# Dominion Strategy Forum

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#### pacovf

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« Reply #825 on: May 09, 2017, 06:38:18 pm »
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Extra credit question I got lately: Find a good (unbinned) Maximum Likelihood Estimator Goodness of Fit.

Too late to get the extra credit, but I would be curious if anyone has any thoughts on this kind of thing.
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#### Titandrake

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« Reply #826 on: May 13, 2017, 03:26:46 am »
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For a somewhat ridiculous extra credit question (I don't think the teacher realized what he was doing), I need to solve this integral:

∫(cos(θ) + 2sin(θ))*sqrt((cos(θ)+2sin(θ))^2-4.75) dθ

It's used for a definite integral that can be calculated numerically (and gives the correct answer), but I'm curious if there's any way to figure out the indefinite integral.  No online calculators have been able to find the answer.  It's probably not an elementary function, but I want to see if anybody can figure it out.

Isn't the square root not always defined? I plotted cos(θ)+2sin(θ) and it definitely crosses 0.
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#### sudgy

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« Reply #827 on: May 13, 2017, 05:14:11 pm »
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For a somewhat ridiculous extra credit question (I don't think the teacher realized what he was doing), I need to solve this integral:

∫(cos(θ) + 2sin(θ))*sqrt((cos(θ)+2sin(θ))^2-4.75) dθ

It's used for a definite integral that can be calculated numerically (and gives the correct answer), but I'm curious if there's any way to figure out the indefinite integral.  No online calculators have been able to find the answer.  It's probably not an elementary function, but I want to see if anybody can figure it out.

Isn't the square root not always defined? I plotted cos(θ)+2sin(θ) and it definitely crosses 0.

You would have to plot (cos(θ)+2sin(θ))^2-4.75, but yeah, it does cross 0.  Like I said, the context was a definite integral that was within the domain of the function.
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#### silverspawn

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« Reply #828 on: May 26, 2017, 04:25:00 pm »
+1

Is there any meta advice which mathy people here would give about how to increase efficiency in solving problems? Anything you've learned over time? E.g. I think I spend too little time thinking broadly and too much following the first approach I've found. Stuff like that.
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#### pacovf

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« Reply #829 on: May 26, 2017, 06:08:49 pm »
+1

I don't know about "real" problems, but as far as homework goes, it's just a bunch of doing stuff until you start recognizing the patterns. Homework thinking is affected by the knowledge that you are expected to be able to do it.

It's sometimes useful to start at the answer, as in "if there was a solution, then to be able to find it I would likely have to do X", and then look into X.
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#### Cuzz

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« Reply #830 on: May 26, 2017, 06:11:18 pm »
+5

I don't know about "real" problems, but as far as homework goes, it's just a bunch of doing stuff until you start recognizing the patterns. Homework thinking is affected by the knowledge that you are expected to be able to do it.

It's sometimes useful to start at the answer, as in "if there was a solution, then to be able to find it I would likely have to do X", and then look into X.

This is the drastic and sometimes frightening distinction between doing math problems and doing math research.
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#### heron

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« Reply #831 on: May 26, 2017, 08:42:29 pm »
+1

Is there any meta advice which mathy people here would give about how to increase efficiency in solving problems? Anything you've learned over time? E.g. I think I spend too little time thinking broadly and too much following the first approach I've found. Stuff like that.

My thoughts:

It is important to reflect after either solving a difficult problem or looking at the solution. Figure out the motivation for the solution, and try to find a framework to view the problem from that makes the solution obvious or natural. When you do lots of problems, you can begin to see when similar frameworks can cause solutions to fall out of problems that might look very different at first.

Also, just do lots of problems.

Most of my experience comes from math olympiad, here are some links from people on solving math olympiad problems (in a broad sense):
https://usamo.wordpress.com/ (couldn't find a specific post that really addressed what you are talking about (although it probably exists) but maybe you will find some things interesting.
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#### pacovf

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« Reply #832 on: May 27, 2017, 12:28:30 am »
+2

I don't know about "real" problems, but as far as homework goes, it's just a bunch of doing stuff until you start recognizing the patterns. Homework thinking is affected by the knowledge that you are expected to be able to do it.

It's sometimes useful to start at the answer, as in "if there was a solution, then to be able to find it I would likely have to do X", and then look into X.

This is the drastic and sometimes frightening distinction between doing math problems and doing math research.

On the other hand, one advantage of math research, compared to homework, is that there's probably only, like, three people in the whole world that understand what you do, so people can't judge you for not being able to solve all questions :p
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#### ConMan

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« Reply #833 on: May 28, 2017, 07:08:37 pm »
+2

When doing maths research, there are two techniques that often come into play:

1. Generalise the problem into something that might be easier to solve, then show that it holds for the specific case you're looking into.
2. Solve a specific instance of the problem, then try to generalise it to the thing you're interested in.
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#### Polk5440

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« Reply #834 on: May 30, 2017, 03:45:25 pm »
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Step A: Identify the big, key theorems, techniques, and ideas that might be relevant to solving a problems.  Study problems that trip you up. Is there often a method used that you often miss (e.g. trigonometric u-substitutions) or a theorem you forget you can apply (e.g. pigeonhole principle)? Study more in areas in which you are having trouble. What are the big ideas?

