Stat formula

[Samadhi]

I know the formulas for permutations and combinations. But suppose you have a bag filled with 4 black marbles and 4 white marbles. What is the formula to figure out the different sequences for drawing them out? Total brain fart and I can't find it in the lame book.

[Bicho the Inhaler]

Do you want the number of distinct sequences? That's the "combinations" formula, i.e., the binomial coefficient (4+4)!/(4!4!) If you had more than 2 colors (say, n), then you would use the multinomial coefficient (a ₁ + ... + a _n )!/(a ₁ !...a _n !).

[Samadhi]

Distinct sequences is permutations. Where the order doesn't matter it's combinations.
The binomial coeficient you stated seems to be correct. I don't seem to be able to find it in the book. I'll look it up (internet) to see why this is the right formula. Thanks

[Bicho the Inhaler]

You've probably discovered why it works, but here's an explanation:

The number of distinct sequences, assuming all the items are distinguishable themselves, is, as you say, the number of permutations (N! for N items). However, if M of the items are identical, then for any given permutation of the N items, permuting those M will not change the sequence. There are always M! ways to permute the M items amongst themselves, i.e., M! ways to permute a given sequence without changing how it looks, so if K is the number of distinct sequences, then K*M! = N!, or K = N!/M!. If there are another P that are identical, then you apply the logic again and divide by P!, and so on.

[Mr Stoofer]

This is a follow-on question from a non-mathematician to all you mathematical wizards.

I have X balls, each of which is unique and distinguishable. I can pick any number of balls from 1 ball to all X balls (but not zero balls). The order is irrelevant. How many different possible choices do I have?

This is actually a real problem I have at work (although the X items are not balls, I just used balls for simplicity).

[Samadhi]

That's combinations you want.

X!/(X-n)!n! Where n is the number of balls you choose.

! means factorial. A factorial is n! = n(n-1).....2*1 except for 0! which is 1.

So, for example, 4! would be 4*3*2*1 or 24.

[Quailman]

So he'd have to evaluate that for each n from 1 through 4, right?

X=4, n=1 --> 4
X=4, n=2 --> 6
X=4, n=3 --> 4
X=4, n=4 --> 1

Total = 15

[extropalopakettle]

[extropalopakettle]

Exercise: Prove:

SUM for n=0 to X of X!/(X-n)!n! equals 2 ^X

.

[Antrax]

Can we use a combinatorical proof?
If so:
Both sides of the equations are solutions to the question "in how many ways can you divide X elements into two sets?"
On the right, every element can be either in group A or group B, so you have two options, X times - 2^x.
On the left, choose i elements without repetitions and without ordering, and the rest go to the other group, and sum on all Is.
_________________
After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[Bicho the Inhaler]

That proof is fine. Here's another:

We know that

(a + b)^n = {sum from k = 0 to n of}(n!/(k!(n-k)!)a^k b^(n-k);

that's why they're called "binomial coefficients."

Take a = b = 1 in the above formula.

[Samadhi]

I thought he was asking the formula for discrete X and n, not the combined set.

[Bicho the Inhaler]

I remembered generalizing that exercise a little while ago, but I couldn't remember what the general formula was. I did manage to reconstruct it, though:

Exercise 2.

extro's formula is actually a special case of the more general identity:

{sum over all integral a ₁ ,...,a _n-1 satisfying a ₀ <= a ₁ <= ... <= a _n-1 <= a _n of}{product from k = 1 to n of}(C(a _k , a _k-1 )) = ?

(you fill in the blank)

Notes:
- C(m, n) is the combinations function m!/(n!(m-n)!).
- the variables a ₀ and a _n appear free in the above identity; the right hand side can depend on them. (Assume 0 <= a ₀ <= a _n .)

This is an example of a post that would be a lot more readable with a LaTeX mod.

[Antrax]

[Bicho the Inhaler]

Here's what I was trying to say:

[Mr Stoofer]

Thanks to all for you guys for yuor help on my problem.

If you actually are interested in the RL problem I have, it is this: I am a lawyer, and my opponent has set out a list of X different things which he says my client has done. One of the possible responses is: "well, none of those things are on their own unlawful". So he says, "yes but you have done all X." But he might not prove that I have done every single one of X. So he accuses my client of (a) every individual act; (b) every combination of 2 or more from the list; (c) all X acts.

My problem was, how many accusations is that in total? And I understand the answer to be 2 ^x -1.

[Antrax]

Yep. Think of it this way - he divides the accusations into two groups: the ones he's going to make and the ones he's not. So that's 2^x. Minus one, because he's not going to accuse your client of nothing.
_________________
After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[extropalopakettle]

[Samadhi]

I just got my midterm back. There were two extra credit quesions and I was the only one who answered both correctly. Here are the questions:

E1: An honest man rolls two dice that cannot be seen by his audience. Someone asks if one of the dice came up 3. Given that the man answers "yes," find the probability that both dice came up 3. Express as a fraction.]
(this is elementary for any GLer)

E2: If an apple a perfect sphere ^* is hanging from a string and three flies land on it, find the probability that all three are on points that are within the same hemisphere.

