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#### Tables

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« Reply #575 on: February 06, 2016, 07:04:44 am »
0

What do you mean by the smallest list? By definition the list must have n elements, so they will all be the same size.
Unless I am looking for the smallest n that you can do this for?

Yeah, I mean the smallest n. Sorry, it was late at night while I worded it.

WW: Your solution to part 1 is correct, part 2 is not. Hint: Don't assume the list must be "symmetric" around the mean
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

#### Limetime

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« Reply #576 on: February 06, 2016, 08:14:07 am »
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What is the probability of a golden Sombrero?(Both opening cards miss shuffle assuming olny two card gained turn 1-2)
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#### Rabid

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« Reply #577 on: February 06, 2016, 08:23:14 am »
+2

What is the probability of a golden Sombrero?(Both opening cards miss shuffle assuming olny two card gained turn 1-2)
2/12 * 1/11 = 1/66
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#### Awaclus

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« Reply #578 on: February 06, 2016, 08:26:35 am »
0

What is the probability of a golden Sombrero?(Both opening cards miss shuffle assuming olny two card gained turn 1-2)

(2*10!)/12! = ~0,015

There are 12! different possible ways to arrange all the cards, and there are 2 ways to arrange the opening buys times the 10! ways to arrange the rest of the cards.

PPE: Or that.
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#### DStu

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« Reply #579 on: February 06, 2016, 01:10:46 pm »
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What do you mean by the smallest list? By definition the list must have n elements, so they will all be the same size.
Unless I am looking for the smallest n that you can do this for?

Yeah, I mean the smallest n. Sorry, it was late at night while I worded it.

WW: Your solution to part 1 is correct, part 2 is not. Hint: Don't assume the list must be "symmetric" around the mean

I can prove that 6 is possible .

2.75, 3.8, 6, 6, 8.7, 8.75
have mean value 6 and variance 6.051
2.75, 4, 6, 6, 8.5, 8.75
have mean value 6 and variance 5.675

so we have the numbers 2.75,6,6,8.75 and 3.8+h,  8.7-h. For h in [0,0.2], the range doesn't change, nor does the mode or median.  Mean doesn't change for any h.  The variance is continuous in h, so by mean value theorem there exists an h \in (0,0.2) such that variance is 6. (The h is probably quite easy to find as variance should be monotonous in h, too (but what for a mathematician would I be if I wouldn't be satisfied with proving the existence, and would continue to find out what the value is. Especially if it already stated that the solution is not unique anyway)
« Last Edit: February 06, 2016, 01:20:48 pm by DStu »
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#### florrat

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« Reply #580 on: February 06, 2016, 03:08:57 pm »
+5

The problem is that the question is ambiguous, since variance can mean two different things, either sN-12 or sN2. See Wolfram Mathworld.

WW assumed that the variance is sN2, and in that case he is right: with n=6 you can get sN2 arbitrarily close to 6, but not exactly 6. His solution for n=7 is correct under this interpretation

Tables and DStu assume the variance is sN-12, and in this case n=6 is possible and optimal.
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#### florrat

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« Reply #581 on: February 06, 2016, 03:21:11 pm »
+1

PS: if the variance means sN-12, an exact solution is
(3,6-\sqrt{6},6,6,6+\sqrt{6},9)
and all solutions for n=6 are of the form
(x,9-x-y,6,6,9-x+y,x+6) for 2+(1/2)\sqrt{2} < x < 4-(1/2)\sqrt{2} and y=\sqrt{-2x^2+12x-12}.

EDIT: typos
« Last Edit: February 06, 2016, 05:09:28 pm by florrat »
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#### DStu

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« Reply #582 on: February 06, 2016, 04:47:12 pm »
+1

The problem is that the question is ambiguous, since variance can mean two different things, either sN-12 or sN2. See Wolfram Mathworld.

WW assumed that the variance is sN2, and in that case he is right: with n=6 you can get sN2 arbitrarily close to 6, but not exactly 6. His solution for n=7 is correct under this interpretation

Tables and DStu assume the variance is sN-12, and in this case n=6 is possible and optimal.

