1

(For context): $X$ is the random variable for number of successes ($x$) out of $n$ independent trials with probability of success $p$ and probability of failure $1-p = q$

The formula for mass my book gives me for the binomial distribution is this: $P(X=x) = {n \choose x}p^xq^{n-x}$

My question is this. How come order doesn't matter? Why do we use ${n \choose x}$ in the above to calculate all the ways we can get $x$ successes out of $n$ independent trials instead of using $nPx$? I.E. Why do we use combination instead of permutation? Can someone explain this to me because it's been really bothering me.

Thank You!

  • Not sure how useful it might be, but I feel it's similar enough to help you out -- https://math.stackexchange.com/questions/3121134/. The rough idea being you have to account for all of the possible arrangements of successes and failures. While each sequence is equally likely, the overall probability of just a given number (order not mattering) is a bit more complicated. – PrincessEev Jan 10 '21 at 07:13
  • You may find an answer at this post. – Will Orrick Jan 10 '21 at 07:17
  • 1
    But the short answer is that order does matter. The binomial coefficients that come up here are really permutations with repeated elements. Think of $n$ letter words in which $x$ letters are $S$ (for success) and $n-x$ letters are $F$ (for failure). But it can also be thought of as a selection problem: how many ways are there of choosing which $x$ of the $n$ trials are the successful ones. – Will Orrick Jan 10 '21 at 07:22
  • I'd advise you to look at the product space of $({0,1},P({0,1}), \tilde{\mathbb P})$ where $ \tilde{\mathbb P}$ is the uniform distribution. Now define a random variable $Y:{0,1}^n\to \mathbb N$ so that $ \tilde{\mathbb P}^{-1}\circ Y^{-1} \equiv \mathbb P \circ X^{-1}$. Now look at which points of the product space actually make up the probability $\mathbb P(X=x)$: Is it all permutations? Or are some missing? – Sudix Jan 10 '21 at 09:08

3 Answers3

0

If we have three events $A,B,C$ where $A,B$ are successes and $C$ is a failure, then $A,B,C$ and $B,A,C$ both represent two successes in the first two trials and a failure in the last trial, so they are considered to be the same.

5201314
  • 2,247
0

Look at a small example. In a sequence of $4$ trials, what are the ways to get $2$ successes? With $S$ for success and $F$ for failure they are $SSFF$, $SFSF$, $SFFS$, $FSSF$, $FSFS$, AND $FFSS$; there aren’t any more. Specifically, there is one way to get $2$ successes for every pair of trials, and there are $\binom42$ pairs of trials.

The successes and failures don’t have individual identities: their only identities are as success or failure and where in the sequence of trials they occur. You might as well imagine that you have $x$ white and $n-x$ black stones, identical in all respects save color, and want to know in how many ways you can line them up in a row. The stones aren’t labelled; all that matters is the color of each stone and its position in the row. For instance, there is only one distinguishable arrangement with all $x$ of the white stones together at the head of the line.

It would be another story if the white stones were labelled $1$ through $x$: then there would be $k$ distinguishable arrangements with all of white stones together at the head of the line. That’s the model that your proposed calculation would be following, and it clearly doesn’t fit what you’re trying to model: the individual successes don’t come with individual labels beyond success in trial number so-and-so.

Brian M. Scott
  • 631,399
  • Thank you for your answer. I understand binomial distribution, but I now reading about Hypergeometric Distribution in my introductory probability textbook. Does the same logic apply to the Hypergeometric Distribution? Thus, that's the reason why we also use combination instead of permutation in the mass formula for Hypergeometric Distribution? Just Curious –  Jan 15 '21 at 03:43
  • 1
    @BillBillwater: Pretty much, yes. It’s less obvious, but it turns out that all of the different ways of getting $k$ successes in $n$ trials are equally likely. If you’re drawing red and blue marbles without replacement, for instance, you might just as well draw $n$ of them at once instead of drawing them one at a time. – Brian M. Scott Jan 15 '21 at 04:07
0

This is tautological, or definitional. You only use the binomial distribution when the order of the successes doesn’t matter because the successes are all interchangeable and independent. If the order of the successes matters, or the successes aren’t interchangeable, or they’re not independent, then you shouldn’t be using the binomial distribution, lol.

Rivers McForge
  • 6,002
  • So how do you calculate possible combinations where order does matter? Given 2048 unique words, and 12 choices with repeats allowed, 2049^12 is the possible phrase combos unordered right? I want to compare those odds to the case where the words must be in alphabetical order. https://bitcoin.stackexchange.com/questions/113432/how-much-entropy-is-lost-alphabetising-your-mnemonics/113440 – HansBKK Apr 27 '22 at 06:07