My extra credit for you: My teacher claims there are two ways to solve E2, both giving the same answer. I disagree with one of the methods. What are the two methods and why would I claim one of them is incorrect?

^* _{substituted to reduce the question}

[Antrax]

E1:
There are a total of 12 die-rolling outcomes for which we'll hear a "yes". Out of those, one is "3, 3". So, 1/12.
E2:
1/2 (let the first two flies define the hemisphere - then the third can land either on it or outside it, and the area is the same).
_________________
After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[Samadhi]

E1: NO.

E2: NO.

*slowly backs away from every poker thought antrax has ever expressed*

[jadesmar]

[extropalopakettle]

#2 - 3 points define a plane that slices the sphere. Assume the does not slice through the center of the sphere. Now slice the sphere through the center with another plane that is parallel to the first plane (defined by the three flies). This divides the sphere into two hemispheres, with all three flies on one of them. Probability 1. If the first assumption is false, then the three flies are on the boundary between two hemispheres - if they are mathematical points on a sphere, probability zero (of the assumption being false - so answer is probability=1).

[Antrax]

~shrugs~
re: E1, it was 8 am my time.
re: E2, I still have no clue what you're on about, but I'm exceedingly weak at anything involving 3d shapes. Also why I barely passed advanced calc. 2
And I still make more than you playing poker
_________________
After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[Antrax]

After reading extro's post thoroughly, I'm convinced his answer is correct, but I still can't get myself to visualise it. I have this mental picture of two flies huddling close together on one side, and the other standing on the other side
_________________
After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[Samadhi]

nah, it doesn't matter where they are. That's a bit more ingenious than I was looking for though.

[extropalopakettle]

How about 4 flies? Naive reasoning for 50% would be that the fourth fly has a 50% chance of being on the hemisphere that the first 3 are on. But there are more than hemispheres that the first three are on. If the first three are very close together, for instance, then there is a very high probability that the fourth will be on some same hemisphere as them. Only when the first three are on a plane that nearly intersects the center of the sphere does the probability of the fourth being on the same or a different hemisphere approach 50%.

[Samadhi]

The two methods she had gave the answer of 1/4

[extropalopakettle]

I think it's a matter of how the question is interpreted.

If you first divide the sphere into two hemispheres (there are an infinite number of ways to do this), then if three flies land, there is a 1/4 chance that all are on the same hemisphere (1/2 chance the second fly is on the same hemisphere as the first fly, and independently 1/2 chance the third fly is).

But if three flies land, they will always be together on some hemisphere (not so for 4 flies).

[Mr Stoofer]

[Samadhi]

Ok, this is annoying.

[The Ragin' South Asian]

isn't there something with z scores or something where you figure out the probability of getting that result with a random sample if the real figure is whatever the final outcome is. and you just choose beforehand what your threshold would be, like if there's less than a 10% chance or whatever you'll conclude it's invalid

[Lucky Wizard]

What RSA said.

Furthermore, if you omit from RSA's statement a couple of bits specific to your question, the resulting statement becomes a statement of how any "Is this due to sampling variability or not?" question is answered.

(By the way, the usual threshold is 5%, though 10% and 1% are sometimes used.)

[Samadhi]

You need the standard deviation for a z score. No sd was given.
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And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.

[Samadhi]

Basically you've conflated some things there. A z score is determined by (value - mean)/(standard deviation)
Something is considered usual if it is within 2 standard deviations (or -2 <= z <= +2)
The empirical rule when dealing with bell-shaped distributions states that about 95% of all values fall withing 2 standard deviations.
_________________
And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.

[Lucky Wizard]

Since each trial (person in this instance) can have two outcomes, this is a binomial distribution. The standard deviation for a binomial distribution is sqrt(n*p*q) where p and q are the probabilities of each outcome, and n is the number of people polled.

The normal distribution works well as an approximation for the binomial distribution, and this approximation is often used when n is large enough to make actually calculating the binomial distribution impractical.

[Samadhi]

[kevinatilusa]

I think the only thing you're disagreeing on is the statistic to be measured.

If we were to ask one person before the election whether they intended to vote Democrat or not, they would say yes with 51% probability.

If we treat "yes" as a 1 and no as a "0", this is a random variable with mean 0.51 and standard deviation sqrt(0.51*0.49)

If we were to ask 400 people their intentions, the mean and standard deviation depend on what your sample statistic is.

If it is the number of people who answered Democrat (as in Lucky Wizard's post), the mean would be 0.51*400=204 and the SD would be sqrt(400*0.51*0.49).

If it is the proportion of people who answered correctly (as you seem to be assuming), the mean would be 0.51 and the SD would be sqrt(0.51*0.49)/sqrt(400).


And not quite every distribution can be approximated by a normal distribution. If your variables are coming from a Cauchy distribution (The ratio of two independent normally distributed variables), all of your sample means will also come from a Cauchy distribution, so you can't approximate them by a normal distribution.

[Samadhi]