I used VARIANCE in LO, it seems to be s_N, and I would argue that this is what should be used.  s_{N-1} you take if you have a random sample from a distribution, to get an unbiased estimator. This is not what we want in this case, we talk about the variance of a list.  This is as if the list is the complete distribution, to get the variance of this one you must take s_N.

Which confuses me now a bit, as I nevertheless "proved" that N=6 is possible. Might have some mistake in there...

Edit: A fuck it, I'm too drunk for this shit. Was s_{N-1} all along in my formula, at least this clears that. My above post is wrong because wrong definition of variance, WW is right, use s_N!
« Last Edit: February 06, 2016, 04:49:09 pm by DStu »
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#### Tables

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« Reply #583 on: February 06, 2016, 05:41:34 pm »
0

The problem is that the question is ambiguous, since variance can mean two different things, either sN-12 or sN2. See Wolfram Mathworld.

WW assumed that the variance is sN2, and in that case he is right: with n=6 you can get sN2 arbitrarily close to 6, but not exactly 6. His solution for n=7 is correct under this interpretation

Tables and DStu assume the variance is sN-12, and in this case n=6 is possible and optimal.

It doesn't matter which definition you use. The minimum n is the same for both definitions.

You only (IIRC) use the sN-12 definition for a sample. This isn't a sample. So for the problem, you can use variance = E((X-µ)2).
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

#### Tables

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« Reply #584 on: February 06, 2016, 06:09:37 pm »
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I just noticed, during completing my proof that you could do it in 6, that I was actually causing the 2nd largest number to increase above the largest number. I was sure I checked this didn't happen yesterday. Sigh.

Yeah I guess 6 isn't actually possible. But at least I have a conclusive proof of it now.
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

#### Tables

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« Reply #585 on: February 06, 2016, 06:25:02 pm »
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Well, okay, so here's what I had in terms of proof for both problems:

I guess I should post my solutions now. Maybe I did make a mistake somewhere, there's people here better at this kind of maths than I am, so whatever.

Easy version: 2, 4, 4, 6 is the list, so n = 4. Mode is 4 (two fours), Median is 4 (4 and 4 in the middle), Mean is 4 (sum is 16, 16/4=4) and range is 4 (6-2 = 4). This is also unique as the solution - you need two or more fours. for the mode. 3-4 fours doesn't work, as then the range or mean will fail. So two fours for the mode. As the median is 4 they have to be the middle numbers. And as the range and mean are also 4, the last two numbers must be equidistant from 4, i.e. one is two less, one or two more.

You can't do it in three numbers, since getting a mode of 3 means two 3s, so for the range your final number must be 0 or 6, which makes your mean 0 or 6. You can't do it in two numbers, as to get a range of 2 and a mean of 2 your numbers must be 1 and 3, which has no mode. One number is has range 0.

Hard version: 6 numbers. Here's an example list: 2.5, 6.5 - sqrt(34)/2, 6, 6, 6.5 + sqrt(34)/2, 8.5.

1-4 numbers can't work, which follows for the working in the easy version. For 4 numbers, there's a unique layout, right? But that doesn't have variance 4 (it has a variance of 8/3). What about 5? Well, we need 5, 5 in there for the median. Let x be the smallest number in the list. Then x + 5 is the largest number (range), and finally 10 - 2x is the final number (mean). Then firstly 4/3 < x < 10/3 in order to ensure 10 - 2x is not outside our range. But then we run into issues. Plugging x, x + 5, 10 - 2x, 5, 5 into our variance equation - E((X-µ)2) gives us:

25 = (x - 5)2 + x2 + (5 - 2x)2

25 = x2 - 10x + 25 + x2 + 25 - 20x + 4x2

25 = 6x2 - 30x + 50

0 = 6x2 - 30x + 25

x ~= 1.057 or x ~= 3.943

But both of these are outside our range of x, since they make the range of the list greater than 5. Therefore, 5 doesn't work.

So now 6. Well, we have a solution, so we're done! Let's find the family of solutions though. We need to have 6, 6 in there because mode. The smallest number we can call x again, and the biggest x + 6 (range). Let's call the 5th number y and then finally the 6th number will be 18 - 2x - y (mean). WLOG let y<=6 (median requires either it or 18 - 2x - y is less than or equal to 6 and the other greater than or equal).

