:xgdLZpNp No.14903 Ignore Report Quick Reply

You get two answers when you square, because of parabolas, and that's why we have to deal with absolute values, too.
John Brookstock - Sun, 20 Sep 2015 15:40:32 EST ID:BE4o4wO4 No.14904 Ignore Report Quick Reply

Probability off by factor if 10 by Faggy Blablingbanks - Mon, 14 Sep 2015 16:01:05 EST ID:gibSXrA4 No.14889 Ignore Report Reply Quick Reply
File: 1442260865295.jpg -(45730B / 44.66KB, 329x347) Thumbnail displayed, click image for full size. 45730
Using two different methods to find probability, I'm getting the same number but off by a factor of 10

>Five cards are dealt from 52-card deck; what is the probability that we draw 3 Aces & 2 Kings?
  • Method 1 (wrong):
4/52 + 3/51 + 2/50 + 4/49 + 3/48 = 9.23e-7

  • Method 2:
[(4 choose 3)*(4 choose 2)] / (52 choose 5) = 9.23e-6, which is the right answer she says.

So why is it that method 1 (which seems more intuitive to me) gives an answer that's off by 10?
2 posts omitted. Click Reply to view.
Hedda Dattingmitch - Tue, 15 Sep 2015 17:55:55 EST ID:i84x+n57 No.14893 Ignore Report Quick Reply
Okay, so I was half asleep last night when I read and replied to your question. Sorry if I came off as a dick; that happens when I'm tired.

The trick you're trying to use in method 1 doesn't work, because you're not treating getting an ace or a king as independent variables. To treat them as independent, you need to account for all possible arrangements. You did this:

(prob of 1st ace)(prob of 2nd ace)(prob of 3rd ace)(prob of 1st king)(prob of 2nd king)

But you didn't count arrangements like:

(prob of 1st ace)(prob of 1st king)(prob of 2nd ace)(prob of 2nd king)(prob of 3rd ace)

There are (5 choose 3) = (5 choose 2) = 10 possible arrangements all with the same probability, so you just need to multiply your result in method 1 by 10 to get the correct total probability.
Nigel Mammlewater - Thu, 17 Sep 2015 13:54:24 EST ID:i84x+n57 No.14899 Ignore Report Quick Reply
Actually, technically you were treating the random variables as independent simply by multiplying them together as you did. The real issue is that method 1 works on the principle that drawing a single card at a time until you have all 5 is the same as drawing all 5 at once. But since you're drawing them one at a time, you have to account for different orderings. Of course, suits don't matter in the ordering. Method 2 doesn't care about orderings; it's analogous to drawing all 5 cards at once. Method 2 just takes the number of possibilities for drawing a 5 card hand matching the conditions and divides by the number of all possible 5 card hands. This explanation should be clearer.

Alice Cunderchut - Thu, 17 Sep 2015 22:05:25 EST ID:qCDKBx4v No.14900 Ignore Report Quick Reply
Thank for the explanations, I didn't realize that order matters when the variables are independent (it almost seems it should be the other way around), but I'll probably stick to method 2, which is the newer one that was taught to me.

Thanks for teaching me!
Alice Duckstone - Fri, 18 Sep 2015 15:32:30 EST ID:i84x+n57 No.14901 Ignore Report Quick Reply
>Thanks for teaching me!
You're welcome.
Angus Sommlehood - Sun, 20 Sep 2015 00:01:12 EST ID:xgdLZpNp No.14902 Ignore Report Quick Reply
This was the meanest thing I've seen from this site in a long time.

Bad vibes, man.
Funny bad vibes, fam.

<<Last Pages Next>>
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Report Post
Please be descriptive with report notes,
this helps staff resolve issues quicker.