Step B: Understand these big, key theorems, techniques, and ideas. And not just in a superficial "that makes sense, let's move on to the next thing!" kind of way. Really internalize them so they become a natural part of your problem solving toolbox.

Then keep going. A -> B -> A -> B ->....
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#### Tables

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« Reply #835 on: June 06, 2017, 05:19:37 pm »
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Request: If you were to roll a pair of unbiased dice (2d6) and total the result, until you rolled the same total 20 times, what are the chances of each value being the one you reach 20 on?

This came from a lesson I was in yesterday on experimental probability. We'd gotten the pupils/groups to roll 2d6 and record their result until they got 20 of the same number, then collated results and plotted them to see the distribution and compare it to the expected result. Most of the groups got fairly reasonable results, several finished with twenty 7's, but one boy managed to get to twenty 10's before anything else, and in fact did so in a remarkably low number of rolls (73 IIRC). The teacher and I were discussing afterwards about how unlikely this was and how you'd calculate it, and neither of us were especially confident in how to do it. I suspected it would be the hypergeometric distribution, but I've honestly barely used that so I'm not sure.

Any takers? If not I'll have to go over to Reddit's theydidthemaths.
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#### Witherweaver

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« Reply #836 on: June 06, 2017, 06:22:05 pm »
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Request: If you were to roll a pair of unbiased dice (2d6) and total the result, until you rolled the same total 20 times, what are the chances of each value being the one you reach 20 on?

This came from a lesson I was in yesterday on experimental probability. We'd gotten the pupils/groups to roll 2d6 and record their result until they got 20 of the same number, then collated results and plotted them to see the distribution and compare it to the expected result. Most of the groups got fairly reasonable results, several finished with twenty 7's, but one boy managed to get to twenty 10's before anything else, and in fact did so in a remarkably low number of rolls (73 IIRC). The teacher and I were discussing afterwards about how unlikely this was and how you'd calculate it, and neither of us were especially confident in how to do it. I suspected it would be the hypergeometric distribution, but I've honestly barely used that so I'm not sure.

Any takers? If not I'll have to go over to Reddit's theydidthemaths.

My thought was to consider the random variables that is the number of rolls (of 2d6) for such an occurrence for different values for the total (2 repetitions, 3 repetitions, etc.).  Start with 2 instead of 20.

I played with this a little bit; each separate "number of rolls until the next one" is not as simple as the regular case (each is geometric and the sum is negative binomial) because success here is not just rolling a fixed value that number of times; it's rolling any value that number of times.

« Last Edit: June 06, 2017, 06:24:02 pm by Witherweaver »
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#### Kirian

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« Reply #837 on: June 06, 2017, 10:03:12 pm »
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OK, I actually need some stats help.

Assume a value is constrained between 0 and 10.  Do we use the normal formulas for sample standard deviation?  Because mean of 9 with SD 3 doesn't work--it suggests the value might be as high as 12.  Is there some extra constraint that should be used here?
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#### pacovf

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« Reply #838 on: June 06, 2017, 10:38:00 pm »
+2

That's not how standard deviation works though. If the distribution is very skewed, you can get the sort of results you mention. EDIT: to expand a bit upon my answer, your interpretation of the standard deviation has the implicit assumption that the distribution is symmetric about the mean, but there's no reason for it to hold true, and the assumption will lead you to spurious results when it doesn't.

@tables: honestly, that sounds like a job for simulations, and I say that as someone that usually enjoys figuring out things analytically. Any formula will be too horrible to use, assuming it exists.
« Last Edit: June 07, 2017, 12:17:54 am by pacovf »
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#### Polk5440

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« Reply #839 on: June 07, 2017, 10:44:40 am »
+1

one boy managed to get to twenty 10's before anything else, and in fact did so in a remarkably low number of rolls (73 IIRC). The teacher and I were discussing afterwards about how unlikely this was and how you'd calculate it, and neither of us were especially confident in how to do it. I suspected it would be the hypergeometric distribution, but I've honestly barely used that so I'm not sure.

First, what's the probability he rolled fairly?

In high school stats we did an experiment. Each student flipped a coin at his desk 10 times and wrote the results on the board. Quite a lot of students (more than half) wrote up all heads or all tails. People were flipping systematically, not randomly. The teacher scrapped the experiment, and had us redo it standing a few feet from a wall and flipping the coin into the wall and letting it fall to the floor before writing heads/tails. The new results were better.

Anyway, it's harder to consistently roll dice, but it's possible.
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#### Mic Qsenoch

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« Reply #840 on: June 07, 2017, 11:49:32 am »
+2

Request: If you were to roll a pair of unbiased dice (2d6) and total the result, until you rolled the same total 20 times, what are the chances of each value being the one you reach 20 on?