Now we plug these into our variance formula and get:

36 = (6-x)2 + (6-y)2 + (6-6)2 + (6-6)2 + (2x + y - 12)2 + (6 - 6 + x)2

Expanding and simplifying that gives us:

0 = 6x2 + 2y2 + 4xy - 60x - 36y + 180

I can't be bothered to rearrange this myself, but it looks like an ellipse or something. Let's have WA rearrange for y.

y = -sqrt(-2x^2 + 12x - 9) - x + 9

Great. So now our list of six numbers is: x, 9 - x - sqrt(-2x^2 + 12x - 9), 6, 6, 9 - x + sqrt(-2x^2 + 12x - 9), x + 6

And here's where I realised the mistake. So... The issue is that, when we start putting bounds on x, we find there's no value which works. It's hard to state exactly since there's a lot of little cases, but basically: Increase x and the 2nd term drops below 3, so we can't do that. However the second term does eventually start increasing, but not enough to overtake x before the 5th term drops below 6. So we can't increase x. Decreasing x causes the same issue, but in reverse. The 5th term starts increasing above x+6 It starts decreasing eventually but not enough before the 2nd term increases above 6. So there's no value of x which satisfies all needed properties, meaning 6 doesn't work.

So... yeah I'm slightly bummed now. Turns out (unsurprisingly) that florrat was correct.
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

#### enfynet

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« Reply #586 on: February 06, 2016, 11:55:23 pm »
0

For Easy Mode, I assumed that _ 1 _ would satisfy the requirements.
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#### Titandrake

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« Reply #587 on: February 07, 2016, 12:56:28 am »
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More of a probability question, but I wondered about this randomly, and think I actually solved it.

For what values of c does there exist a non-constant random variable X such that for all positive integers k, E[X^k] = c?
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#### DStu

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« Reply #588 on: February 07, 2016, 03:45:47 am »
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More of a probability question, but I wondered about this randomly, and think I actually solved it.

For what values of c does there exist a non-constant random variable X such that for all positive integers k, E[X^k] = c?

I would guess just 1 and 0

With Jensen's inequality (as x->x^\frac{k+1}k is convex), you have E[X^{k+1}] >= E[X^k]^{\frac{k+1}k}, that excludes c<1.  For 0<c<1, you take E[X^{k-1}] <= E[X^k]^\frac{k-1}k.  0 and 1 are constant under this mapping, so no problem there. Negative c can't be because of even k's.

Edit: Of course 0 also doesn't work because of non-constant X, so at least second moment is larger than 0.
Edit2: and of course 1 obviously works with 2*Bernoulli(1/2).
Edit3: Ok moments, not centralized moments. Then take a slightly different distribution, but still a Bernoulli variant, just center it yourself
« Last Edit: February 07, 2016, 03:51:36 am by DStu »
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#### Titandrake

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« Reply #589 on: February 07, 2016, 04:06:37 am »
+1

Hm, I have an example with c = 1/2.

With just moments, I think 2 * Bernoulli(1/2) doesn't work, since E[X] = 1 and E[X^2] = 2.
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#### DStu

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« Reply #590 on: February 07, 2016, 04:18:35 am »
0

Yeah, I think my proof was for with absolute values. So E[|X|^k].

Otherwise, x^\alpha is not really concave for \alpha<1.

c>1 should still work on first glance.
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#### DStu

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« Reply #591 on: February 07, 2016, 04:20:08 am »
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Without absolute values and with real moments, you
get any 0<c<1 just by Bernoulli(c)...

Edit: Hmm, no negative values in here, somehow my reasoning for 0<c<1 above does not work. Not completely surprising though, didn't completely think it through, was more an analogy for c>1, might have some wrong sign or inequality in there....
« Last Edit: February 07, 2016, 04:23:04 am by DStu »
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#### Titandrake

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« Reply #592 on: February 07, 2016, 04:32:09 am »
+1

I got the same final answer, slightly differently but I think your reasoning works.

I think your reasoning E[X^{k+1}] >= ( E[X^k] )^{k+1/k} restricts you to c <= 1, because you can still have 1 >= 1^{k+1/k}. Although actually that doesn't fix the problem you pointed out, I'd need to think about it more carefully.

Edit: decided to add in the argument I used. I'm not as familiar with reasoning the way you did, so I did something a lot more gimmicky.

Consider the moment generating function E[e^{tX}] = 1 + t E[X] / 1! + t^2 E[X^2] / 2! + t^3 E[X^3] / 3! + and so on. If E[X^k] = c, this works out to

E[e^{tX}] = (c + c t/1! + c t^2/2! + ...) + (1 - c) = c e^t + (1 - c)

which is the MGF for Bernoulli(c), giving 0 < c < 1. Can't have c < 0 because of even k, can't have c = 0 because then X = 0 everywhere, can't have c = 1 because X = 1 everywhere, can't have c > 1 because Var(X) = E[X^2] - (E[X])^2 = c - c^2 would be negative.
« Last Edit: February 07, 2016, 04:37:51 am by Titandrake »
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#### DStu

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« Reply #593 on: February 07, 2016, 04:44:49 am »
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I think your variance argument is basically Jensen in the special case k=2, so not really more gimmickly, more like a bit more on point...

Edit: I think my inequalities above are the both in the wrong direction, probably you can them get right in the case c>1 (and using k+2 instead of k+1), but not in c<1.
« Last Edit: February 07, 2016, 04:47:53 am by DStu »
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#### Tables

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« Reply #594 on: February 07, 2016, 07:01:59 am »
0

For Easy Mode, I assumed that _ 1 _ would satisfy the requirements.

1 has a range of 0, so it fails.
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

#### enfynet

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« Reply #595 on: February 07, 2016, 09:21:24 am »
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For Easy Mode, I assumed that _ 1 _ would satisfy the requirements.

1 has a range of 0, so it fails.
Ah, yeah I guess I glossed over that one.
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#### Joseph2302

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« Reply #596 on: February 07, 2016, 04:14:19 pm »
+2

For Easy Mode, I assumed that _ 1 _ would satisfy the requirements.
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#### Tables

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« Reply #597 on: February 07, 2016, 05:05:43 pm »
+3

For Easy Mode, I assumed that _ 1 _ would satisfy the requirements.

Yes, that is the solution, although moat moat moat moat moat, moat Moat moat moat note moat. Mote mote mote, mote Mote meat moat moat:

1) moat

2) moat

3) moat moat

Moat, moat moat, moatmoat = moat - moatmoatmoat. Moat moat, moat meat moat moat.
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...spin-offs are still better for all of the previously cited reasons.
But not strictly better, because the spinoff can have a different cost than the expansion.

#### silverspawn

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« Reply #598 on: February 07, 2016, 05:15:03 pm »
0

For Easy Mode, I assumed that _ 1 _ would satisfy the requirements.

Yes, that is the solution, although moat moat moat moat moat, moat Moat moat moat note moat. Mote mote mote, mote Mote meat moat moat:

1) moat

2) moat

3) moat moat

Moat, moat moat, moatmoat = moat - moatmoatmoat. Moat moat, moat meat moat moat.

makes me wonder

For which moat ∈ ℝ does the equation moatmoat = moat - moatmoatmoat yield true?

I don't think I'm good enough at maths to calculate that. I guess it can only work for a moat ∈ (0, 1) or maybe also a moat ∈ (-1, 0) ?

Oort Cloud
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#### sudgy

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« Reply #599 on: February 07, 2016, 05:29:31 pm »
0

For Easy Mode, I assumed that _ 1 _ would satisfy the requirements.

Yes, that is the solution, although moat moat moat moat moat, moat Moat moat moat note moat. Mote mote mote, mote Mote meat moat moat:

1) moat

2) moat

3) moat moat

Moat, moat moat, moatmoat = moat - moatmoatmoat. Moat moat, moat meat moat moat.

makes me wonder

For which moat ∈ ℝ does the equation moatmoat = moat - moatmoatmoat yield true?

I don't think I'm good enough at maths to calculate that. I guess it can only work for a moat ∈ (0, 1) or maybe also a moat ∈ (-1, 0) ?

Oort Cloud

It doesn't look like it, even in the complex numbers.  Here's a link.
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