This came from a lesson I was in yesterday on experimental probability. We'd gotten the pupils/groups to roll 2d6 and record their result until they got 20 of the same number, then collated results and plotted them to see the distribution and compare it to the expected result. Most of the groups got fairly reasonable results, several finished with twenty 7's, but one boy managed to get to twenty 10's before anything else, and in fact did so in a remarkably low number of rolls (73 IIRC). The teacher and I were discussing afterwards about how unlikely this was and how you'd calculate it, and neither of us were especially confident in how to do it. I suspected it would be the hypergeometric distribution, but I've honestly barely used that so I'm not sure.

Any takers? If not I'll have to go over to Reddit's theydidthemaths.

Here's sim results for 10 million trials:
#      %         Avg. number of rolls before reaching 10 20
2   0.00065   59.15
3   0.1233   56.25
4   1.59693   52.93
5   7.36613   50.21
6   20.17438   47.95
7   41.47858   46.09
8   20.17635   47.96
9   7.36619   50.20
10   1.59349   52.96
11   0.12337   56.01
12   0.00063   58.43

I guess you can sort of evaluate the convergence by comparing the ideally symmetric results.

I don't know how many trials you had, but getting a 1.6% result isn't too crazy. His number of turns to get it is above average too. That value is constrained because of how you choose the final result, it's only selected if it reaches 20 before all the more likely combinations, so you don't actually deviate too far from the averages of the very likely results. I guess the distribution would be more telling than the average. It was all a huge mistake.
« Last Edit: June 07, 2017, 07:44:09 pm by Mic Qsenoch »
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#### Polk5440

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« Reply #841 on: June 07, 2017, 04:35:29 pm »
+2

I actually get much higher average number of rolls and slightly different probabilities. I ran a smaller sim (100k; which explains lack of 2 and 12 results), but that alone shouldn't explain the differences.

#     %     Avg. number of rolls before reaching 20
2     -        -
3     0.0    140
4     0.3    123
5     4.2    114
6     19.1  109
7     52.8  104
8     19.1  109
9     4.2    115
10    0.3   122
11    0.0   143
12    -       -
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#### Mic Qsenoch

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« Reply #842 on: June 07, 2017, 06:05:42 pm »
+2

I actually get much higher average number of rolls and slightly different probabilities. I ran a smaller sim (100k; which explains lack of 2 and 12 results), but that alone shouldn't explain the differences.

#     %     Avg. number of rolls before reaching 20
2     -        -
3     0.0    140
4     0.3    123
5     4.2    114
6     19.1  109
7     52.8  104
8     19.1  109
9     4.2    115
10    0.3   122
11    0.0   143
12    -       -

Most likely I goofed.
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#### Rabid

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« Reply #843 on: June 07, 2017, 07:35:32 pm »
+2

Very rough estimate is:
6/36 rolls are 7's.
So about 20*6 = 120 rolls to get 20 sevens.
I would think 104 is more likely than 46
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#### Mic Qsenoch

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« Reply #844 on: June 07, 2017, 07:43:14 pm »
+1

Yeah, my numbers are counting to 10, not 20, whoops

I get Polk's #s with 20, up to whatever little deviations.

I say lock the kid up and throw away the key.
« Last Edit: June 07, 2017, 07:50:51 pm by Mic Qsenoch »
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#### liopoil

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« Reply #845 on: June 07, 2017, 08:49:49 pm »
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Yeah, my numbers are counting to 10, not 20, whoops

I get Polk's #s with 20, up to whatever little deviations.

I say lock the kid up and throw away the key.
Even 0.3% isn't absurd though - or is it the speed with which he got it?
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#### pacovf

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« Reply #846 on: June 07, 2017, 09:46:28 pm »
+2

Depends on how many groups there were, but 0.3% sounds unlikely. That's a full order of magnitude above the usual p-value required to reject the null-hypothesis :p Somebody with more time than me could calculate how many groups of students you would need for that not to be significant (where "that" is the most unlikely result you get out of repeating the experience n times).

Honestly, if you asked me to roll a dice ~120 times, I would have probably gotten bored of it way before then.
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#### liopoil

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« Reply #847 on: June 07, 2017, 09:58:59 pm »
+1

Depends on how many groups there were, but 0.3% sounds unlikely. That's a full order of magnitude above the usual p-value required to reject the null-hypothesis :p Somebody with more time than me could calculate how many groups of students you would need for that not to be significant (where "that" is the most unlikely result you get out of repeating the experience n times).

Honestly, if you asked me to roll a dice ~120 times, I would have probably gotten bored of it way before then.
Well, the probability is actually more like ~0.006, and so I think we want the minimum number n such that (1 - 0.994^n) > 0.05, which is n = 9.
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#### pacovf

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« Reply #848 on: June 07, 2017, 10:34:24 pm »
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Ha, you are right. I keep forgetting that calculating the CDF of "worst/best of n trials" is simpler than my intuition thinks.

I assume the probabilities look worse when taking into account the number of rolls reported, though.
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#### Witherweaver

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« Reply #849 on: June 08, 2017, 08:23:46 am »
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Yeah, my numbers are counting to 10, not 20, whoops

I get Polk's #s with 20, up to whatever little deviations.

I say lock the kid up and throw away the key.

Huh, interesting.  What happens when you go to 30, 40, 50?  I would guess that it would approach a stable distribution.
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